学步园

同HDU 1853 & 3488，最小费用圈覆盖。

只不过这一次是无向边，所以在建图时建立双向边即可。

图的完备匹配实际上就是n个环的并。

/*HDU 1853*/
#include <iostream>
#include <cstdlib>
#include <cstdio>
#include <string>
#include <cstring>
#include <cmath>
#include <vector>
#include <queue>
#include <algorithm>
#include <map>
using namespace std;

const int maxn = 1010;
const int INF = 0x3f3f3f3f;

int n, m;

int W[maxn][maxn];
int Lx[maxn], Ly[maxn];

int Left[maxn];
bool S[maxn], T[maxn];

bool match(int i)
{
    S[i] = 1;
    for(int j = 1; j <= m; j++) if(Lx[i]+Ly[j] == W[i][j] && !T[j])
    {
        T[j] = 1;
        if(!Left[j] || match(Left[j]))
        {
            Left[j] = i;
            return 1;
        }
    }
    return 0;
}

void update()
{
    int a = INF;
    for(int i = 1; i <= n; i++) if(S[i])
    for(int j = 1; j <= m; j++) if(!T[j])
        a = min(a, Lx[i]+Ly[j]-W[i][j]);
        
    for(int i = 1; i <= n; i++)
    {
        if(S[i]) Lx[i] -= a;
    }
    for(int j = 1; j <= m; j++)
    {
        if(T[j]) Ly[j] += a;
    }
}

void KM()
{
    memset(Left, 0, sizeof(Left));
    memset(Lx, 0, sizeof(Lx));
    memset(Ly, 0, sizeof(Ly));
    
    for(int i = 1; i <= n; i++)
    {
        for(int j = 1; j <= m; j++)
            Lx[i] = max(Lx[i], W[i][j]);
    }
    for(int i = 1; i <= n; i++)
    {
        for(;;)
        {
            for(int j = 1; j <= m; j++) S[j] = T[j] = 0;
            if(match(i)) break; else update();
        }
    }
}

inline void readint(int &x)
{
    char c;
    c = getchar();
    while(!isdigit(c)) c = getchar();
    
    x = 0;
    while(isdigit(c)) x = x*10+c-'0', c = getchar();
}

inline void writeint(int x)
{
    if(x > 9) writeint(x/10);
    putchar(x%10+'0');
}

int N, M;

void read_case()
{
	readint(N), readint(M);
	n = m = 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++)
	{
		int u, v, w;
		readint(u), readint(v), readint(w);
		W[u][v] = W[v][u] = max(W[u][v], -w);
	}
}

int cal()
{
	int ans = 0;
	
	for(int i = 1; i <= m; i++)
	{
		if(Left[i] != 0 && W[Left[i]][i] != -INF) ans -= W[Left[i]][i];
		else return -1;
	}
	return ans;
}

int times;

void solve()
{
	read_case();
	
	KM();
	
	int ans = cal();
	if(ans == -1) printf("Case %d: NO\n", ++times);
	else { printf("Case %d: ", ++times); writeint(ans), puts(""); }
}

int main()
{
	int T;
	times = 0;
	for(readint(T); T > 0; T--)
	{
		solve();
	}
	return 0;
}