hdu 2433 Travel 最短路

/*
题目大意：
去掉给定的边，看每一个点是否能从别的点到达！
如果能够到达，则求出对于每一个点到其他所有的点最短距离之和，将这些和相加就是最后的结果
解法：
对每个顶点求一次最短路，然后用sum[i]将各个顶点的最短路的和保存起来！
对将要删除的每个顶点，如果求最短路u能够到达v,则再就一遍v,分别记为sum_u,sum_v
那么答案即为：sum[1]+...+sum
+sum_u+sum_v-sum[u]-sum[v];
*/
#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <cmath>
#include <algorithm>
#include <queue>
using namespace std;
const int N = 250;
const int INF=99999999;
struct Node
{
int v,next,w;
bool operator < (const Node &a) const
{
return w > a.w;
}
} Edge[2000002],t1,t2;
int dis
,vis
,head
,cnt,n,m,sum
,num[3002];
void addEdge(int u,int v,int w)
{
Edge[cnt].v = v;
Edge[cnt].next = head[u];
Edge[cnt].w = w;
head[u]=cnt++;
}
void dijstra(int  st)
{
priority_queue<Node> q;
memset(vis,0,sizeof(vis));
for(int i=0;i<=n+1;i++)dis[i]=INF;
for(int i = head[st] ; i != -1 ; i = Edge[i].next)
{
int v = Edge[i].v;
if(Edge[i].w<dis[v])
{
dis[v] = Edge[i].w;
t1.w = dis[v];
t1.v =  v;
q.push(t1);
}
}
dis[st]=0;//此句没加错了2次，囧~~~
vis[st] = 1;
while(!q.empty())
{
t1 = q.top();
q.pop();
int u = t1.v;
if(vis[u]) continue;
vis[u] = 1;
for(int i = head[u]; i != -1; i = Edge[i].next)
{
int v = Edge[i].v;
if(!vis[v] && dis[v]>dis[u]+Edge[i].w)
{
dis[v] =dis[u]+Edge[i].w;
t2.v = v;
t2.w = dis[v];
q.push(t2);
}
}
}
return;
}
int main()
{
int i,j;
while(scanf("%d%d",&n,&m)!=EOF)
{
memset(head,-1,sizeof(head));
memset(sum,0,sizeof(sum));
memset(num,-1,sizeof(num));
cnt=0;
int ans=0,res;
int u,v;
for(j=1; j<=m; j++)
{
cin>>u>>v;
num[j]=cnt;
addEdge(u,v,1);
addEdge(v,u,1);
}
for(i=1;i<=n;i++)
{
dijstra(i);
for(j=1;j<=n;j++)sum[i]+=dis[j];
ans+=sum[i];
}
for(i=1;i<=m;i++)
{
if(ans>=INF)
{
printf("INF\n");continue;
}
Edge[num[i]^1].w=Edge[num[i]].w=INF;
res=0;
dijstra(Edge[num[i]].v);
for(j=1;j<=n;j++)res+=dis[j];
if(res>=INF)
{
printf("INF\n");
}else
{
dijstra(Edge[num[i]^1].v);
for(j=1;j<=n;j++)res+=dis[j];
if(res>=INF)printf("INF\n");
else printf("%d\n",ans+res-(sum[Edge[num[i]].v]+sum[Edge[num[i]^1].v]));
}
Edge[num[i]^1].w=Edge[num[i]].w=1;
}
}
return 0;
}
/*
5 4
5 1
1 3
3 2
5 4
2 2
1 2
1 2

*/