P3254 圆桌问题 网络流

P3254 圆桌问题

  1 #include <bits/stdc++.h>
  2 using namespace std;
  3 const int maxn = 1005, inf = 0x3f3f3f;
  4 struct Edge {
  5     int from, to, cap, flow;
  6 };
  7 
  8 struct Dinic {
  9     int n, m, s, t;
 10     vector<Edge> edges;
 11     vector<int> G[maxn];
 12     bool vis[maxn];
 13     int d[maxn];
 14     int cur[maxn];
 15 
 16     void AddEdge(int from, int to, int cap) {
 17         edges.push_back((Edge){from, to, cap, 0});
 18         edges.push_back((Edge){to, from, 0, 0});
 19         m = edges.size();
 20         G[from].push_back(m-2);
 21         G[to].push_back(m-1);
 22     }
 23     bool bfs() {
 24         memset(vis, 0, sizeof(vis));
 25         queue<int> que;
 26         que.push(s);
 27         d[s] = 0;
 28         vis[s] = true;
 29         while (!que.empty()) {
 30             int x = que.front(); que.pop();
 31             for (int i = 0; i < G[x].size(); ++i) {
 32                 Edge& e = edges[G[x][i]];
 33                 if (!vis[e.to] && e.cap > e.flow) {
 34                     vis[e.to] = true;
 35                     d[e.to] = d[x] + 1;
 36                     que.push(e.to);
 37                 }
 38             }
 39         }
 40         return vis[t];
 41     }
 42     int dfs(int x, int a) {
 43         if (x == t || a == 0) return a;
 44         int flow = 0, f;
 45         for (int& i = cur[x]; i < G[x].size(); ++i) {
 46             Edge& e = edges[G[x][i]];
 47             if (d[x] + 1 == d[e.to] && (f = dfs(e.to, min(a, e.cap-e.flow))) > 0) {
 48                 e.flow += f;
 49                 edges[G[x][i]^1].flow -= f;
 50                 flow += f;
 51                 a -= f;
 52                 if (a == 0) break;
 53             }
 54         }
 55         return flow;
 56     }
 57     int maxflow(int s, int t) {
 58         this->s = s; this->t = t;
 59         int flow = 0;
 60         while (bfs()) {
 61             memset(cur,0,sizeof(cur));
 62             flow += dfs(s,inf);
 63         }
 64         return flow;
 65     }
 66 }dinic;
 67 
 68 int r[maxn], c[maxn];
 69 int main() {
 70     int m, n; scanf("%d%d",&m,&n);
 71     int s = m+n+1, t = m+n+2, sum = 0;
 72     for (int i = 1; i <= m; ++i) {
 73         scanf("%d",&r[i]);
 74         sum += r[i];
 75     }
 76     for (int i = 1; i <= n; ++i) scanf("%d",&c[i]);
 77 
 78     for (int i = 1; i <= m; ++i) {
 79         dinic.AddEdge(s,i,r[i]);
 80     }
 81     for (int i = 1; i <= n; ++i) {
 82         dinic.AddEdge(i+m,t,c[i]);
 83     }
 84     for (int i = 1; i <= m; ++i) {
 85         for (int j = 1; j <= n; ++j) {
 86             dinic.AddEdge(i,j+m,1);
 87         }
 88     }
 89     int ans = dinic.maxflow(s,t);
 90     if (ans != sum) puts("0");
 91     else {
 92         puts("1");
 93         for (int i = 1; i <= m; ++i) {
 94             for (int j = 0; j < dinic.edges.size(); ++j) {
 95                 if (dinic.edges[j].from == i && dinic.edges[j].flow == 1) {
 96                     printf("%d ",dinic.edges[j].to-m);
 97                 }
 98             }
 99             putchar('\n');
100         }
101     }
102     return 0;
103 }

 

posted @ 2019-11-05 22:10  麻辣猪仔  阅读(136)  评论(0)    收藏  举报