网络流最大流dinic
hdu 6214
#include <bits/stdc++.h> #include<cstdio> #include<cstring> #include<queue> #include<cmath> #define TS printf("!!!\n") using namespace std; const int INF = 0x3f3f3f3f; const int MAXN = 201; //点数的最大值 const int MAXM = 10500; //边数的最大值 struct Node { int from, to, next; int cap; } edge[MAXM]; int tol; int dep[MAXN];//dep为点的层次 int head[MAXN]; int n; void init() { tol = 0; memset(head, -1, sizeof(head)); } void addedge(int u, int v, int w) //第一条变下标必须为偶数 { edge[tol].from = u; edge[tol].to = v; edge[tol].cap = w; edge[tol].next = head[u]; head[u] = tol++; edge[tol].from = v; edge[tol].to = u; edge[tol].cap = 0; edge[tol].next = head[v]; head[v] = tol++; } int BFS(int start, int end) { int que[MAXN]; int front, rear; front = rear = 0; memset(dep, -1, sizeof(dep)); que[rear++] = start; dep[start] = 0; while (front != rear) { int u = que[front++]; if (front == MAXN) { front = 0; } for (int i = head[u]; i != -1; i = edge[i].next) { int v = edge[i].to; if (edge[i].cap > 0 && dep[v] == -1) { dep[v] = dep[u] + 1; que[rear++] = v; if (rear >= MAXN) { rear = 0; } if (v == end) { return 1; } } } } return 0; } int dinic(int start, int end) { int res = 0; int top; int stack[MAXN];//stack为栈,存储当前增广路 int cur[MAXN];//存储当前点的后继 while (BFS(start, end)) { memcpy(cur, head, sizeof(head)); int u = start; top = 0; while (1) { if (u == end) { int min = INF; int loc; for (int i = 0; i < top; i++) if (min > edge[stack[i]].cap) { min = edge[stack[i]].cap; loc = i; } for (int i = 0; i < top; i++) { edge[stack[i]].cap -= min; edge[stack[i] ^ 1].cap += min; } res += min; top = loc; u = edge[stack[top]].from; } for (int i = cur[u]; i != -1; cur[u] = i = edge[i].next) if (edge[i].cap != 0 && dep[u] + 1 == dep[edge[i].to]) { break; } if (cur[u] != -1) { stack[top++] = cur[u]; u = edge[cur[u]].to; } else { if (top == 0) { break; } dep[u] = -1; u = edge[stack[--top]].from; } } } return res; } int main() { int time; cin >> time; int st, tp; int u, v, w; while (time--) { init(); int n, m; scanf("%d %d", &n, &m); scanf("%d %d", &st, &tp); //TS; for (int i = 1; i <= m; i++) { scanf("%d %d %d", &u, &v, &w); addedge(u, v, w * 10000 + 1); } //TS; cout << dinic(st,tp) % 10000 << endl; } }