参考:
https://blog.csdn.net/txl199106/article/details/64441994
分析:
该算法是用bfs求出是否有路从s到t, 然后建立反向边(关于反向边), 最终求出答案, 复杂度(mn)
#include<bits/stdc++.h> using namespace std; const int MAXN = 10000 + 7; const int MAXM = 100000 + 7; const int MAX_INT = (1 << 30); struct Edge { int v, nxt, w; }; struct Node { int v, id; }; int n, m, ecnt, S , T; // n , m , S , T 点,边,源点,汇点 bool vis[MAXN]; int head[MAXN]; Node pre[MAXN]; Edge edge[2 * MAXM]; void init() { ecnt = 0; memset(edge, 0, sizeof(edge)); memset(head, -1, sizeof(head)); } void addEdge(int u, int v, int w) { edge[ecnt].v = v; edge[ecnt].w = w; edge[ecnt].nxt = head[u]; head[u] = ecnt++; } bool bfs(int s, int t) { queue <int> que; memset(vis, 0, sizeof(vis)); memset(pre, -1, sizeof(pre)); pre[s].v = s; vis[s] = true; que.push(s); while(!que.empty()) { int u = que.front(); que.pop(); for(int i = head[u]; i + 1; i = edge[i].nxt) { int v = edge[i].v; if(!vis[v] && edge[i].w) { pre[v].v = u; pre[v].id = i; vis[v] = true; if(v == t) return true; que.push(v); } } } return false; } int EK(int s, int t) { int ans = 0; while(bfs(s, t)) { int mi = MAX_INT; for(int i = t; i != s; i = pre[i].v) { mi = min(mi, edge[pre[i].id].w); } for(int i = t; i != s; i = pre[i].v) { edge[pre[i].id].w -= mi; edge[pre[i].id ^ 1].w += mi; //异或1的用处是 偶数+1, 奇数-1 (两条正反向边是相邻的 } ans += mi; } return ans; } int main() { freopen("1.txt","r", stdin); init(); ios::sync_with_stdio(false); cin >> n >> m >> S >> T; for(int i = 0; i < m; i++) { int u, v, w; cin >> u >> v >> w; addEdge(u, v, w); addEdge(v, u, 0); } // 调用 cout << EK(S, T) << " "; } // 加边