C++博客-心無雜念的空白-文章分类-Algorithm

Timus 1208 Legendary Teams Contest(题解)

Lshain — Thu, 21 Jul 2011 09:44:00 GMT

http://acm.timus.ru/problem.aspx?space=1&num=1208

题目大意：给出n个队,n<=18, 每队包含三个人名，求n个队中最多选出几个队，选出的队中的人名不重复。
思路：i队和j队，如果有相同名字的队员，则连一无向边i-j，题目变为求给定无向图G的最大独立集，由于n不大，直接建图暴力枚举的。O(n*2^n)复杂度。

Code:

#include
#include
#include

struct Node{
char member[3][21];
}teamSet[20];
int n;
unsigned int graph[21][21];
unsigned int teams[20];

unsigned int isOk(int I, int J){
   int i = 0;
   int j = 0;
   unsigned int flag = 0;
   for(i = 0; i < 3 && !flag; i++){
      for(j = 0; j < 3; j++){
         if(!strcmp(teamSet[I].member[i], teamSet[J].member[j])){
            flag = 1;
            break;
         }
      }
   }

return flag;
}

int Check(unsigned int team){
   unsigned int c = 0;
   unsigned int tempSet = 0;
   int nCount = 0;
   int num = n;
   int isValid = 1;
   c = 0x00020000;
   while(num++ < 18){
      c >>= 1;
   }
   for(num = n; c > 0; c >>= 1, num--){
      if(c&team){
         nCount++;
         if(tempSet&c){
            isValid = 0;
            break;
         }
         tempSet |= teams[n - num + 1];
      }
   }

return (1 == isValid) ? nCount : -1;
}

int main(int argc, char* argv[], char* env[])
{
   int i = 0;
   int j = 0;
   int t = 0;
   int maxTeams = 0;
   unsigned int c = 0;
   unsigned int tempN = 0;
   while(scanf("%d", &n) != EOF){
      for(i = 1; i <= n; i++){
         scanf("%s%s%s", teamSet[i].member[0],
         teamSet[i].member[1], teamSet[i].member[2]);
      }
      for(i = 1; i <= n; i++){
         graph[i][i] = 0;
         for(j = i + 1; j <= n; j++){
            graph[i][j] = graph[j][i] = isOk(i, j);
         }
      }
      for(i = 1; i <= n; i++){
         teams[i] = graph[i][j];;
         for(j = 2; j <= n; j++){
            teams[i] <<= 1;
            teams[i] |= graph[i][j];
         }
      }
      maxTeams = 0;
      tempN = 1 << n;
      for(c = 1; c < tempN; c++){
         t = Check(c);
         if(t > maxTeams){
            maxTeams = t;
         }
      }
      printf("%d\n", maxTeams);
   }
   return 0;
}

Lshain 2011-07-21 17:44 发表评论

Timus 1353. Milliard Vasya's Function(题解)

Lshain — Thu, 21 Jul 2011 06:11:00 GMT

1353. Milliard Vasya's Function

Time Limit: 1.0 second
Memory Limit: 16 MB

Vasya is the beginning mathematician. He decided to make an important contribution to the science and to become famous all over the world. But how can he do that if the most interesting facts such as Pythagor’s theorem are already proved? Correct! He is to think out something his own, original. So he thought out the Theory of Vasya’s Functions. Vasya’s Functions (VF) are rather simple: the value of the N^th VF in the point S is an amount of integers from 1 to N that have the sum of digits S. You seem to be great programmers, so Vasya gave you a task to find the milliard VF value (i.e. the VF with N = 10⁹) because Vasya himself won’t cope with the task. Can you solve the problem?

Input

Integer S (1 ≤ S ≤ 81).

Output

The milliard VF value in the point S.

Sample

input	output
1	10

题意：输入s，输出自然数1...N中各位数字之和等于s的数的个数。
解题思路：
DP: F[i][j]表示i位数中，数字之和为j的数的个数，F[i][j] = F[i - 1][k]; (k = j....j-9 && k > 0);
递推方程由：F[i][j] = F[i - 1][j]个数尾部补0 + F[i - 1][j - k]个数头部补k，(k = 1...9, && j - k > 0)，而来。
Code:

#include

int F[10][82];
int count[82];

void init(){
   int i = 0;
   int j = 0;
   int k = 0;
   count[0] = 1;
   for(i = 0; i < 10; i++){
      for(j = 0; j < 82; j++){
         F[i][j] = 0;
      }
   }
   for(i = 0; i <= 9; i++){
      F[1][i] = 1;
   }
   for(i = 2; i < 10; i++){
      for(j = 1; j < 82; j++){
         for(k = j; k > 0 && k >= j - 9; k--){
            F[i][j] += F[i - 1][k];
         }
      }
   }
   count[1] = 10;
   for(j = 2; j < 82; j++){
      count[j] = 0;
      for(i = 1; i < 10; i++){
         count[j] += F[i][j];
      }
   }
}

int main(int argc, char* argv[], char* env[])
{
   int n = 0;
   init();
   while(scanf("%d", &n) != EOF){
      printf("%d\n", count[n]);
   }
   return 0;
}

Lshain 2011-07-21 14:11 发表评论

Timus 1260 Nudnik Photographer （题解 + 递推）

Lshain — Wed, 20 Jul 2011 09:51:00 GMT

http://acm.timus.ru/problem.aspx?space=1&num=1260

思路：假设F[n]表示n个数的满足题意的合法排列数，那么F[n] = F[n - 1] + F[n -3] + 1;
         F[n - 1] 是这样来的：F[n]，1排第一，第2个位置只能排2,3. 假设第2个位置排2 那么序列就是12..... = 1, (2....)序列。2.... = F[n - 1];
         F[n - 3] + 1 是这样来的：F[n]，1排第一，第2个位置只能排2,3. 假设第2个位置排3 那么序列就是13 ..... = 13(....)2或者132(....)序列。
               13(....)2只能有一种，因为4只能排2前面， 5只能排3后面。。。。就如这样：13 5764 2
               132(....) = 132(4......)，(4....) = F[n-3];
Code:

#include

int F[56];
void init(){
   int i = 0;
   F[0] = 0;
   F[1] = 1;
   F[2] = 1;
   for(i = 3; i <= 55; i++){
      F[i] = F[i - 1] + F[i - 3] + 1;
   }
}

int main(int argc, char* argv[], char* env[])
{
   int n = 0;
   init();
   while(scanf("%d", &n) != EOF){
      printf("%d\n", F[n]);
   }
   return 0;
}

Lshain 2011-07-20 17:51 发表评论

Timus 1078. Segments (spfa（最长路）)

Lshain — Wed, 20 Jul 2011 02:59:00 GMT

http://acm.timus.ru/problem.aspx?space=1&num=1078

主要思路：i包含j连一有向边i->j，这样问题就变成了给一个有向无环图(DAG)，求其最长路，直接spfa就OK啦！

Code
#include

#define MAXN 501

int graph[MAXN][MAXN];
int rd[MAXN];
int dist[MAXN];
int pre[MAXN];
int mark[MAXN];
int seg[MAXN][2];
int maxLen;
int n;

int isContains(int I, int J){
   int reVal = 0;
   if(seg[I][0] < seg[J][0] && seg[I][1] > seg[J][1]){
      reVal = 1;
   }
   if(seg[J][0] < seg[I][0] && seg[J][1] > seg[I][1]){
      reVal = -1;
   }

return reVal;
}

void spfa(){
   int i = 0;
   int I = 0;
   int t = 0;
   int qHead = 0;
   int qTail = 0;
   int Q[MAXN] = {0};

   for(i = 0; i < n; i++){
      mark[i] = 0;
      pre[i] = -1;
      dist[i] = -MAXN;
   }
   for(i = 0; i < n; i++){
      if(!rd[i]){
         Q[qHead++] = i;
         mark[i] = 1;
         dist[i] = 1;
      }
   }
   do{
      qHead--;
      if(qHead < 0){
         qHead = n - 1;
      }
      I = Q[qHead];
      mark[I] = 0;
      for(i = 0; i < n; i++){
         if(graph[I][i] && dist[I] + 1 > dist[i]){
            pre[i] = I;
            dist[i] = dist[I] + 1;
            if(!mark[i]){
               mark[i] = 1;
               t = qHead - 1;
               if(t < 0)t = n - 1;
               if(qHead != qTail
                     && dist[i] < dist[Q[t]]){
                     Q[qHead++] = i;
                     if(qHead >= n){
                        qHead = 0;
                     }
               }
               else{
                     qTail--;
                     if(qTail < 0){
                        qTail = n - 1;
                     }
                     Q[qTail] = i;
               }
            }
         }
      }
   }while(qHead != qTail);

   maxLen = -1;
   for(i = 0, I = -1; i < n; i++){
      if(dist[i] > maxLen){
         maxLen = dist[i];
         I = i;
      }
   }
   printf("%d\n", maxLen);
   t = I;
   i = 0;
   while(t != -1){
      if(i){
         printf(" ");
      }
      i = 1;
      printf("%d", t + 1);
      t = pre[t];
   }
   printf("\n");
}

int main(int argc, char* argv[], char* env[])
{
   int i = 0;
   int j = 0;
   int t = 0;
   while(scanf("%d", &n) != EOF){
      for(i = 0; i < n; i++){
         scanf("%d%d", &seg[i][0], &seg[i][1]);
      }
      for(i = 0; i < n; i++){
         rd[i] = 0;
         graph[i][i] = 0;
         for(j = i + 1; j < n; j++){
            t = isContains(i, j);
            graph[i][j] = t > 0 ? 1 : 0;
            graph[j][i] = t < 0 ? 1 : 0;
         }
      }
      for(j = 0; j < n; j++){
         for(i = 0; i < n; i++){
            rd[j] += graph[i][j];
         }
      }
      spfa();
   }
   return 0;
}

Lshain 2011-07-20 10:59 发表评论

最大连续子段和 + 最大子矩阵和

Lshain — Wed, 20 Jul 2011 00:58:00 GMT

#include

int numberMatrix[101][101];
int subNumberSum[101][101];
int numbers[101];
int n;

int maxSubSum(int* numbers, int n){//计算最大子序列和
      int i = 0;
      int sum = 0;
      int maxSum = numbers[0];
      for(i = 0; i < n; i++){
         sum += numbers[i];
         if(sum > maxSum){
            maxSum = sum;
         }
         if(sum < 0){
            sum = 0;
         }
      }

return maxSum;
}

int maxSubMatrixSum(int numberMatrix[101][101], int n){//计算最优子矩阵
      int i = 0;
      int j = 0;
      int k = 0;
      int tempSum = 0;
      int maxSum = 0;
      for(i = 1; i <= n; i++){//初始化第零行，全为0
         subNumberSum[0][i] = 0;
      }
      for(i = 1; i <= n; i++){//计算subNumberSum
            for(j = 1; j <= n; j++){
                  subNumberSum[i][j] = subNumberSum[i - 1][j] + numberMatrix[i][j];
            }
      }
      maxSum = subNumberSum[1][1];
      for(i = n; i >= 1; i--){//枚举计算maxSubSum
            for(j = i - 1; j >= 0; j--){
                  for(k = 1; k <= n; k++){
                        numbers[k - 1] = subNumberSum[i][k] - subNumberSum[j][k];
                  }
                  tempSum = maxSubSum(numbers, n);
                  if(tempSum > maxSum){
                        maxSum = tempSum;
                  }
            }
       }

return maxSum;
}

int main(int argc, char* argv[], char* env[])
{
      int i = 0;
      int j = 0;
      while(scanf("%d", &n) != EOF){
            for(i = 1; i <= n; i++){
                  for(j = 1; j <= n; j++){
                         scanf("%d", &numberMatrix[i][j]);
                  }
            }
            printf("%d\n", maxSubMatrixSum(numberMatrix, n));
      }

return 0;
}

Lshain 2011-07-20 08:58 发表评论

dij+heap | spfa + slf

Lshain — Tue, 19 Jul 2011 02:53:00 GMT

Source Code

Problem: 1511		User: LUCKLY
Memory: 51100K		Time: 6266MS
Language: C++		Result: Accepted

Source Code

#include

#define MAXINT	(((long long)1)<<60)
#define SWAP(x, y) do{ if((x)!=(y)){(x)^=(y); (y)^=(x); (x)^=(y);}}while(0);
#define	MAXM	1000001
#define MAXN	1000001

struct Edg{
	int n;
	int p;
	int next;
}EE[MAXM<<2];

int head1[MAXN];
int head2[MAXN];
long long dist[MAXN];
long long ans;
int top;
int n;
int m;

long long heap[MAXN];
int heap_size;
int id[MAXN];

void swim(int u){
	int parent = u >> 1;
	while(parent > 0){
		if(dist[heap[u]] < dist[heap[parent]]){
			SWAP(id[heap[parent]], id[heap[u]]);
			SWAP(heap[parent], heap[u]);
			u = parent;
			parent = u >> 1;
		}
		else{
			break;
		}
	}
}

void sink(int u){
	int child = u << 1;
	while(child < heap_size){
		if(child + 1 < heap_size && dist[heap[child]] < dist[heap[child + 1]]){
			child++;
		}
		if(dist[heap[child]] < dist[heap[u]]){
			SWAP(id[heap[child]], id[heap[u]]);
			SWAP(heap[child], heap[u]);
			u = child;
			child = u << 1;
		}
		else{
			break;
		}
	}
}

int getMin(){
	int reVal = -1;
	if(heap_size > 1){
		reVal = heap[1];
		heap_size--;
		if(heap_size > 1){
			SWAP(id[heap[1]], id[heap[heap_size]]);
			SWAP(heap[1], heap[heap_size]);
			sink(1);
		}
	}
	return reVal;
}

inline int nextInt(){
	char ch = getchar();
	int x = 0;
	while(ch < '0' || ch > '9'){
		ch = getchar();
	}
	while(ch >= '0' && ch <= '9'){
		x = x * 10 + ch - 48;
		ch = getchar();
	}
	return x;
}

void dij(int* head){
	int i = 0;
	int minP = 0;
	int I = 0;
	for(i = 1; i <= n; i++){
		dist[i] = MAXINT;
		heap[i] = i;
		id[i] = i;
	}
	dist[1] = 0;
	heap_size = n + 1;
	while((I = getMin()) != -1){
		minP = dist[I];
		i = head[I];
		while(i != -1){
			I = EE[i].n;
			if(minP + EE[i].p < dist[I]){
				dist[I] = minP + EE[i].p;
				swim(id[I]);
			}
			i = EE[i].next;
		}
	}
	for(i = 1; i <= n; i++){
		ans += dist[i];
	}
}

void buildEdg(int I, int J, int p, int* head){
	EE[top].n = J;
	EE[top].p = p;
	EE[top].next = head[I];
	head[I] = top++;
}

int main(int argc, char* argv[], char* env[])
{
	int i = 0;
	int I = 0;
	int J = 0;
	int P = 0;
	int caseNum = 0;
	while(scanf("%d", &caseNum) != EOF){
		while(caseNum-- > 0){
			n = nextInt();
			m = nextInt();
			top = 0;
			for(i = 1; i <= n; i++){
				head1[i] = head2[i] = -1;
			}
			while(m-- > 0){
				I = nextInt();	
				J = nextInt();
				P = nextInt();
				buildEdg(I, J, P, head1);
				buildEdg(J, I, P, head2);
			}
			ans = 0;
			dij(head1);
			dij(head2);
			printf("%lld\n", ans);
		}
	}
	return 0;
}

Source Code

Problem: 1511		User: LUCKLY
Memory: 47188K		Time: 3688MS
Language: C++		Result: Accepted

Source Code

#include

#define MAXINT	(((long long)1)<<60)
#define	MAXM	1000001
#define MAXN	1000001

struct Edg{
	int n;
	int p;
	int next;
}EE[MAXM<<2];

int head1[MAXN];
int head2[MAXN];
long long dist[MAXN];
int spfa_que[MAXN];
int mark[MAXN];
long long ans;
int top;
int spfa_que_head;
int spfa_que_tail;
int n;
int m;

inline int nextInt(){
	char ch = getchar();
	int x = 0;
	while(ch < '0' || ch > '9'){
		ch = getchar();
	}
	while(ch >= '0' && ch <= '9'){
		x = x * 10 + ch - 48;
		ch = getchar();
	}
	return x;
}

inline void buildEdg(int I, int J, int p, int* head){
	EE[top].n = J;
	EE[top].p = p;
	EE[top].next = head[I];
	head[I] = top++;
}

void spfa_slf(int* head){
	int i = 0;
	int I = 0;
	int q = 0;
        int t = 0;
	for(i = 1; i <= n; i++){
		dist[i] = MAXINT;
		mark[i] = 0;
	}
	dist[1] = 0;
	spfa_que_head = 0;
	spfa_que_tail = 0;
	spfa_que[spfa_que_head++] = 1;
	mark[1] = 1;
	do{
		spfa_que_head--;
		if(spfa_que_head < 0){
			spfa_que_head = n - 1;
		}
		I = spfa_que[spfa_que_head];
		mark[I] = 0;
		q = head[I];
		while(q != -1){
			i = EE[q].n;
			if(dist[I] + EE[q].p < dist[i]){
				dist[i] = dist[I] + EE[q].p;
				if(!mark[i]){
					mark[i] = 1;
                                        t = spfa_que_head - 1;
                                        if(t < 0)t = n - 1;
					if(spfa_que_head != spfa_que_tail 
						&& dist[i] < dist[spfa_que[t]]){
						spfa_que[spfa_que_head++] = i;
						if(spfa_que_head >= n){
							spfa_que_head = 0;
						}
					}
					else{
						spfa_que_tail--;
						if(spfa_que_tail < 0){
							spfa_que_tail = n - 1;
						}
						spfa_que[spfa_que_tail] = i;
					}
				}
			}
			q = EE[q].next;
		}
	}while(spfa_que_head != spfa_que_tail);
	for(i = 1; i <= n; i++){
		ans += dist[i];
	}
}

int main(int argc, char* argv[], char* env[])
{
	int i = 0;
	int I = 0;
	int J = 0;
	int P = 0;
	int caseNum = 0;
	while(scanf("%d", &caseNum) != EOF){
		while(caseNum-- > 0){
			n = nextInt();
			m = nextInt();
			top = 0;
			for(i = 1; i <= n; i++){
				head1[i] = head2[i] = -1;
			}
			while(m-- > 0){
				I = nextInt();	
				J = nextInt();
				P = nextInt();
				buildEdg(I, J, P, head1);
				buildEdg(J, I, P, head2);
			}
			ans = 0;
			spfa_slf(head1);
			spfa_slf(head2);
			printf("%lld\n", ans);
		}
	}
	return 0;
}

Lshain 2011-07-19 10:53 发表评论

图论专题二分图(转)

Lshain — Mon, 18 Jul 2011 06:20:00 GMT

摘要: 转自http://blog.163.com/zjut_nizhenyang/blog/static/169570029201011285564684/ 图论专题二分图二分图：二分图是这样一个图，它的顶点可以分类两个集合X和Y，所有的边关联的两个顶点恰好一个属于集合X，另一个属于集合Y。二分图匹配：给定一个二分图G，在G的一个子图M中，M的边集中的任意两条边都不依附于同一个顶... 阅读全文

Lshain 2011-07-18 14:20 发表评论

Timus 1056. Computer Net 树的中心

Lshain — Mon, 20 Jun 2011 09:30:00 GMT

http://acm.timus.ru/problem.aspx?space=1&num=1056

1056. Computer Net

Time Limit: 2.0 second
Memory Limit: 16 MB

Background

Computer net is created by consecutive computer plug-up to one that has already been connected to the net. Each new computer gets an ordinal number, but the protocol contains the number of its parent computer in the net. Thus, protocol consists of several numbers; the first of them is always 1, because the second computer can only be connected to the first one, the second number is 1 or 2 and so forth. The total quantity of numbers in the protocol is N − 1 (N is a total number of computers). For instance, protocol 1, 1, 2, 2 corresponds to the following net:

1 - 2 - 5
|   |
3   4

The distance between the computers is the quantity of mutual connections (between each other) in chain. Thus, in example mentioned above the distance between computers #4 and #5 is 2, and between #3 and #5 is 3.

Definition. Let the center of the net be the computer which has a minimal distance to the most remote computer. In the shown example computers #1 and #2 are the centers of the net.

Problem

Your task is to find all the centers using the set protocol.

Input

The first line of input contains an integer N, the quantity of computers (2 ≤ N ≤ 10000). Successive N − 1 lines contain protocol.

Output

Output should contain ordinal numbers of the determined net centers in ascending order.

Sample

input	output
5 1 1 2 2	1 2

给定一棵树，求该树的中心。
中心定义：到其它点的距离最大值最小
树的中心即为树中最长链的中心，2次dfs即可

Code
#include
struct Node{
   int n;
   int next;
}EE[20001];

int ind;
int n;
int top;
int head[10001];
int check[10001];
int pre[10001];
int maxDep;
void dfs(int u, int t){
   int q;
   check[u] = 1;
   if(t > maxDep){
      maxDep = t;
      ind = u;
   }
   q = head[u];
   while(q != -1){
      if(!check[EE[q].n]){
         pre[EE[q].n] = u;
         dfs(EE[q].n, t + 1);
      }
      q = EE[q].next;
   }
}

int main(int argc, char* argv[], char* env[])
{
   int i = 0;
   int j = 0;
   while(scanf("%d", &n) != EOF){
      for(i = 1; i <= n; i++){
         head[i] = -1;
         check[i] = 0;
      }
      top = 0;
      for(i = 2; i <= n; i++){
         scanf("%d", &j);
         EE[top].n = i;
         EE[top].next = head[j];
         head[j] = top++;
         EE[top].n = j;
         EE[top].next = head[i];
         head[i] = top++;
      }
      maxDep = -1;
      dfs(1, 1);
      for(i = 1; i <= n; i++){
         check[i] = 0;
         pre[i] = -1;
      }
      maxDep = -1;
      dfs(ind, 1);
      i = ind;
      top = 0;
      while(i != -1){
         check[top++] = i;
         i = pre[i];
      }
      if(top&1){
         printf("%d\n", check[top>>1]);
      }
      else{
         printf("%d %d\n", check[(top>>1) - 1], check[top>>1]);
      }
   }
   return 0;
}

Lshain 2011-06-20 17:30 发表评论

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