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;
}

HDU 3416 Marriage Match IV（最短路+最大流）