USACO 4.2.1 PROB Drainage Ditches

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标准网络流

注意：

1、这个题中，流单向

2、环可以不考虑

3、如果有多条 u-->v,直接相加就可以

/*ID : lancelo2PROG : ditchLANG : C++*/ #include <fstream>#include <iostream>#include <algorithm>#include <queue>#include <cstring>#define INT_MAX 100000000using namespace std;const int MAXN = 200;int n, p[MAXN+1], map[MAXN+1][MAXN+1];int N, M;bool ek_bfs(int s, int e){queue<int> q;bool flag[MAXN+1];memset(flag, 0, sizeof(flag));memset(p, -1, sizeof(p));q.push(s);flag[s] = true;while(!q.empty()){int n = q.front();if(n == e)return true;q.pop();for(int i = 1; i <= e; i++){if(map[n][i] > 0 && !flag[i]){flag[i] = true;p[i] = n;q.push(i);}}}return false;}int ek(int s, int e){int u, flow_ans = 0, mn;while(ek_bfs(s, e)){mn = INT_MAX;u = e;while(p[u] != -1){mn = min(mn, map[p[u]][u]);u = p[u];}flow_ans += mn;u = e;while(p[u] != -1){map[p[u]][u] -= mn;map[u][p[u]] += mn;u = p[u];}  }return flow_ans;}int main(){ofstream fout("ditch.out");ifstream fin("ditch.in");memset(map, 0, sizeof(map));fin >> N >> M;int u, v, w;for(int i = 0; i < N; i++){fin >> u >> v >> w;if(map[u][v]){map[u][v] += w;}else{map[u][v] = w;}}int ans = ek(1, M);fout << ans << endl;}