InputThere are several test cases in the input. You should process to the end of file (EOF).
The first line of each test case contains two integers N (N ≤ 100) and M, indicating the number of cities and the number of roads. The M lines followed, each of them contains three numbers A, B, and C, indicating that there is a road from city A to city B, whose length is C. (1 ≤ A,B ≤ N, A ≠ B, 1 ≤ C ≤ 1000).

6 9
1 2 5
2 3 5
3 1 10
3 4 12
4 1 8
4 6 11
5 4 7
5 6 9
6 5 4
6 5
1 2 1
2 3 1
3 4 1
4 5 1
5 6 1

Sample Output

42
-1


        
 

Hint

 In the first sample, there are two cycles, (1->2->3->1) and (6->5->4->6) whose length is 20 + 22 = 42. 
        
 
 

題意：輸入n個頂點m條邊，讓這些頂點形成環，至少包括兩個頂點.輸出最小的完備匹配和，如果不存在，輸出-1。

思路：把一個點一分為二，就形成了兩個集合，求最小匹配，則將權值取負後求最大匹配，最後返回負值。如果圖由多個有向環構成，則一定存在完備匹配，如果有權值最大的完備匹配存在，則一定是這個圖的最優完備匹配的值，如果有任何一個匹配邊是初始化的邊權值，說明這個完備匹配不存在。

之前嫌棄題水，現在又寫的頭疼，今天一天到現在才出了3題，自己真的好菜啊，什麽時候能像師父那樣厲害就好了，感覺算法真的需要自己理解了才能夠應對各種變形，就這道題就超時了

5遍，我怎麽知道要判斷重邊啊~~~

#include<stdio.h>
#include<string.h>
#define N 210
#define INF 0x3f3f3f3f

int map[N][N];
int ans,d;
int w[N][N],lx[N],ly[N];
int linker[N],visx[N],visy[N];
int n,m,nx,ny;

int dfs(int x)
{
    int y,tmp;
    visx[x] = 1;
    for(y = 1; y <= ny; y ++)
    {
        if(!visy[y])
        {
            tmp = lx[x] + ly[y] - w[x][y];
            if(!tmp)
            {
                visy[y] = 1;
                if(linker[y]==-1||dfs(linker[y]))
                {
                    linker[y] = x;
                    return 1;
                }
            }
            else if( d > tmp)
                d = tmp;
        }
    }
    return 0;
}

int KM()
{
    int x,y,i,j,sum;
    memset(linker,-1,sizeof(linker));
    memset(ly,0,sizeof(ly));
    for(x = 1; x <= nx; x ++)
        for(y = 1,lx[x] = -INF; y <= ny; y ++)
            if(lx[x] < w[x][y])
                lx[x] = w[x][y];
                
    for(x = 1; x <= nx; x ++)
    {
        while(1)
        {
            d = INF;
            memset(visx,0,sizeof(visx));
            memset(visy,0,sizeof(visy));
            if(dfs(x))
                break;
            for(i = 1; i <= nx; i ++)
                if(visx[i])
                    lx[i] -= d;
            for(i = 1; i <= ny; i ++)
                if(visy[i])
                    ly[i] += d;
        }
    }
    sum = 0;
    for(i = 1; i <= ny; i ++)
    {
        if(w[linker[i]][i] != -INF)
            sum += w[linker[i]][i];
        else
            return -1;
    }
    return -sum;
}

int main()
{
    int t1,t2,t3;
    while(scanf("%d%d",&n,&m)!=EOF)
    {
        nx = ny = n;
        for(int i = 1; i <= n; i ++)
            for(int j = 1; j <= n; j ++)
                w[i][j] = -INF;
        for(int i = 1; i <= m; i ++)
        {
            scanf("%d%d%d",&t1,&t2,&t3);
            if(-t3 > w[t1][t2])//重復輸入取小值 
                w[t1][t2] = -t3;
        }
        ans = KM();
        printf("%d\n",ans);    
    }
    return 0;
}

【二分圖匹配入門專題1】M - Cyclic Tour hdu1853【km算法--判斷自重邊】