HDU 3251 Being a Hero（最小割）

题目地址
题意：你是一个英雄，你们国家的总统要给你f个城市中的若干个城市。但是要求你拥有的城市不能与首都1有通路，所以你要毁掉一些道路，但是每条道路会有一个毁掉需要的代价，每个城市有利润，求最大的利润，以及删除的边的编号。
思路：这就是要把这个图分为2个集合，求最小的边要割多少，这样看就是最小割的题目了，因为最大流等于最小割（证明戳这里）。先把你能的到的f个城市能获得的利润和求出来，然后减去最小割就是答案了。
吐槽：看了2个小时代码，硬是最后才看到自己链式前向星又忘记开2倍空间，导致TLE了。

#include <iostream>#include <cstring>#include <string>#include <queue>#include <vector>#include <map>#include <set>#include <stack>#include <cmath>#include <cstdio>#include <algorithm>#include <iomanip>#define N 4010#define M 100010 #define LL __int64#define inf 0x3f3f3f3f#define lson l,mid,ans<<1#define rson mid+1,r,ans<<1|1#define getMid (l+r)>>1#define movel ans<<1#define mover ans<<1|1using namespace std;const LL mod = 1000000007;int head[N], level[N];int n, m, cnt, f;struct node {    int to;    int cap;//剩余流量    int next;    int id;//边的编号}edge[2 * M];//这辈子都要记得链式前向星开二倍空间bool vis[N];struct Dinic {    void init() {        memset(vis, false, sizeof(vis));        memset(head, -1, sizeof(head));        cnt = 0;    }    void add(int u, int v, int cap, int id) {//有向图        edge[cnt].to = v, edge[cnt].cap = cap, edge[cnt].next = head[u], edge[cnt].id = id, head[u] = cnt++;        edge[cnt].to = u, edge[cnt].cap = 0, edge[cnt].next = head[v], edge[cnt].id = -1, head[v] = cnt++;//反向边    }    bool bfs(int s, int t) {//建立分层图        memset(level, -1, sizeof(level));        queue<int>q;        level[s] = 0;//源点的层次最高        q.push(s);        while (!q.empty()) {            int u = q.front();            q.pop();            for (int i = head[u]; i != -1; i = edge[i].next) {                int v = edge[i].to;                if (edge[i].cap > 0 && level[v] < 0) {                    level[v] = level[u] + 1;                    q.push(v);                }            }        }        return level[t] != -1;    }    int dfs(int u, int t, int num) {//找增广路        if (u == t) {//找到了汇点返回当前的最小值，在这条路径上分别减去最小值            return num;        }        for (int i = head[u]; i != -1; i = edge[i].next) {            int v = edge[i].to;            if (edge[i].cap > 0 && level[u] < level[v]) {                int d = dfs(v, t, min(num, edge[i].cap));                if (d > 0) {                    edge[i].cap -= d;                    edge[i ^ 1].cap += d;//反向边加值                    return d;                }            }        }        level[u] = -1;        return 0;    }    int minflow(int s, int t) {//源点和汇点        int sum = 0, num;        while (bfs(s, t)) {            while (num = dfs(s, t, inf), num > 0) {//当前层次图不断的找增广路                sum += num;            }        }        return sum;    }}dc;void dfs(int u) {    for (int i = head[u]; i != -1; i = edge[i].next) {        int v = edge[i].to;        if (!vis[v] && edge[i].cap > 0) {            vis[v] = true;            dfs(v);        }    }}int main() {    cin.sync_with_stdio(false);    int T, Case = 1;    int a, b, c;    cin >> T;    while (T--) {        in >> n >> m >> f;        dc.init();        dc.add(0, 1, inf, -1);        for (int i = 0; i < m; i++) {            cin >> a >> b >> c;            dc.add(a, b, c, i + 1);        }        int sum = 0;        for (int i = 0; i < f; i++) {            cin >> a >> b;            dc.add(a, n + 1, b, -1);            sum += b;        }        cout << "Case " << Case++ << ": " << sum - dc.minflow(0, n + 1) << endl;        vis[1] = true;        dfs(1);        vector<int>v;        for (int i = 0; i < cnt; i += 2) {            if (vis[edge[i ^ 1].to] && !vis[edge[i].to] && edge[i].id != -1) {                v.push_back(edge[i].id);            }        }        cout << v.size();        for (int i = 0; i < v.size(); i++) {            cout << " " << v[i] + 1;//因为是边号从一开始所以要加一        }        cout << endl;    }    return 0;}