Aizu 2304 Reverse Roads

原题链接：http://judge.u-aizu.ac.jp/onlinejudge/description.jsp?id=2304

题意：

给你一个网络，其中每条边的容量是1，你可以通过调整边的方向来获得更大的流量，现在问你能获得的最大流量是多少。并且输出更改方向的边的编号。

题解：

就每条边弄成无向的，并且标记一下是否是原始边，然后跑一发Dinic即可。然后在残余网络上寻找解即可。

代码：

#include<iostream>

#include<stack>

#include<vector>

#include<cstring>

#include<string>

#include<algorithm>

#include<cmath>

#include<cstdio>

#include<queue>

#define MAX_S (1<<10)+10

#define MAX_V 500

#define MAX_N MAX_V

#define INF 1000009

using namespace std;

struct edge {

    int to, cap, rev;

    bool isRev;

    bool isOri;

    int id;

    edge(int t, int c, int r, bool ir, bool io,int iid)

            : to(t), cap(c), rev(r), isRev(ir), isOri(io),id(iid) { }

    edge() { }

};

template <class T>

inline bool scan_d(T &ret)

{

    char c;

    int sgn;

    if(c=getchar(),c==EOF) return ; //EOF

    while(c!=' -' &&(c<'' ||c>'' )) c=getchar();

    sgn=(c==' -' )?-:;

    ret=(c==' -' )?:(c-'' );

    while(c=getchar(),c>='' &&c<='' ) ret=ret*+(c-'' );

    ret*=sgn;

    return ;

}

vector<edge> G[MAX_N];

int level[MAX_V];

int iter[MAX_V];

void init(int totNode) {

    for (int i = ; i <= totNode; i++)

        G[i].clear();

    memset(level, , sizeof(level));

    memset(iter, , sizeof(iter));

}

void add_edge(int from,int to,int cap,bool io,int id) {

    G[from].push_back(edge (to, cap, G[to].size(),,io,id));

    G[to].push_back(edge (from, , G[from].size() - ,,io,id));

}

void bfs(int s) {

    queue<int> que;

    memset(level, -, sizeof(level));

    level[s] = ;

    que.push(s);

    while (!que.empty()) {

        int v = que.front();

        que.pop();

        for (int i = ; i < G[v].size(); i++) {

            edge &e = G[v][i];

            if (e.cap >  && level[e.to] < ) {

                level[e.to] = level[v] + ;

                que.push(e.to);

            }

        }

    }

}

int dfs(int v,int t,int f) {

    if (v == t)return f;

    for (int &i = iter[v]; i < G[v].size(); i++) {

        edge &e = G[v][i];

        if (e.cap >  && level[v] < level[e.to]) {

            int d = dfs(e.to, t, min(f, e.cap));

            if (d > ) {

                e.cap -= d;

                G[e.to][e.rev].cap += d;

                return d;

            }

        }

    }

    return ;

}

int max_flow(int s,int t) {

    int flow = ;

    for (; ;) {

        bfs(s);

        if (level[t] < )return flow;

        memset(iter, , sizeof(iter));

        int f;

        while ((f = dfs(s, t, INF)) > ) {

            flow += f;

        }

    }

}

int S,T;

int N,M;

vector<int> ans;

int main() {

    scanf("%d%d", &N, &M);

    for (int i = ; i < M; i++) {

        int u, v;

        scanf("%d%d", &u, &v);

        add_edge(u, v, , , i + );

        add_edge(v, u, , , i + );

    }

    scanf("%d%d", &S, &T);

    int f = max_flow(S, T);

    printf("%d\n", f);

    for (int i = ; i <= N; i++)

        for (int j = ; j < G[i].size(); j++)

            if (G[i][j].isRev ==  && G[i][j].isOri ==  && G[i][j].cap == )

                ans.push_back(G[i][j].id);

    printf("%d\n", ans.size());

    for (int i = ; i < ans.size(); i++)

        printf("%d\n", ans[i]);

    return ;

}