HDU 3416 Marriage Match IV（最短路+最大流）

HDU 3416 Marriage Match IV

题目链接

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;}