[SWUSTOJ1737] 太空飞行计划问题（最大权闭合子图，记录路径）

题目链接：https://www.oj.swust.edu.cn/problem/show/1737

很经典的建图，但是需要记录路径。

vis数组标记点是否被扩展，在每次dinic通过bfs扩展的时候假如被扩展的点，最后一次bfs随后不再有增广路便是最终结果。

#include <bits/stdc++.h>
using namespace std;

typedef struct Edge {
    int u, v, w, ww, next;
}Edge;

const int inf = 0x7f7f7f7f;
const int maxn = 3010;
const int maxm = maxn << 2;
int cnt, dhead[maxn];
int cur[maxn], dd[maxn];
bool vis[maxn];
Edge dedge[maxm];
int S, T, N;

void init() {
    memset(dhead, -1, sizeof(dhead));
    for(int i = 0; i < maxn; i++) dedge[i].next = -1;
    S = 0; cnt = 0;
}

void adde(int u, int v, int w, int c1=0) {
    dedge[cnt].u = u; dedge[cnt].v = v; dedge[cnt].w = w; dedge[cnt].ww = w;
    dedge[cnt].next = dhead[u]; dhead[u] = cnt++;
    dedge[cnt].u = v; dedge[cnt].v = u; dedge[cnt].w = c1; dedge[cnt].ww = c1;
    dedge[cnt].next = dhead[v]; dhead[v] = cnt++;
}

bool bfs(int s, int t, int n) {
    memset(vis, 0, sizeof(vis));
    queue<int> q;
    for(int i = 0; i < n; i++) dd[i] = inf;
    dd[s] = 0;
    q.push(s);
    while(!q.empty()) {
        int u = q.front(); q.pop();
        for(int i = dhead[u]; ~i; i = dedge[i].next) {
            if(dd[dedge[i].v] > dd[u] + 1 && dedge[i].w > 0) {
                dd[dedge[i].v] = dd[u] + 1;
                vis[dedge[i].v] = 1;
                if(dedge[i].v == t) return 1;
                q.push(dedge[i].v);
            }
        }
    }
    return 0;
}

int dinic(int s, int t, int n) {
    int st[maxn], top;
    int u;
    int flow = 0;
    while(bfs(s, t, n)) {
        for(int i = 0; i < n; i++) cur[i] = dhead[i];
        u = s; top = 0;
        while(cur[s] != -1) {
            if(u == t) {
                int tp = inf;
                for(int i = top - 1; i >= 0; i--) {
                    tp = min(tp, dedge[st[i]].w);
                }
                flow += tp;
                for(int i = top - 1; i >= 0; i--) {
                    dedge[st[i]].w -= tp;
                    dedge[st[i] ^ 1].w += tp;
                    if(dedge[st[i]].w == 0) top = i;
                }
                u = dedge[st[top]].u;
            }
            else if(cur[u] != -1 && dedge[cur[u]].w > 0 && dd[u] + 1 == dd[dedge[cur[u]].v]) {
                st[top++] = cur[u];
                u = dedge[cur[u]].v;
            }
            else {
                while(u != s && cur[u] == -1) {
                    u = dedge[st[--top]].u;
                }
                cur[u] = dedge[cur[u]].next;
            }
        }
    }
    return flow;
}

int n, m;
char tmp[maxm];
vector<int> s;

int main() {
    // freopen("in", "r", stdin);
    int pay;
    while(~scanf("%d%d",&n,&m)) {
        init();
        S = 0, T = m + n + 1, N = T + 1;
        int pos = 0;
        for(int i = 1; i <= n; i++) {
            scanf("%d",&pay);
            pos += pay;
            adde(S, i, pay);
            char p = getchar();
            while((p = getchar()) != '
') {
                pay = p - '0';
                while((p = getchar()) && p >= '0' && p <= '9') pay = pay * 10 + p - '0';
                adde(i, pay+n, inf);
                if(p == '
') break;
            }
        }
        for(int i = 1; i <= m; i++) {
            scanf("%d", &pay);
            adde(n+i, T, pay);
        }
        int ret = pos - dinic(S, T, N);
        s.clear();
        for(int i = 1; i <= n; i++) {
            if(vis[i]) s.push_back(i);
        }
        printf("%d", s[0]);
        for(int i = 1; i < s.size(); i++) printf(" %d", s[i]);
        printf("
");
        s.clear();
        for(int i = n+1; i <= n+m; i++) {
            if(vis[i]) s.push_back(i-n);
        }
        printf("%d", s[0]);
        for(int i = 1; i < s.size(); i++) printf(" %d", s[i]);
        printf("
");
        printf("%d
", ret);
    }
    return 0;
}