C++博客-Smile-文章分类-其他题目

hunnu 10982 优先队列的使用

Smile3 — Fri, 26 Aug 2011 16:48:00 GMT

这个现场没有写的原因是优先队列不太熟练，并且情况比较多，这里wa了一次是因为初始化错了，逻辑思路还是比较清晰的。

#include
#include
#include
#include
using namespace std;

int const max_n = 1010;
struct node{//buy 从大到小
        int price;
        int num;
        friend bool operator < ( node n1, node n2 ){
               if( n1.price < n2.price )
                   return true;
               if( n1.price == n2.price )
                   if( n1.num > n2.num )
                       return true;
               return false;
        }
        node(int a,int b ):price(a),num(b){}
};
struct node1{//sell 从小到大
        int price;
        int num;
        friend bool operator < ( node1 n1, node1 n2 ){
               if( n1.price > n2.price )
                   return true;
               if( n1.price == n2.price )
                   if( n1.num > n2.num )
                       return true;
               return false;
        }
        node1(int a,int b ):price(a),num(b){}
};
int main(){//ask:high need bid：sell low ldprice：目前成交价格
    int t;
    scanf("%d",&t);
    while( t-- ){
           int n;
           scanf("%d",&n);
           char a[1000],b[1000],c[1000];
           int dprice,dnum;
           int ldprice = 0,ask=0,bid=0;
           priority_queue< node,vector > buy;
           priority_queue< node1,vector > sell;
           while( n-- ){
                  //ldprice = 0;
                  ask = 0;
                  bid = 0;
                  scanf("%s %d %s %s %d",a,&dnum,&b,&c,&dprice);
                  if( strcmp(a,"buy") == 0 ){
                      if( !sell.empty() )
                          if( dprice < (sell.top()).price ){
                              buy.push(node(dprice,dnum));
                              ask = (buy.top()).price;
                              bid = (sell.top()).price;
                          }
                          else{
                               node1 s1(0,0);
                               while( !sell.empty() && dnum > 0 && (sell.top()).price <=dprice ){
                                      s1 = sell.top();
                                      sell.pop();
                                      if( s1.num<=dnum ){
                                          dnum -= s1.num;
                                          s1.num = 0;
                                      }
                                      else{
                                            s1.num -= dnum;
                                            dnum = 0;
                                      }
                                      ldprice = s1.price;
                               }
                               if( dnum > 0 )
                                   buy.push(node(dprice,dnum));
                               if( s1.num >0 )
                                   sell.push(s1);
                               if( !buy.empty() )
                                   ask = (buy.top()).price;
                               else
                                   ask = 0;
                               if( !sell.empty() )
                                   bid = (sell.top()).price;
                               else
                                   bid = 0;
                               //ldprice = dprice;
                      }
                      else{
                           buy.push(node(dprice,dnum));
                           ask = (buy.top()).price;
                           bid = 0;
                      }
                  }
                  else{
                        if( buy.empty() ){
                            sell.push(node1(dprice,dnum));
                            ask = 0;
                            bid = (sell.top()).price;
                        }
                        else{
                              if( dprice > (buy.top()).price ){
                                  sell.push(node1(dprice,dnum));
                                  ask = (buy.top()).price;
                                  bid = (sell.top()).price;
                              }
                              else{
                                    node b1(0,0);
                                    while( !buy.empty() && dnum > 0 && dprice <= (buy.top()).price){
                                           b1 = buy.top();
                                           buy.pop();
                                           if( b1.num<=dnum ){
                                               dnum -= b1.num;
                                               b1.num = 0;
                                           }
                                           else{
                                                 b1.num -= dnum;
                                                 dnum = 0;
                                           }
                                           //ldprice = b1.price;
                                    }
                               if( dnum > 0 )
                                   sell.push(node1(dprice,dnum));
                               if( b1.num >0 )
                                   buy.push(b1);
                               if( !buy.empty() )
                                   ask = (buy.top()).price;
                               else
                                   ask = 0;
                               if( !sell.empty() )
                                   bid = (sell.top()).price;
                               else
                                   bid = 0;
                               ldprice = dprice;
                               }
                        }
                  }
           if( bid == 0 )
               printf("- ");
           else
               printf("%d ",bid);
           if( ask == 0 )
               printf("- ",ask);
           else
               printf("%d ",ask);
           if( ldprice == 0 )
               printf("-\n");
           else
               printf("%d\n",ldprice);
           }
    }
    return 0;
}

Smile3 2011-08-27 00:48 发表评论

树状数组 japan

Smile3 — Tue, 09 Aug 2011 13:50:00 GMT

#include
#include
#include

using namespace std;

int const max_n = 1010;

typedef struct node{
    int x;
    int y;
}node;

node a[max_n*max_n];
int num[max_n];
int b[max_n*max_n];
__int64 ans;
int t,n,m,k,max_y;

bool cmp(node na,node nb){
    if(na.x        return true;
    if( na.x == nb.x )
        if( na.y            return true;
    return false;
}

int lowbit( int t){//设t的末尾有k个0，返回值为2的k次方
    return t & (t^(t-1));
}

void modify(int pos,int add){//修改操作
    while ( pos <= max_y) {//变量范围为1至最大的值
       num[pos] += add;
       pos += lowbit(pos);
   }
}

int getsum(int end)
{
   int sum=0;
   while (end>0) {
       sum+=num[end];
       end-=lowbit(end);
   }
   return sum;
}

int main(){
    while( ~scanf("%d",&t) ){
        int id = 0;
        while(t--){
           scanf("%d%d%d",&n,&m,&k);
           max_y = -1;
           for( int i=0; i               scanf("%d%d",&a[i].x,&a[i].y);
               if( max_y < a[i].y )
                   max_y = a[i].y;
           }
           sort(a,a+k,cmp);
           for( int i=0; i               b[i+1] = a[i].y;
           ans = 0;
           memset(num,0,sizeof(num));
           for( int i=1; i<=k; i++ ){
               modify(b[i],1);//将b[i]之后的加1操作，表示比它小
               ans += i - getsum(b[i]);//getsum得到是比b[i]小的个数。
           }
           printf("Test case %d: %I64d\n",++id,ans);
       }
   }
    return 0;
}

Smile3 2011-08-09 21:50 发表评论

归并排序求逆序数 hunnu Japan 10888

Smile3 — Mon, 08 Aug 2011 14:36:00 GMT

这个题目是逆序数的变形，主要我存在的问题是思维的问题，先按x从小到大排序，如果x相等，再按y从小到大排序，我觉得我应该能想到怎么做，但是主要原因是我没有积极去思考，我发现每次我都是这样，有很强的依赖感和懒惰感，其实好多题目按照常规思路去想，加一点优化，或者利用现有的知识其实我是能够解决的，比如说比赛的时候我就过分依赖队友，我觉得这样不仅会让自己的水平得不到提高，更重要的是自己会丧失队友对我的信任，其实我是能够做出来的，比如说上次的动归，那还算有点难度，但是我还是想出来，现场也不是很多人想出来了，我下次会选择一个题目，然后仔细的思考，争取做出来，先想暴力的思路是什么，再优化，如果我连暴力都不敢暴力的话，我想我们怎么能够想出更复杂的算法去解决这个问题，所以我想对自己说的是，一要相信自己，不要去依赖别人，而是要积极思考，哪怕是最难的题目，也可以先用最暴力的方法解，但是如果不尝试的话，我想这是根本就不可能解出来的，所以一定要开动脑筋想，先想暴力，再想优化。

#include
#include

using namespace std;
typedef struct node{
   int x;
   int y;
}node;
__int64 ans;
int const max_n = 1000010;
node a[max_n];
int b[max_n];
int c[max_n];
int n,m,k;

bool cmp(node na,node nb){
   if( na.x < nb.x)
       return true;
   if( na.x > nb.x )
       return false;
   if( na.y < nb.y)
       return true;
   return false;
}

void merge_sort(int *A,int x,int y,int *T){//A数组x——y的左闭又开区间的归并算法
   if( y-x >1 ){
       //printf("%I64d\n",ans);
       int m = x +(y-x)/2;//此处不用（x+y）/2避免越界
       int p = x,q=m,i=x;
       merge_sort(A,x,m,T);
       merge_sort(A,m,y,T);
       while( p           if( q>=y || ( p               T[i++] = A[p++];
           else{
               T[i++] = A[q++];
               ans += m-p;
           }
       }
       for( i=x; i           A[i] = T[i];
       //printf("x = %d y = %d ans = %I64d\n",x,y,ans);
   }
}

int main(){
   int t;
   while( 1 == scanf("%d",&t) ){
       int id = 0;
       while( t-- ){
           scanf("%d%d%d",&n,&m,&k);
           for( int i=0; i               scanf("%d%d",&a[i].x,&a[i].y);
           sort(a,a+k,cmp);
           ans = 0;
           for( int i=0; i               b[i+1] = a[i].y;
           b[k+1] = 10000;
           merge_sort(b,1,k+1,c);
           printf("Test case %d: %I64d\n",++id,ans);
       }
   }
   return 0;
}

Smile3 2011-08-08 22:36 发表评论

hnu 12057 Cutting the Cake

Smile3 — Mon, 01 Aug 2011 09:35:00 GMT

  1 #include <cstdio>
  2 #include <cmath>
  3
  4 double const EPS = 1e-9;
  5
  6 struct point_d{
  7     double x, y;
  8     point_d(){}
  9     point_d(point_d const &p): x(p.x), y(p.y){}
10     point_d(double const &dx, double const &dy): x(dx), y(dy){}
11     point_d operator=(point_d const point){
12         x = point.x;
13         y = point.y;
14         return *this;
15     }
16 };
17
18 struct vector_d : public point_d{
19     vector_d(){}
20     vector_d(point_d const &p0, point_d const &p1){
21         x = p1.x - p0.x;
22         y = p1.y - p0.y;
23     }
24     vector_d operator=(vector_d const v){
25         x = v.x;
26         y = v.y;
27         return *this;
28     }
29 };
30
31 struct K{
32     double k0, k1;
33     int flag;
34     K():flag(1){}
35     struct K operator=(struct K const k){
36         k0 = k.k0; k1 = k.k1; flag = k.flag;
37         return *this;
38     }
39 };
40
41 struct line_d{
42     double a, b, c;
43 };
44
45 double distance(point_d p0, point_d p1){
46     return sqrt((p1.y - p0.y) * (p1.y - p0.y) + (p1.x - p0.x) * (p1.x - p0.x));
47 }
48
49 K func(double ls, double e0, double e1){
50     K k0, k1, k2;
51     k2.flag = 0;
52     if(ls * ls / 4.0 <= 2.0 * e0 * e1 - EPS)
53         return k2;
54     double r2 = sqrt(ls * ls / 4.0 - 2.0 * e0 * e1);
55     k0.k1 = (ls / 2.0 - r2) / 2.0 / e0;
56     k1.k1 = (ls / 2.0 + r2) / 2.0 / e0;
57     k0.k0 = (ls / 2.0 - k0.k1 * e0) / e1;
58     k1.k0 = (ls / 2.0 - k1.k1 * e0) / e1;
59     if(0 <= k0.k0 + EPS && k0.k0 <= 1.0 + EPS &&
60        0 <= k0.k1 + EPS && k0.k1 <= 1.0 + EPS)
61         return k0;
62     if(0 <= k1.k0 + EPS && k1.k0 <= 1.0 + EPS &&
63        0 <= k1.k1 + EPS && k1.k1 <= 1.0 + EPS)
64         return k1;
65     return k2;
66 }
67
68 line_d get_line(point_d p, point_d q){
69     line_d line;
70     line.a = -(q.y - p.y);
71     line.b = q.x - p.x;
72     line.c = line.a * q.x + line.b * q.y;
73     if(line.a < 0)
74         line.a *= -1.0, line.b *= -1.0, line.c *= -1.0;
75     double d = sqrt(line.a * line.a + line.b * line.b);
76     line.a /= d; line.b /= d; line.c /= d;
77     line.a += EPS; line.b += EPS; line.c += EPS;
78     return line;
79 }
80
81
82 point_d get_point(point_d start, vector_d vec, double k){
83     return point_d(start.x + vec.x * k, start.y + vec.y * k);
84 }
85
86 int main(void){
87     int t, cas;
88     point_d a, b, c, p, q;
89     double ls, eab, eac, ebc;
90     K k;
91     line_d line;
92
93     scanf("%d", &t);
94     while(t--){
95         scanf("%d%lf%lf%lf%lf%lf%lf", &cas, &a.x, &a.y, &b.x, &b.y,\
96                                       &c.x, &c.y);
97         eab = distance(a, b);
98         eac = distance(a, c);
99         ebc = distance(b, c);
100         ls = eab + eac + ebc;
101
102         k = func(ls, eab, eac);// a, bc
103         if(k.flag){
104             p = get_point(a, vector_d(a, b), k.k1);//ab
105             q = get_point(a, vector_d(a, c), k.k0);//ac
106             line = get_line(p, q);
107             printf("%d %.5lf %.5lf %.5lf\n", cas, line.a, line.b, line.c);
108             continue;
109         }
110         k = func(ls, eab, ebc);// b, a,c
111         if(k.flag){
112             p = get_point(b, vector_d(b, a), k.k1);//ba
113             q = get_point(b, vector_d(b, c), k.k0);//bc
114             line = get_line(p, q);
115             printf("%d %.5lf %.5lf %.5lf\n", cas, line.a, line.b, line.c);
116             continue;
117         }
118         k = func(ls, eac, ebc);//c, ab
119         if(k.flag){
120             p = get_point(c, vector_d(c, a), k.k1);//ca
121             q = get_point(c, vector_d(c, b), k.k0);//cb
122             line = get_line(p, q);
123             printf("%d %.5lf %.5lf %.5lf\n", cas, line.a, line.b, line.c);
124             continue;
125         }
126     }
127
128     return 0;
129 }

Smile3 2011-08-01 17:35 发表评论

xtu 转会

Smile3 — Fri, 15 Jul 2011 13:32:00 GMT

摘要: Code highlighting produced by Actipro CodeHighlighter (freeware)http://www.CodeHighlighter.com/--> 1 #include 2 #include &... 阅读全文

Smile3 2011-07-15 21:32 发表评论

poj 2777 Count Color

Smile3 — Tue, 31 May 2011 13:44:00 GMT

第一个线段树搞了好多天啊！！！！！！！！！！！！！！！！！
线段树就是个二叉树非叶子节点都有两个儿子。
线段树就是每个节点都是一条线段神马的父亲线段（非叶子）包含儿子线段。

这个题目就是更新的时候要注意
0 == color 就是代表有多种颜色在该线段上

1 #include <cstdio>
2 #include <cstring>
3 #include <algorithm>
4
5 int const max_length = 400005;
6 int const max_color  = 40;
7 struct Node{
8     int left, right;
9     int color;
10     void set(int _left, int _right, int _color = 1){
11         left = _left;
12         right = _right;
13         color = _color;
14     }
15 }st[max_length];
16 int used[max_color];
17
18 void init(int t, int left, int right);
19 void color(int t, int left, int right, int color);
20 void in_bucket(int t, int left, int right);
21
22 int main(void){
23     int l, t, o;
24     char buff[10];
25     int left, right, _color;
26
27     while(3 == scanf("%d%d%d", &l, &t, &o)){
28         init(1, 1, l + 1);
29         for(int i = 0; i < o; ++i){
30             scanf("%s", buff);
31             if('C' == buff[0]){
32                 scanf("%d%d%d", &left, &right, &_color);
33                 if(left > right)
34                     std::swap(left, right);
35                 color(1, left, right + 1, _color);
36             } else if('P' == buff[0]){
37                 scanf("%d%d", &left, &right);
38                 memset(used, 0, sizeof(used));
39                 in_bucket(1, left, right + 1);
40                 printf("%d\n", static_cast<int>(std::count(used, used + max_color, 1)));
41             }
42         }
43     }
44
45     return 0;
46 }
47
48 void init(int t, int left, int right){
49     st[t].set(left, right);
50     int mid = (left + right) >> 1;
51     if(mid == left)
52         return ;
53     init(t << 1, left, mid);
54     init(t << 1 | 1, mid, right);
55 }
56
57 void color(int t, int left, int right, int _color){
58     if(st[t].color == _color)
59         return ;
60     if(st[t].left == left && st[t].right == right){
61         st[t].color = _color;
62         return ;
63     }
64     if(0 != st[t].color){
65         st[t << 1].color = st[t].color;//保持非涂色区域本色(首先木有写着里无数WA啊)
66         st[t << 1 | 1].color = st[t].color;//保持非涂色区域本色
67         st[t].color = 0;
68     }
69     int mid =(st[t].left + st[t].right) >> 1;
70     if(mid >= right)
71         color(t << 1, left, right, _color);//在左儿子
72     else if(mid <= left)
73         color(t << 1 | 1, left, right, _color);//在右儿子
74      else {
75         color(t << 1, left, mid, _color);//在左儿子
76         color(t << 1 | 1, mid, right, _color);//在右儿子
77     }
78 }
79
80 void in_bucket(int t, int left, int right){
81     if(st[t].color > 0){
82         used[st[t].color] = 1;
83         return ;
84     }
85     int mid = (st[t].left + st[t].right) >> 1;
86     if(right <= mid)
87         in_bucket(t << 1, left, right);
88     else if(left >= mid)
89         in_bucket(t << 1 | 1, left, right);
90     else {
91         in_bucket(t << 1, left, mid);
92         in_bucket(t << 1 | 1, mid, right);
93     }
94 }

Smile3 2011-05-31 21:44 发表评论