POJ 3114 Countries in War（强连通+最短路）

POJ 3114 Countries in War

题目链接

题意：给定一个有向图。强连通分支内传送不须要花费，其它有一定花费。每次询问两点的最小花费

思路：强连通缩点后求最短路就可以

代码：

#include <cstdio>
#include <cstring>
#include <vector>
#include <queue>
#include <stack>
#include <algorithm>
using namespace std;

const int MAXNODE = 505;
const int MAXEDGE = 255005;

typedef int Type;
const Type INF = 0x3f3f3f3f;

struct Edge {
	int u, v;
	Type dist;
	Edge() {}
	Edge(int u, int v, Type dist) {
		this->u = u;
		this->v = v;
		this->dist = 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;
	Edge 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] = Edge(u, v, dist);
		next[m] = first[u];
		first[u] = m++;
	}

	int dijkstra(int s, int t) {
		priority_queue<HeapNode> Q;
		for (int i = 0; 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 (u == t) return d[t];
			if (done[u]) continue;
			done[u] = true;
			for (int i = first[u]; i != -1; i = next[i]) {
				Edge& 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));
				}
			}
		}
		return -1;
	}
} gao;

const int N = 505;

int n, m;
vector<Edge> g[N];

int pre[N], dfn[N], dfs_clock, sccn, sccno[N];
stack<int> S;

void dfs_scc(int u) {
	pre[u] = dfn[u] = ++dfs_clock;
	S.push(u);
	for (int i = 0; i < g[u].size(); i++) {
		int v = g[u][i].v;
		if (!pre[v]) {
			dfs_scc(v);
			dfn[u] = min(dfn[u], dfn[v]);
		} else if (!sccno[v]) dfn[u] = min(dfn[u], pre[v]);
	}
	if (pre[u] == dfn[u]) {
		sccn++;
		while (1) {
			int x = S.top(); S.pop();
			sccno[x] = sccn;
			if (x == u) break;
		}
	}
}

void find_scc() {
	dfs_clock = sccn = 0;
	memset(sccno, 0, sizeof(sccno));
	memset(pre, 0, sizeof(pre));
	for (int i = 1; i <= n; i++)
		if (!pre[i]) dfs_scc(i);
}

int main() {
	while (~scanf("%d%d", &n, &m) && n) {
		for (int i = 1; i <= n; i++) g[i].clear();
		int u, v, w;
		while (m--) {
			scanf("%d%d%d", &u, &v, &w);
			g[u].push_back(Edge(u, v, w));
		}
		find_scc();
		gao.init(n);
		for (int u = 1; u <= n; u++) {
			for (int j = 0; j < g[u].size(); j++) {
				int v = g[u][j].v;
				if (sccno[u] == sccno[v]) continue;
				gao.add_Edge(sccno[u] - 1, sccno[v] - 1, g[u][j].dist);
			}
		}
		int q;
		scanf("%d", &q);
		while (q--) {
			scanf("%d%d", &u, &v);
			int tmp = gao.dijkstra(sccno[u] - 1, sccno[v] - 1);
			if (tmp == -1) printf("Nao e possivel entregar a carta
");
			else printf("%d
", tmp);
		}
		printf("
");
	}
	return 0;
}