Aizu 2304 Reverse Roads(无向流)

把有向图修改成无向图,并保证每条边的流量守恒并满足有向容量(即abs(flow(u,v) - flow(v,u)) <= 1)满足限制。

得到最大流,根据残流输出答案。

因为最后少了' '而WA...

#include<bits/stdc++.h>
using namespace std;
const int N = 301,M = N*(N-1);
int n,m;
int hd[N],nx[M],to[M],cap[M],ect;

inline void addEdge(int u,int v,int c = 1)
{
    nx[ect] = hd[u];
    to[ect] = v;
    cap[ect] = c;
    hd[u] = ect++;
}

int S,T;
int vis[N],clk;
int lv[N],q[N];
int cur[N];

bool bfs()
{
    int l = 0,r = 0;
    clk++;
    q[r++] = S; vis[S] = clk; lv[S] = 0;
    while(l<r){
        int u = q[l++];
        for(int i = hd[u]; ~i; i = nx[i]){
            int v = to[i];
            if(vis[v] != clk && cap[i]>0 ){
                vis[v] = clk;
                lv[v] = lv[u]+1;
                q[r++] = v;
            }
        }
    }
    return vis[T] == clk;
}

int aug(int u,int a)
{
    if(u == T||!a) return a;
    int flow = 0,f;
    for(int &i = cur[u]; ~i; i = nx[i]){
        int v = to[i];
        if(lv[v] == lv[u]+1 && (f = aug(v,min(cap[i],a)))>0){
            cap[i] -= f; cap[i^1] += f;
            a -= f; flow += f;
            if(!a) break;
        }
    }
    return flow;
}
const int INF = 0x3f3f3f3f;
int MaxFlow()
{
    int flow = 0;
    while(bfs()){
        memcpy(cur,hd,sizeof(hd));
        flow += aug(S,INF);
    }
    return flow;
}

//#define LOCAL
int main()
{
#ifdef LOCAL
    freopen("in.txt","r",stdin);
#endif
    scanf("%d%d",&n,&m);
    memset(hd,-1,sizeof(hd));
    for(int i = 0; i < m; i++){
        int u,v; scanf("%d%d",&u,&v);
        addEdge(v,u);
        addEdge(u,v);
    }
    scanf("%d%d",&S,&T);
    printf("%d
",MaxFlow());
    vector<int> ans;
    for(int i = 0; i < m; i++){
        if(!cap[i<<1]) ans.push_back(i+1);
    }
    printf("%d
",(int)ans.size());
    for(auto a:ans){
        printf("%d
",a);
    }
    return 0;
}
原文地址:https://www.cnblogs.com/jerryRey/p/4851357.html