线性规划与网络流24题

  1 #include <iostream>
  2 #include <cstdio>
  3 #include <cstring>
  4 #include <algorithm>
  5 #include <queue>
  6 using namespace std;
  7 const int INF = 1e9;
  8 const int maxn = 2e5 + 10;
  9 int lv[maxn], it[maxn];
 10 int cnt, h[maxn];
 11 
 12 struct edge
 13 {
 14     int to, pre, cap;
 15 } e[maxn<<1];
 16 
 17 void init()
 18 {
 19     memset(h, -1, sizeof(h));
 20     cnt = 0;
 21 }
 22 
 23 void add(int from, int to, int cap)
 24 {
 25     e[cnt].pre = h[from];
 26     e[cnt].to = to;
 27     e[cnt].cap = cap;
 28     h[from] = cnt;
 29     cnt++;
 30 }
 31 
 32 void ad(int from, int to, int cap)
 33 {
 34     add(from, to, cap);
 35     add(to, from, 0);
 36 }
 37 
 38 void bfs(int s)
 39 {
 40     memset(lv, -1, sizeof(lv));
 41     queue<int> q;
 42     lv[s] = 0;
 43     q.push(s);
 44     while(!q.empty())
 45     {
 46         int v = q.front(); q.pop();
 47         for(int i = h[v]; i >= 0; i = e[i].pre)
 48         {
 49             int cap = e[i].cap, to = e[i].to;
 50             if(cap > 0 && lv[to] < 0)
 51             {
 52                 lv[to] = lv[v] + 1;
 53                 q.push(to);
 54             }
 55         }
 56     }
 57 }
 58 
 59 int dfs(int v, int t, int f)
 60 {
 61     if(v == t) return f;
 62     for(int &i = it[v]; i >= 0; i = e[i].pre)
 63     {
 64         int &cap = e[i].cap, to = e[i].to;
 65         if(cap > 0 && lv[v] < lv[to])
 66         {
 67             int d = dfs(to, t, min(f, cap));
 68             if(d > 0)
 69             {
 70                 cap -= d;
 71                 e[i^1].cap += d;
 72                 return d;
 73             }
 74         }
 75     }
 76     return 0;
 77 }
 78 
 79 int Dinic(int s, int t)
 80 {
 81     int flow = 0;
 82     while(1)
 83     {
 84         bfs(s);
 85         if(lv[t] < 0) return flow;
 86         memcpy(it, h, sizeof(it));
 87         int f;
 88         while((f = dfs(s, t, INF)) > 0) flow += f;
 89     }
 90 }
 91 
 92 int main(void)
 93 {
 94     init();
 95     int m, n;
 96     scanf("%d %d", &m, &n);
 97     while(1)
 98     {
 99         int u, v;
100         scanf("%d %d", &u, &v);
101         if(u == -1) break;
102         ad(u, v, 1);
103     }
104     int S = n + 1, T = S + 1;
105     for(int i = 1; i <= m; i++) ad(S, i, 1);
106     for(int i = m + 1; i <= n; i++) ad(i, T, 1);
107     int ans = Dinic(S, T);
108     if(!ans) puts("No Solution!");
109     else
110     {
111         printf("%d\n", ans);
112         for(int i = 1; i <= m; i++)
113             for(int j = h[i]; j != -1; j = e[j].pre)
114                 if(e[j].to <= n && !e[j].cap) printf("%d %d\n", i, e[j].to);
115     }
116     return 0;
117 }