Sabotage 【UVA - 10480】【最大流割边】

题目链接

很容易会想到是最大流建边，但是同样的这里有坑点，就是有的人去输出边的时候，去把残余网络的流为0的边给输出了，其实不然，我们应当输出的是那些最后跑到深度为0的不能再走下去的点，只要把他们割了，就一定会是最优的解。

#include <iostream>
#include <cstdio>
#include <cmath>
#include <string>
#include <cstring>
#include <algorithm>
#include <limits>
#include <vector>
#include <stack>
#include <queue>
#include <set>
#include <map>
#define lowbit(x) ( x&(-x) )
#define pi 3.141592653589793
#define e 2.718281828459045
#define INF 0x3f3f3f3f
#define HalF (l + r)>>1
#define lsn rt<<1
#define rsn rt<<1|1
#define Lson lsn, l, mid
#define Rson rsn, mid+1, r
#define QL Lson, ql, qr
#define QR Rson, ql, qr
#define myself rt, l, r
using namespace std;
typedef unsigned long long ull;
typedef long long ll;
const int maxN = 107, maxE = 2e3 + 7, S = 1, T = 2;
int N, M, cnt, head[maxN], cur[maxN];
struct Eddge
{
    int nex, to, flow, old;
    Eddge(int a=-1, int b=0, int c=0):nex(a), to(b), flow(c), old(c) {}
}edge[maxE];
inline void addEddge(int u, int v, int flow)
{
    edge[cnt] = Eddge(head[u], v, flow);
    head[u] = cnt++;
}
inline void _add(int u, int v, int flow) { addEddge(u, v, flow); addEddge(v, u, 0); }
int deep[maxN];
queue<int> Q;
inline bool bfs()
{
    memset(deep, 0, sizeof(deep)); deep[S] = 1;
    while(!Q.empty()) Q.pop();
    Q.push(S);
    while(!Q.empty())
    {
        int u = Q.front(); Q.pop();
        for(int i=head[u], v, f; ~i; i=edge[i].nex)
        {
            v = edge[i].to; f = edge[i].flow;
            if(f && !deep[v])
            {
                deep[v] = deep[u] + 1;
                Q.push(v);
            }
        }
    }
    return deep[T];
}
inline int dfs(int u, int dist)
{
    if(u == T) return dist;
    for(int &i=cur[u], v, f; ~i; i=edge[i].nex)
    {
        v = edge[i].to; f = edge[i].flow;
        if(f && deep[v] == deep[u] + 1)
        {
            int di = dfs(v, min(dist, f));
            if(di)
            {
                edge[i].flow -= di;
                edge[i^1].flow += di;
                return di;
            }
        }
    }
    return 0;
}
inline void Dinic()
{
    while(bfs())
    {
        for(int i=1; i<=N; i++) cur[i] = head[i];
        while(dfs(S, INF)) ;
    }
}
bool used[55][55];
inline void Out(int u, int fa)
{
    for(int i=head[u], v, f; ~i; i=edge[i].nex)
    {
        v = edge[i].to; f = edge[i].flow;
        if(v == fa) continue;
        if(!deep[v] && edge[i].old)
        {
            if(used[u][v]) continue;
            used[u][v] = used[v][u] = true;
            printf("%d %d\n", u, v);
        }
        else if(edge[i].old > f) { Out(v, u); }
    }
}
inline void init()
{
    cnt = 0;
    memset(head, -1, sizeof(head));
    memset(used, false, sizeof(used));
}
int main()
{
    int Cas = 0;
    while(scanf("%d%d", &N, &M) && (N || M))
    {
        if(Cas++) printf("\n");
        init();
        for(int i=1, u, v, w; i<=M; i++)
        {
            scanf("%d%d%d", &u, &v, &w);
            _add(u, v, w);
            _add(v, u, w);
        }
        Dinic();
        Out(S, -1);
    }
    return 0;
}

原文地址：https://www.cnblogs.com/WuliWuliiii/p/10987052.html