【強連通縮點+最短路】 POJ 3114 Countries in War

先強連通縮點處理處DAG圖，然後再DAG圖上求兩點間最短路就行了。。。用floyd算法會逾時。。

#include <iostream>  
#include <queue>  
#include <stack>  
#include <map>  
#include <set>  
#include <bitset>  
#include <cstdio>  
#include <algorithm>  
#include <cstring>  
#include <climits>  
#include <cstdlib>
#include <cmath>
#include <time.h>
#define maxn 505
#define maxm 250005
#define eps 1e-10
#define mod 1000000009
#define INF 99999999
#define lowbit(x) (x&(-x))  
//#define lson o<<1, L, mid  
//#define rson o<<1 | 1, mid+1, R  
typedef long long LL;
//typedef int LL;
using namespace std;

int H[maxn], NEXT[maxm], V[maxm], W[maxm];
int h[maxn], next[maxm], v[maxm], w[maxm];
int dfn[maxn], dis[maxn], low[maxn];
int ins[maxn], id[maxn], done[maxn];
int n, m, scc_cnt, top;
stack<int> s;
struct node
{
	int x, d;
	bool operator < (const node &a) const {
		return a.d<d;
	}
}temp, now;
priority_queue<node> q;

void init(void)
{
	top = scc_cnt = 0;
	memset(h, -1, sizeof h);
	memset(H, -1, sizeof H);
	memset(ins, 0, sizeof ins);
	memset(dfn, 0, sizeof dfn);
}
void read(void)
{
	int a, b, c, cnt = 0;
	while(m--) {
		scanf("%d%d%d", &a, &b, &c);
		NEXT[cnt] = H[a], H[a] = cnt, V[cnt] = b, W[cnt] = c, cnt++;
	}
}
void tarjan(int u)
{
	dfn[u] = low[u] = ++top;
	s.push(u), ins[u] = 1;
	for(int e = H[u]; ~e; e = NEXT[e])
		if(!dfn[V[e]]) {
			tarjan(V[e]);
			low[u] = min(low[u], low[V[e]]);
		}
		else if(ins[V[e]]) low[u] = min(low[u], dfn[V[e]]);
	if(dfn[u] == low[u]) {
		int tmp = s.top(); s.pop(), ins[tmp] = 0;
		++scc_cnt;
		while(tmp != u) {
			id[tmp] = scc_cnt;
			tmp = s.top();
			s.pop();
			ins[tmp] = 0;
		}
		id[tmp] = scc_cnt;
	}
}
void narrow(void)
{
	int cnt = 0;
	for(int i = 1; i <= n; i++)
		for(int e = H[i]; ~e; e = NEXT[e])
			next[cnt] = h[id[i]], h[id[i]] = cnt, v[cnt] = id[V[e]], w[cnt] = W[e], cnt++;
}
void dijstra(int s)
{
	for(int i = 0; i <= scc_cnt; i++) dis[i] = INF;
	memset(done, 0, sizeof done);
	dis[s] = 0, now.x = s, now.d = 0;
	q.push(now);
	while(!q.empty()) {
		now = q.top();
		q.pop();
		if(done[now.x]) continue;
		done[now.x] = 1;
		for(int e = h[now.x]; ~e; e = next[e])
			if(v[e] != now.x)
				if(dis[v[e]] > dis[now.x] + w[e]) {
					dis[v[e]] = dis[now.x] + w[e];
					temp.x = v[e];
					temp.d = dis[v[e]];
					q.push(temp);
				}
	}
}
void work(void)
{
	int a, b, i;
	for(i = 1; i <= n; i++)
		if(!dfn[i]) tarjan(i);
	narrow();
	scanf("%d", &m);
	while(m--) {
		scanf("%d%d", &a, &b);
		a = id[a], b = id[b];
		if(a == b) {
			printf("0\n");
			continue;
		}
		dijstra(a);
		if(dis[b] != INF) printf("%d\n", dis[b]);
		else printf("Nao e possivel entregar a carta\n");
	}
	printf("\n");
}
int main(void)
{
	while(scanf("%d%d", &n, &m)&&n!=0) {
		init();
		read();
		work();
	}
	return 0;
}