POJ 3204 Ikki‘s Story I - Road Reconstruction
题意:给定一个有向图,求出最大流后,问哪些边增加容量后,可以使最大流增加
思路:对于一个可以增加的,必然原来就是满流,并且从源点到汇点,的一条路径上,都是还有残留容量的,这样只要从源点和汇点分别出发dfs一遍,标记掉经过点,然后枚举满流边,如果两端都是标记过的点,这个边就是可以增加的
代码:
#include <cstdio> #include <cstring> #include <queue> #include <algorithm> using namespace std; const int MAXNODE = 505; const int MAXEDGE = 100005; typedef int Type; const Type INF = 0x3f3f3f3f; struct Edge { int u, v; Type cap, flow; int cnt; Edge() {} Edge(int u, int v, Type cap, Type flow) { this->u = u; this->v = v; this->cap = cap; this->flow = flow; this->cnt = 0; } }; struct Dinic { int n, m, s, t; Edge edges[MAXEDGE]; int first[MAXNODE]; int next[MAXEDGE]; bool vis[MAXNODE]; Type d[MAXNODE]; int cur[MAXNODE]; vector<int> cut; void init(int n) { this->n = n; memset(first, -1, sizeof(first)); m = 0; } void add_Edge(int u, int v, Type cap) { edges[m] = Edge(u, v, cap, 0); next[m] = first[u]; first[u] = m++; edges[m] = Edge(v, u, 0, 0); next[m] = first[v]; first[v] = m++; } bool bfs() { memset(vis, false, sizeof(vis)); queue<int> Q; Q.push(s); d[s] = 0; vis[s] = true; while (!Q.empty()) { int u = Q.front(); Q.pop(); for (int i = first[u]; i != -1; i = next[i]) { Edge& e = edges[i]; if (!vis[e.v] && e.cap > e.flow) { vis[e.v] = true; d[e.v] = d[u] + 1; Q.push(e.v); } } } return vis[t]; } Type dfs(int u, Type a) { if (u == t || a == 0) return a; Type flow = 0, f; for (int &i = cur[u]; i != -1; i = next[i]) { Edge& e = edges[i]; if (d[u] + 1 == d[e.v] && (f = dfs(e.v, min(a, e.cap - e.flow))) > 0) { e.flow += f; edges[i^1].flow -= f; flow += f; a -= f; if (a == 0) break; } } return flow; } Type Maxflow(int s, int t) { this->s = s; this->t = t; Type flow = 0; while (bfs()) { for (int i = 0; i < n; i++) cur[i] = first[i]; flow += dfs(s, INF); } return flow; } void MinCut() { cut.clear(); for (int i = 0; i < m; i += 2) { if (vis[edges[i].u] && !vis[edges[i].v]) cut.push_back(i); } } bool mark[MAXNODE][2]; void find(int u, int tp) { mark[u][tp] = true; for (int i = first[u]; i + 1; i = next[i]) { int v = edges[i].v; if (mark[v][tp]) continue; if (tp == 0 && i % 2) continue; if (tp && i % 2 == 0) continue; if (edges[i^tp].cap == edges[i^tp].flow) continue; find(v, tp); } } int solve() { Maxflow(0, n - 1); memset(mark, false, sizeof(mark)); find(0, 0); find(n - 1, 1); int ans = 0; for (int i = 0; i < m; i += 2) if (edges[i].cap == edges[i].flow && mark[edges[i].u][0] && mark[edges[i].v][1]) ans++; return ans; } } gao; int n, m; int main() { while (~scanf("%d%d", &n, &m)) { gao.init(n); int u, v, w; while (m--) { scanf("%d%d%d", &u, &v, &w); gao.add_Edge(u, v, w); } printf("%d\n", gao.solve()); } return 0; }
POJ 3204 Ikki's Story I - Road Reconstruction(最大流)
时间: 2024-11-10 01:28:15