1 #include <iostream> 2 #include <cstdio> 3 #include <algorithm> 4 #include <vector> 5 #include <queue> 6 using namespace std; 7 #define pb push_back 8 9 const int N = 1e4 + 10; 10 const int M = 1e5 + 10; 11 const int INF = 2e9; 12 struct edge{ 13 int to, nxt, flow; 14 }e[M << 1]; 15 int head[N], d[N], cur[N]; 16 queue<int > que; 17 int n, m, s, t, tot; 18 19 //dinic算法 复杂度O(n^2*m) 二部图时O(n^0.5*m) 20 //bfs分层 增广路只能从上一层走到下一层 21 //dfs求流 得到该增广路的最小允许流 22 23 inline void add(int u, int v, int flow){ 24 e[tot].to = v; e[tot].flow = flow; 25 e[tot].nxt = head[u]; head[u] = tot++; 26 e[tot].to = u; e[tot].flow = 0; 27 e[tot].nxt = head[v]; head[v] = tot++; 28 } 29 30 inline void init(){ 31 for(int i = 1; i <= n; ++i) cur[i] = head[i]; 32 while(!que.empty()) que.pop(); 33 for(int i = 1; i <= n; ++i) d[i] = 0; 34 } 35 36 bool bfs(int s, int t){ 37 d[s] = 1; 38 que.push(s); 39 int to, flow; 40 while(!que.empty()){ 41 int now = que.front(); 42 que.pop(); 43 for(int o = head[now]; ~o; o = e[o].nxt){ 44 to = e[o].to; 45 flow = e[o].flow; 46 if(!d[to] && flow){ 47 d[to] = d[now] + 1; 48 que.push(to); 49 } 50 } 51 } 52 if(d[t]) return true; 53 else return false; 54 } 55 56 int dfs(int now, int t, int F){ 57 if(now == t) return F; 58 59 int sum = 0; 60 int to, flow, tmp; 61 //当前弧优化,因为是dfs,如果遍历了当前边,当它使用之后,之后再用到这个边很大概率效果也是和之前一样,减少了一些时间开销 62 for(int& o = cur[now]; ~o; o = e[o].nxt){ 63 to = e[o].to; 64 flow = e[o].flow; 65 if(flow && d[to] == d[now] + 1){ 66 tmp = dfs(to, t, min(F - sum, flow)); 67 e[o].flow -= tmp; e[o^1].flow += tmp; 68 sum += tmp; 69 if(sum == F) return sum; 70 } 71 } 72 73 return sum; 74 } 75 76 void dinic(int s,int& sum_flow){ 77 while(bfs(s, t)){ 78 sum_flow += dfs(s, t, INF); 79 init(); 80 } 81 } 82 83 void solve(){ 84 85 while(~scanf("%d%d%d%d", &n, &m, &s, &t)){ 86 int u, v, w; 87 for(int i = 1; i <= n; ++i) head[i] = -1; tot = 0; 88 for(int i = 1; i <= m; ++i){ 89 scanf("%d%d%d", &u, &v, &w); 90 add(u, v, w); 91 } 92 int sum_flow = 0; 93 dinic(s, sum_flow); 94 //printf("sum_flow = %d\n", sum_flow); 95 printf("%d\n", sum_flow); 96 } 97 } 98 99 int main(){ 100 101 solve(); 102 103 return 0; 104 }