POJ 1816 思维建图 + dinic + 枚举

思维建图的过程参考了这篇博客：参考

然后就是套Dinic模板，但是题目让我们求的是可以去掉那个顶点，以前求最小割的做法是正向DFS加反向DFS找出最小割，但是DFS找不到点，所以我们只能逐个枚举，反复套dinic模板。代码由详细注释

#include <iostream>
#include <cstring>
#include <cstdio>
#include <queue>
#define INF 0x3f3f3f3f
using namespace std;
const int maxv = 1010;
const int maxe = 200010;
int mmap[maxv][maxv], mmap2[maxv][maxv];
int n, src, dst;
/*******************Dinic模板**********************/
struct Edge {//因为有的时候（比如求最小割）外部需要用到遍信息，为了方便，把边放到class外面
    int from, to, cap, next;
} edge[maxe];
int cnt, head[maxv];
void edgeinit() {
    cnt = 0;
    memset(head, -1, sizeof head);
}
void addedge(int from, int to, int cap) {
    edge[cnt].from = from;
    edge[cnt].to = to;
    edge[cnt].cap = cap;
    edge[cnt].next = head[from];
    head[from] = cnt++;
    ;
    edge[cnt].from = to;
    edge[cnt].to = from;
    edge[cnt].cap = 0;
    edge[cnt].next = head[to];
    head[to] = cnt++;
}
/***************上面为建边，下面为Dinic**********************/
class Dinic {
private:
    int s, t, n, depth[maxv], cur[maxe]; //cur为当前狐优化
public:
    void init(int _s, int _t) {//这个模板只需要接受两个参数，一个是源点一个是汇点
        s = _s, t = _t;
    }
    int Dfs(int u, int dist) {
        if (u == t)
            return dist;
        for (int& i = cur[u]; i != -1; i = edge[i].next) {//注意这里的取址符
            if (depth[edge[i].to] == depth[u] + 1 && edge[i].cap != 0) {
                int delta = Dfs(edge[i].to, min(dist, edge[i].cap));
                if (delta > 0) {
                    edge[i].cap -= delta;
                    edge[i ^ 1].cap += delta;
                    return delta;
                }
            }
        }
        return false;
    }
    int Bfs() {
        queue<int> Q;
        memset(depth, 0, sizeof depth);
        depth[s] = 1;
        Q.push(s);
        while (!Q.empty()) {
            int u = Q.front();
            Q.pop();
            for (int i = head[u]; i != -1; i = edge[i].next) {
                if (edge[i].cap > 0 && depth[edge[i].to] == 0) {
                    depth[edge[i].to] = depth[u] + 1;
                    Q.push(edge[i].to);
                }
            }
        }
        if (depth[t] > 0)
            return true;
        return false;
    }
    int maxFlow() {
        int ans = 0;
        while (Bfs()) {
            for (int i = 0; i < cnt; i++)
                cur[i] = head[i];//cur是当前弧度优化，可以优化dfs的速度，但是dfs完后要初始化
            while (int delta = Dfs(s, INF))
                ans += delta;
        }
        return ans;
    }
} DC;
/*******************Dinic模板**********************/
void buildGraph() {
    edgeinit();
    for (int i = 1; i <= n; i++)
        for (int j = 1; j <= n; j++)
            if (i != j) {
                if (mmap[i][j] == 1)
                    addedge(n + i, j, INF); //因为是无向图，所以我们这里把点拆为两个点
            } else {
                if (i == src || i == dst)
                    addedge(i, n + i, INF); //源点和汇点不可能拆为割边，所以我们把这里设置为INF
                else
                    addedge(i, n + i, 1); //之前我们拆点的时候设置为INF，这里链接拆点就设置为1
            }
}
int main(void) {
    int res[maxv];
    while (cin >> n >> src >> dst) {
        for (int i = 1; i <= n; i++)
            for (int j = 1; j <= n; j++)
                cin >> mmap[i][j];
        if (mmap[src][dst] == 1) {
            cout << "NO ANSWER!" << endl;
            continue;
        }
        buildGraph();

        DC.init(src, n + dst);
        int maxflow = DC.maxFlow();

        cout << maxflow << endl;
        if (maxflow == 0)
            continue;

        int total = 0;
        for (int i = 1; i <= n; i++) { //枚举去掉每一个人，但是不包括s和t，如果maxflow发生变化total++
            if (i == src || i == dst)
                continue;
            for (int j = 1; j <= n; j++)
                for (int k = 1; k <= n; k++) {
                    mmap2[j][k] = mmap[j][k];
                    if (j == i || k == i)
                        mmap[j][k] = 0;
                }
            buildGraph();
            int temp = DC.maxFlow();
            if (maxflow > temp) {
                maxflow--;
                res[total] = i;
                total++;
            }
            else {
                for (int j = 1; j <= n; j++)
                    for (int k = 1; k <= n; k++)
                        mmap[j][k] = mmap2[j][k];
            }
            if (temp == 0)
                break;
        }
        for (int i = 0; i < total; i++)
            cout << res[i] << " \n"[i == total - 1];
    }
    return 0;
}