HDU1853-Cyclic Tour KM

Description
There are N cities in our country, and M one-way roads connecting them. Now Little Tom wants to make several cyclic tours, which satisfy that, each cycle contain at least two cities, and each city belongs to one cycle exactly. Tom wants the total length of all the tours minimum, but he is too lazy to calculate. Can you help him?

Input
There 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).

Output
Output one number for each test case, indicating the minimum length of all the tours. If there are no such tours, output -1.

Sample Input
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

题意：给你N个城市M条路，求若干个环，保证每个城市在且仅在某一个环里，输出最小代价。

思路：想到每个环的每个节点都保证了出度=入度=1，就可以转化为二分图匹配了~。x点集和y点集都是给出的点集，求一遍KM即可。

#include<bits/stdc++.h>using namespace std;#define M 310#define inf 0x3f3f3f3fint nx,ny,n,m; int link[M];   int visx[M],visy[M];int w[M][M],lx[M],ly[M],slack[M];int ans[M];    int DFS(int x)    {        visx[x] = 1;        for (int y = 1;y <= ny;y ++)        {            if (visy[y])                continue;            int t = lx[x] + ly[y] - w[x][y];            if (t < 1e-5)                   {                visy[y] = 1;                if (link[y] == -1||DFS(link[y]))                {                    link[y] = x;                    return 1;                }            }            else if (slack[y] > t)                  slack[y] = t;        }        return 0;    }    void KM()    {        int i,j;        memset (link,-1,sizeof(link));        memset (ly,0,sizeof(ly));        memset(lx,0,sizeof(lx));        for (i = 1;i <= nx;i ++)                       for (j = 1,lx[i] = -inf;j <= ny;j ++)                if (w[i][j] > lx[i])                    lx[i] = w[i][j];        for (int x = 1;x <= nx;x ++)        {            for (i = 1;i <= ny;i ++)                slack[i] = inf;            while (1)            {                memset (visx,0,sizeof(visx));                memset (visy,0,sizeof(visy));                if (DFS(x))                         break;                  int a=inf;                for(int i=1;i<=nx;i++)if(visx[i])                for(int j=1;j<=ny;j++)if(!visy[j])                a=min(a,lx[i]+ly[j]-w[i][j]);                for(int i=1;i<=nx;i++){                   if(visx[i])lx[i]-=a;                   if(visy[i])ly[i]+=a;                }            }        }        int res = 0,cnt = 0;        for(int i = 1;i <= ny;i++)        {            if(link[i] > -1 && w[link[i]][i] != -inf)            res+=w[link[i]][i],cnt++;        }        if(cnt != ny) printf("-1\n");        else printf("%d\n",-res);    }int main(){    freopen("h1853.in","r",stdin);    while(~scanf("%d%d",&n,&m))    {        for(int i = 0;i <= n;i++)        {            for(int j = 0;j <= n;j++)            {                w[i][j] = -inf;            }        }        for(int i = 0;i < m;i++)        {            int u,v,w1;            scanf("%d%d%d",&u,&v,&w1);            if(-w[u][v] > w1) w[u][v] = -w1;        }        nx = ny = n;        KM();    }   return 0;}