poj Countries in War(强连通+最短路)

                                                                      Countries in War

题目链接：Click Here~

题目分析：

     有是一道考英语阅读理解的题。给你N个点，E个关系。如果，m个点直接形成了回路，则这m个点之间的花费是为0的。这里我们就可以想到用强连通的缩点算法了。然后，缩点后再建图，求解最短路。一开始没太在意时间。用了一个floyd TLE了。后来改成spfa就过了。改题的AC的过程真艰辛啊，虽然不是什么难题，但是调试代码用了一堆时间。看来代码能力还是好弱啊。

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

const int maxn = 500+5;
const int INF = 1e8;
int n,m,graph[maxn][maxn],d[maxn][maxn];
int pre[maxn],sccno[maxn],lowlink[maxn];
int scc_cnt,dfs_clock;
vector<int> G[maxn];
stack<int> S;
void Init()
{
while(!S.empty())S.pop();
for(int i = 0;i < maxn;++i)
G[i].clear();
for(int i = 0;i < maxn;++i)
for(int j = 0;j < maxn;++j)
d[i][j] =(i==j?0:INF),graph[i][j] = (i==j?0:INF);
}
void dfs(int u)
{
pre[u] = lowlink[u] = ++dfs_clock;
S.push(u);
for(int i = 0;i < (int)G[u].size();++i){
int v = G[u][i];
if(!pre[v]){
dfs(v);
lowlink[u] = min(lowlink[u],lowlink[v]);
}
else
if(!sccno[v])
lowlink[u] = min(lowlink[u],pre[v]);
}
if(pre[u]==lowlink[u]){
int x;
scc_cnt++;
do
{
x = S.top();
S.pop();
sccno[x] = scc_cnt;
}while(x!=u); //attention!!!
}
}
void find_scc()
{
scc_cnt = dfs_clock = 0;
memset(pre,0,sizeof(pre));
memset(sccno,0,sizeof(sccno));
for(int i = 1;i <= n;++i)
if(!pre[i])dfs(i);
}
int Spfa(int s,int e)
{
queue<int> Q;
int dist[maxn];
bool inq[maxn];
memset(inq,0,sizeof(inq));
for(int i = 0;i <= n;++i)
dist[i] = INF;
dist[s] = 0;
Q.push(s);
while(!Q.empty()){
int u = Q.front();
Q.pop();
inq[u] = false;
for(int i = 1;i <= n;++i)if(d[u][i]!=INF){
if(dist[i] > dist[u]+d[u][i]){
dist[i] = dist[u]+d[u][i];
if(!inq[i]){
inq[i] = true;
Q.push(i);
}
}
}
}
return dist[e];
}
int main()
{
// freopen("Input.txt","r",stdin);
while(scanf("%d%d",&n,&m),(n||m))
{
Init();
int u,v,c;
while(m--){
scanf("%d%d%d",&u,&v,&c);
G[u].push_back(v);
graph[u][v] = c;
}
find_scc();
for(int u = 1;u <= n;++u) //缩点
for(int i = 0;i < (int)G[u].size();++i){
int v = G[u][i];
if(sccno[u]!=sccno[v]){
if(d[sccno[u]][sccno[v]] > graph[u][v])
d[sccno[u]][sccno[v]] = graph[u][v];
}
}
int q;
scanf("%d",&q);
while(q--){
scanf("%d%d",&u,&v);
int ans = Spfa(sccno[u],sccno[v]);
if(ans==INF)
puts("Nao e possivel entregar a carta");
else
printf("%d\n",ans);
}
puts("");
}
return 0;
}