poj 3114 Countries in War(强连通分量缩点+spfa求最短路)

http://poj.org/problem?id=3114

题意：给出m条边及其权值，询问x y ，如果x,y在同一个强连通分量中，输出0，否则输出两点所在强连通分量的最短路径。

思路：tarjan缩点 + 最短路。

思路不难，关键是仔细。

#include <stdio.h>
#include <string.h>
#include <algorithm>
#include <vector>
#include <stack>
#include <queue>
using namespace std;

const int maxn = 505;
const int INF = 0x3f3f3f3f;

struct node
{
	int v,w;
};

vector <struct node> edge[maxn];//原图
vector <struct node> edge2[maxn];//缩点后的图
stack<int>st;
int n,m,k;
int vis[maxn],instack[maxn],dfn[maxn],low[maxn],dep;
int scc,set[maxn];
int dis[maxn];

void init()
{
	for(int i = 1; i <= n; i++)
	{
		edge[i].clear();
		edge2[i].clear();
	}
	memset(vis,0,sizeof(vis));
	memset(instack,0,sizeof(instack));
	memset(dfn,0,sizeof(vis));
	memset(low,0,sizeof(vis));
	while(!st.empty()) st.pop();
	dep = 0;
	scc = 0;
}

void tarjan(int u)
{
	dfn[u] = low[u] = ++dep;
	st.push(u);
	instack[u] = 1;
	vis[u] = 1;

	for(int i = 0; i < (int)edge[u].size(); i++)
	{
		int v = edge[u][i].v;
		if(!vis[v])
		{
			tarjan(v);
			low[u] = min(low[u],low[v]);
		}
		else if(instack[v])
			low[u] = min(low[u],dfn[v]);
	}

	if(low[u] == dfn[u])
	{
		scc += 1;//强连通分量数加1
		while(1)
		{
			int t = st.top();
			st.pop();
			instack[t] = 0;
			set[t] = scc;//记录t属于哪个强连通分量
			if(t == u)
				break;
		}
	}
}
//缩点建图
void creat_DAG()
{
	for(int u = 1; u <= n; u++)
	{
		for(int i = 0; i < (int)edge[u].size(); i++)
		{
			int v = edge[u][i].v;
			int w = edge[u][i].w;
			if(set[u] != set[v])
				edge2[ set[u] ].push_back((struct node){set[v], w});
		}
	}
}

void spfa(int s)
{
	queue<int>que;
	int inque[maxn];
	while(!que.empty()) que.pop();
	memset(inque,0,sizeof(inque));
	for(int i = 1; i <= scc; i++)
		dis[i] = INF;

	dis[s] = 0;
	inque[s] = 1;
	que.push(s);

	while(!que.empty())
	{
		int u = que.front();
		que.pop();
		inque[u] = 0;

		for(int i = 0; i < (int)edge2[u].size(); i++)
		{
			int v = edge2[u][i].v;
			if(dis[v] > edge2[u][i].w + dis[u])
			{
				dis[v] = dis[u] + edge2[u][i].w;
				if(!inque[v])
				{
					inque[v] = 1;
					que.push(v);
				}
			}
		}
	}
}

int main()
{
	while(~scanf("%d %d",&n,&m))
	{
		if(n == 0 && m == 0) break;
		init();

		int u,v,w;
		while(m--)
		{
			scanf("%d %d %d",&u,&v,&w);
			edge[u].push_back((struct node){v,w});
		}

		for(int i = 1; i <= n; i++)
			if(!vis[i])
				tarjan(i);

		creat_DAG();

		scanf("%d",&k);
		while(k--)
		{
			scanf("%d %d",&u,&v);
			int uu = set[u];
			int vv = set[v];
			if(uu == vv)
			{
				printf("0\n");
				continue;
			}
			if(edge2[uu].size() == 0)
			{
				printf("Nao e possivel entregar a carta\n");
				continue;
			}

			spfa(uu);
			if(dis[vv] == INF)
				printf("Nao e possivel entregar a carta\n");
			else printf("%d\n",dis[vv]);
		}
		printf("\n");
	}
	return 0;
}