HDU 3416 Marriage Match IV
题意:给一个有向图,给定起点终点,问最多多少条点可以重复,边不能重复的最短路
思路:边不能重复,以为着每个边的容量就是1了,最大流问题,那么问题只要能把最短路上的边找出来,跑一下最大流即可,判断一条边是否是最短路上的边,就从起点和终点各做一次dijstra,求出最短路距离后,如果一条边满足d1[u] + d2[v] + w(u, v) == Mindist,那么这条边就是了
代码:
#include <cstdio> #include <cstring> #include <queue> #include <algorithm> using namespace std; const int MAXNODE = 1005; const int MAXEDGE = 200005; typedef int Type; const Type INF = 0x3f3f3f3f; struct Edge { int u, v; Type cap, flow; Edge() {} Edge(int u, int v, Type cap, Type flow) { this->u = u; this->v = v; this->cap = cap; this->flow = flow; } }; 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); } } } gao; struct Edge2 { int u, v; Type dist; Edge2() {} Edge2(int u, int v, Type dist) { this->u = u; this->v = v; this->dist = dist; } void read() { scanf("%d%d%d", &u, &v, &dist); } }; struct HeapNode { Type d; int u; HeapNode() {} HeapNode(Type d, int u) { this->d = d; this->u = u; } bool operator < (const HeapNode& c) const { return d > c.d; } }; struct Dijkstra { int n, m; Edge2 edges[MAXEDGE]; int first[MAXNODE]; int next[MAXEDGE]; bool done[MAXNODE]; Type d[MAXNODE]; void init(int n) { this->n = n; memset(first, -1, sizeof(first)); m = 0; } void add_Edge(int u, int v, Type dist) { edges[m] = Edge2(u, v, dist); next[m] = first[u]; first[u] = m++; } void dijkstra(int s) { priority_queue<HeapNode> Q; for (int i = 1; i <= n; i++) d[i] = INF; d[s] = 0; memset(done, false, sizeof(done)); Q.push(HeapNode(0, s)); while (!Q.empty()) { HeapNode x = Q.top(); Q.pop(); int u = x.u; if (done[u]) continue; done[u] = true; for (int i = first[u]; i != -1; i = next[i]) { Edge2& e = edges[i]; if (d[e.v] > d[u] + e.dist) { d[e.v] = d[u] + e.dist; Q.push(HeapNode(d[e.v], e.v)); } } } } } gao2; const int N = 1005; const int M = 100005; int T, n, m, sd[N], td[N], s, t, Min; Edge2 es[M]; int main() { scanf("%d", &T); while (T--) { scanf("%d%d", &n, &m); gao2.init(n); for (int i = 0; i < m; i++) { es[i].read(); gao2.add_Edge(es[i].u, es[i].v, es[i].dist); } scanf("%d%d", &s, &t); gao2.dijkstra(s); Min = gao2.d[t]; for (int i = 1; i <= n; i++) sd[i] = gao2.d[i]; gao2.init(n); for (int i = 0; i < m; i++) gao2.add_Edge(es[i].v, es[i].u, es[i].dist); gao2.dijkstra(t); for (int i = 1; i <= n; i++) td[i] = gao2.d[i]; gao.init(n + 1); for (int i = 0; i < m; i++) { if (sd[es[i].u] + td[es[i].v] + es[i].dist == Min) gao.add_Edge(es[i].u, es[i].v, 1); } printf("%d\n", gao.Maxflow(s, t)); } return 0; }
时间: 2024-10-25 04:31:53