版权声明：get busy living...or get busy dying https://blog.csdn.net/qq_41444888/article/details/89363807
https://cn.vjudge.net/contest/68127#problem/Q
KM求解最小权的方法：把所有权变都设为负值，这样最后求解的时候就可以求解最小权了
这道题把每个边的两个点分到二分图两个部中，因为路径都是环，所以直接匹配就可以
#include<iostream>
#include<cstring>
#include<cstdio>
#include<algorithm>
#include<vector>
#include<set>
#include<map>
#include<queue>
#include<cmath>
#define ll long long
#define mod 1000000007
#define inf 0x3f3f3f3f
using namespace std;
const int maxn = 305;
int n, m;
int nx, ny;//两边的点数
int g[maxn][maxn];
int link[maxn], lx[maxn], ly[maxn]; // y中各店匹配状态， x，y中的点标号
bool visx[maxn], visy[maxn];
int slack[maxn];//被匹配者被匹配上需要的最小期望值
bool dfs(int x)
{
visx[x] = 1;
for(int y = 1; y <= ny; y ++)
{
if(visy[y]) continue;
int tmp = lx[x] + ly[y] - g[x][y];
if(tmp == 0)
{
visy[y] = 1;
if(link[y] == -1 || dfs(link[y]))
{
link[y] = x;
return 1;
}
}
else if(slack[y] > tmp)
slack[y] = tmp;
}
return 0;
}
int km()
{
memset(link, -1, sizeof(link));
memset(ly, 0, sizeof(ly));
for(int i = 1; i <= nx; i ++)
{
lx[i] = - inf;
for(int j = 1; j <= ny; j ++)
{
if(g[i][j] > lx[i])
lx[i] = g[i][j];
}
}
for(int x = 1; x <= nx; x ++)
{
for(int 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 d = inf;
for(int i = 1; i <= ny; i ++)
if(! visy[i] && d > slack[i])
d = slack[i];
for(int i = 1; i <= nx; i ++)
if(visx[i])
lx[i] -= d;
for(int i = 1; i <= ny; i ++)
if(visy[i]) ly[i] += d;
else slack[i] -= d;
}
}
int ans = 0;
for(int i = 1; i <= ny; i ++)
if(link[i] != -1)
ans += g[link[i]][i];
return ans;
}
int main()
{
int t;
scanf("%d", &t);
while(t --)
{
scanf("%d%d", &n, &m);
for(int i = 1; i <= n; i ++)
for(int j = 1; j <= n; j ++)
g[i][j] = -inf;
int u, v, w;
for(int i = 1; i <= m; i ++)
{
scanf("%d%d%d", &u, &v, &w);
if(g[u][v] < -w)
g[u][v] = -w;
}
nx = ny = n;
km();
int ans = 0;
for(int i = 1; i <= n; i ++)
ans += -g[link[i]][i];
printf("%d\n", ans);
}
return 0;
}