[PKU][POJ][3686][THE WINDY'S]

Source Code

• Source Code

# include <cstdio>
# include <cstring>
# include <cstdlib>
using namespace std;
# define max(a,b) ((a)>(b)?(a):(b))
# define min(a,b) ((a)<(b)?(a):(b))
# define N 15
int map[N][N],match[N],ln[N],rn[N],n,c;
bool l[N],r[N];
bool dfs(int pos)
{
	//if(l[pos]) return false;
	l[pos]=true;
	for(int i=0;i<n;i++)
		if(!r[i]&&ln[pos]+rn[i]==map[pos][i])
		{
			r[i]=true;
			if(match[i]==-1||dfs(match[i]))
			{
				match[i]=pos;
				return true;
			}

		}
		return false;
}
void adjust()
{
	int minnum=0xfffffff;
	for(int i=0;i<n;i++)
		if(l[i])
			for(int j=0;j<n;j++)
				if(!r[j])
					minnum=min(minnum,ln[i]+rn[j]-map[i][j]);
	for(int i=0;i<n;i++)
	{
		if(l[i]) ln[i]-=minnum;
		if(r[i]) rn[i]+=minnum;
	}
}
void search(int now)
{
	if(now==n)
	{
		int ret[N];
		for(int i=0;i<n;i++)
			ret[match[i]]=i;
		printf("Best Pairing %d\n",++c);
		for(int i=0;i<n;i++)
			printf("Supervisor %d with Employee %d\n",i+1,ret[i]+1);
		//system("pause");
	}
	for(int i=0;i<n;i++)
		if(match[i]==-1&&ln[now]+rn[i]==map[now][i])
		{
			match[i]=now;
			search(now+1);
			match[i]=-1;
		}
}
int main()
{
	int test;
	scanf("%d",&test);
	for(int tt=1;tt<=test;tt++)
	{
		scanf("%d",&n);
		memset(map,0,sizeof(map));
		for(int i=0;i<n;i++)
			for(int j=0;j<n;j++)
			{
				int t;
				scanf("%d",&t);
				map[t-1][i]-=j;
			}
		for(int i=0;i<n;i++)
			for(int j=0;j<n;j++)
			{
				int t;
				scanf("%d",&t);
				map[i][t-1]-=j;
			}
		
		memset(match,-1,sizeof(match));
		memset(rn,0,sizeof(rn));
		for(int i=0;i<n;i++)
		{
			int maxnum=-0xfffffff;
			for(int j=0;j<n;j++)
				maxnum=max(maxnum,map[i][j]);
			ln[i]=maxnum;
		}
		double ans=0;
		for(int i=0;i<n;i++)
		{
			memset(l,false,sizeof(l));
			memset(r,false,sizeof(r));
			while(!dfs(i))
			{
				adjust();
				memset(l,false,sizeof(l));
				memset(r,false,sizeof(r));
			}
		}
		for(int i=0;i<n;i++)
			ans-=map[match[i]][i];
		printf("Data Set %d, Best average difference: %.6f\n",tt,ans/n/2.0);
		memset(match,-1,sizeof(match));
		c=0;
		search(0);
		printf("\n");
	}
	return 0;
}

Here is a procedure's pseudocode:

	   go(int dep, int n, int m)
begin
output the value of dep.
if dep < m and x[a[dep]] + x[b[dep]] != c[dep] then go(dep + 1, n, m)
end

In this code n is an integer. a, b, c and x are 4 arrays of integers. The index of array always starts from 0. Array a and b consist of non-negative integers smaller than n. Array x consists of only 0 and 1. Array c consists of only 0, 1 and 2. The lengths of array a, b and c are m while the length of array x is n.

Given the elements of array a, b, and c, when we call the procedure go(0, n , m) what is the maximal possible value does the procedure output?

Input

There are multiple test cases. The first line of input is an integer T (0 < T ≤ 100), indicating the number of test cases. Then T test cases follow. Each case starts with a line of 2 integers n and m (0 < n ≤ 200, 0 < m ≤ 10000). Then m lines of 3 integers follow. The i-th(1 ≤ i ≤ m) line of them are a_i-1 ,b_i-1 and c_i-1 (0 ≤ a_i-1, b_i-1 < n, 0 ≤ c_i-1 ≤ 2).

Output

For each test case, output the result in a single line.

Sample Input

3
2 1
0 1 0
2 1
0 0 0
2 2
0 1 0
1 1 2

Sample Output

1
1
2

解法：

看到x为二进制数组就很敏感了，然后又看到c的值只能取0,1,2。很快想到了2-sat

t1=a[i],t2=b[i]

如果x[t1]+x[t2]!=0，那么就是说t1的0和t2的0冲突，添加t10->t21和t20->t11 

如果x[t1]+x[t2]!=2，那么就是说t1的1和t2的1冲突，添加t11->t20和t21->t10 

如果x[t1]+x[t2]!=1，那么就是说t1的0和t2的1冲突，以及t1的1和t2的0冲突，添加t10->t20和t21->t11 以及t11->t21和t20->t10 
贴代码
 1# include <cstdio>
 2# include <vector>
 3# include <cstring>
 4using namespace std;
 5int n,m;
 6vector<int> g[500];
 7int a[10005],b[10005],c[10005];
 8int low[500],dfn,color[500],stack[500],top,co;
 9void tarjan(int pos)
10{
11   int now=dfn++;
12   low[pos]=now;
13   stack[++top]=pos;
14   for(int i=0;i<g[pos].size();i++)
15   {
16      if(low[g[pos][i]]==-1) tarjan(g[pos][i]);
17      if(low[g[pos][i]]<low[pos]) low[pos]=low[g[pos][i]];
18   }
19   if(low[pos]>=now)
20   {
21       do
22       {
23          color[stack[top]]=co;
24          low[stack[top]]=2*n;
25       }while(stack[top--]!=pos);
26       co++;
27   }
28   
29}
30bool chk(int num)
31{
32   for(int i=0;i<(n<<1);i++)
33      g[i].clear();
34   for(int i=0;i<=num;i++)
35      switch(c[i])
36      {
37         case 0:
38              g[a[i]*2].push_back(b[i]*2+1);
39              g[b[i]*2].push_back(a[i]*2+1);
40              break;
41         case 1:
42              g[a[i]*2].push_back(b[i]*2);
43              g[b[i]*2+1].push_back(a[i]*2+1);
44              g[a[i]*2+1].push_back(b[i]*2+1);
45              g[b[i]*2].push_back(a[i]*2);
46              break;
47         case 2:
48              g[a[i]*2+1].push_back(b[i]*2);
49              g[b[i]*2+1].push_back(a[i]*2);
50              break;
51      };
52   memset(low,-1,sizeof(low));
53   dfn=co=0;
54   memset(color,-1,sizeof(color));
55   for(int i=0;i<2*n;i++)
56     if(low[i]==-1)
57     {
58       top=-1;
59       tarjan(i);
60     }
61   for(int i=0;i<n;i++)
62     if(color[2*i]==color[2*i+1])
63       return false;
64   return true;
65}
66int main()
67{
68    int test;
69    scanf("%d",&test);
70    while(test--)
71    {
72        scanf("%d%d",&n,&m);
73        for(int i=0;i<m;i++)
74          scanf("%d%d%d",a+i,b+i,c+i);
75        int s=0,e=m-1;
76        while(s<=e)
77        {
78           int mid=(s+e)>>1;
79           if(chk(mid)) s=mid+1;
80           else e=mid-1;
81        }
82        printf("%d\n",s);
83    }
84    return 0;
85}
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