UVA - 10480 Sabotage 最小割，输出割法_dengpangbu3703的博客-程序员宅基地

UVA - 10480 Sabotage

题意：现在有n个城市，m条路，现在要把整个图分成2部分，编号1，2的城市分成在一部分中，拆开每条路都需要花费，现在问达成目标的花费最少要隔开那几条路。

题解：建图直接按给你的图建一下，然后呢跑一下最大流，我们就知道了最小割是多少，答案就是最小割了  。

现在要求输出割法。我们从s开始往前跑，如果某条正向边有流量，我们就按着这条边继续往外走，知道无法再走，把所有经历过的点都染一下色。最后看所有的边，是不是有一头是染色了，另一头没有染色，如果是，这条边就是割边，输出就好了。

代码：

#include<bits/stdc++.h>
using namespace std;
#define Fopen freopen("_in.txt","r",stdin); freopen("_out.txt","w",stdout);
#define LL long long
#define ULL unsigned LL
#define fi first
#define se second
#define pb emplace_back
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
#define lch(x) tr[x].son[0]
#define rch(x) tr[x].son[1]
#define max3(a,b,c) max(a,max(b,c))
#define min3(a,b,c) min(a,min(b,c))
typedef pair<int,int> pll;
const int inf = 0x3f3f3f3f;
const LL INF = 0x3f3f3f3f3f3f3f3f;
const LL mod =  (int)1e9+7;
const int N = 200;
const int M = N*N;
int head[N], deep[N], cur[N];
int w[M], to[M], nx[M];
int tot;
void add(int u, int v, int val){
    w[tot]  = val; to[tot] = v;
    nx[tot] = head[u]; head[u] = tot++;

    w[tot] = 0; to[tot] = u;
    nx[tot] = head[v]; head[v] = tot++;
}
int bfs(int s, int t){
    queue<int> q;
    memset(deep, 0, sizeof(deep));
    q.push(s);
    deep[s] = 1;
    while(!q.empty()){
        int u = q.front();
        q.pop();
        for(int i = head[u]; ~i; i = nx[i]){
            if(w[i] > 0 && deep[to[i]] == 0){
                deep[to[i]] = deep[u] + 1;
                q.push(to[i]);
            }
        }
    }
    return deep[t] > 0;
}
int Dfs(int u, int t, int flow){
    if(u == t) return flow;
    for(int &i = cur[u]; ~i; i = nx[i]){
        if(deep[u]+1 == deep[to[i]] && w[i] > 0){
            int di = Dfs(to[i], t, min(w[i], flow));
            if(di > 0){
                w[i] -= di, w[i^1] += di;
                return di;
            }
        }
    }
    return 0;
}

int n, m;
int Dinic(int s, int t){
    int ans = 0, tmp;
    while(bfs(s, t)){
        for(int i = 1; i <= n; i++) cur[i] = head[i];
        while(tmp = Dfs(s, t, inf)) ans += tmp;
    }
    return ans;
}
int vis[N];
int top;
int ans[M][2];
void init(){
    memset(head, -1, sizeof(head));
    memset(vis, 0, sizeof(vis));
    top = tot = 0;
}
void solve(int u){
    vis[u] = 1;
    for(int i = head[u]; ~i; i = nx[i]){
        if(vis[to[i]]) continue;
        if(w[i]){
            solve(to[i]);
        }
    }
}
int main(){
    while(~scanf("%d%d", &n, &m) && n + m){
        int u, v, w;
        init();
        for(int i = 1; i <= m; i++){
            scanf("%d%d%d", &u, &v, &w);
            add(u, v, w);
            add(v, u, w);
        }
        Dinic(1, 2);
        solve(1);
        for(int i = 1; i <= n; i++){
            for(int j = head[i]; ~j; j = nx[j]){
                if(!(j&1)){
                    v = to[j];
                    if(vis[i] != vis[v] && i < v)
                        printf("%d %d\n", i, v);
                }
            }
        }
        puts("");
    }
    return 0;
}