[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。

  1 #include <bits/stdc++.h>
  2 using namespace std;
  3
  4 typedef struct Edge {
  5     int u, v, w, ww, next;
  6 }Edge;
  7
  8 const int inf = 0x7f7f7f7f;
  9 const int maxn = 3010;
 10 const int maxm = maxn << 2;
 11 int cnt, dhead[maxn];
 12 int cur[maxn], dd[maxn];
 13 bool vis[maxn];
 14 Edge dedge[maxm];
 15 int S, T, N;
 16
 17 void init() {
 18     memset(dhead, -1, sizeof(dhead));
 19     for(int i = 0; i < maxn; i++) dedge[i].next = -1;
 20     S = 0; cnt = 0;
 21 }
 22
 23 void adde(int u, int v, int w, int c1=0) {
 24     dedge[cnt].u = u; dedge[cnt].v = v; dedge[cnt].w = w; dedge[cnt].ww = w;
 25     dedge[cnt].next = dhead[u]; dhead[u] = cnt++;
 26     dedge[cnt].u = v; dedge[cnt].v = u; dedge[cnt].w = c1; dedge[cnt].ww = c1;
 27     dedge[cnt].next = dhead[v]; dhead[v] = cnt++;
 28 }
 29
 30 bool bfs(int s, int t, int n) {
 31     memset(vis, 0, sizeof(vis));
 32     queue<int> q;
 33     for(int i = 0; i < n; i++) dd[i] = inf;
 34     dd[s] = 0;
 35     q.push(s);
 36     while(!q.empty()) {
 37         int u = q.front(); q.pop();
 38         for(int i = dhead[u]; ~i; i = dedge[i].next) {
 39             if(dd[dedge[i].v] > dd[u] + 1 && dedge[i].w > 0) {
 40                 dd[dedge[i].v] = dd[u] + 1;
 41                 vis[dedge[i].v] = 1;
 42                 if(dedge[i].v == t) return 1;
 43                 q.push(dedge[i].v);
 44             }
 45         }
 46     }
 47     return 0;
 48 }
 49
 50 int dinic(int s, int t, int n) {
 51     int st[maxn], top;
 52     int u;
 53     int flow = 0;
 54     while(bfs(s, t, n)) {
 55         for(int i = 0; i < n; i++) cur[i] = dhead[i];
 56         u = s; top = 0;
 57         while(cur[s] != -1) {
 58             if(u == t) {
 59                 int tp = inf;
 60                 for(int i = top - 1; i >= 0; i--) {
 61                     tp = min(tp, dedge[st[i]].w);
 62                 }
 63                 flow += tp;
 64                 for(int i = top - 1; i >= 0; i--) {
 65                     dedge[st[i]].w -= tp;
 66                     dedge[st[i] ^ 1].w += tp;
 67                     if(dedge[st[i]].w == 0) top = i;
 68                 }
 69                 u = dedge[st[top]].u;
 70             }
 71             else if(cur[u] != -1 && dedge[cur[u]].w > 0 && dd[u] + 1 == dd[dedge[cur[u]].v]) {
 72                 st[top++] = cur[u];
 73                 u = dedge[cur[u]].v;
 74             }
 75             else {
 76                 while(u != s && cur[u] == -1) {
 77                     u = dedge[st[--top]].u;
 78                 }
 79                 cur[u] = dedge[cur[u]].next;
 80             }
 81         }
 82     }
 83     return flow;
 84 }
 85
 86 int n, m;
 87 char tmp[maxm];
 88 vector<int> s;
 89
 90 int main() {
 91     // freopen("in", "r", stdin);
 92     int pay;
 93     while(~scanf("%d%d",&n,&m)) {
 94         init();
 95         S = 0, T = m + n + 1, N = T + 1;
 96         int pos = 0;
 97         for(int i = 1; i <= n; i++) {
 98             scanf("%d",&pay);
 99             pos += pay;
100             adde(S, i, pay);
101             char p = getchar();
102             while((p = getchar()) != ‘\n‘) {
103                 pay = p - ‘0‘;
104                 while((p = getchar()) && p >= ‘0‘ && p <= ‘9‘) pay = pay * 10 + p - ‘0‘;
105                 adde(i, pay+n, inf);
106                 if(p == ‘\n‘) break;
107             }
108         }
109         for(int i = 1; i <= m; i++) {
110             scanf("%d", &pay);
111             adde(n+i, T, pay);
112         }
113         int ret = pos - dinic(S, T, N);
114         s.clear();
115         for(int i = 1; i <= n; i++) {
116             if(vis[i]) s.push_back(i);
117         }
118         printf("%d", s[0]);
119         for(int i = 1; i < s.size(); i++) printf(" %d", s[i]);
120         printf("\n");
121         s.clear();
122         for(int i = n+1; i <= n+m; i++) {
123             if(vis[i]) s.push_back(i-n);
124         }
125         printf("%d", s[0]);
126         for(int i = 1; i < s.size(); i++) printf(" %d", s[i]);
127         printf("\n");
128         printf("%d\n", ret);
129     }
130     return 0;
131 }