传送门:2014广州区区域赛K题
题意:
求从点1到点n存在通路,现在可以选择删除其中一个点(不能是1或者n),问能否使得点1与点n之间不可达,若能输出Inf,若不能求出删除一个点之后从点1到点n的最短路的最大值
解法:
n的数据范围最大只有30,因此可以枚举删除的点然后进行多次最短路即可
/******************************************************
* File Name: k.cpp
* Author: kojimai
* Create Time: 2014年12月03日 星期三 19时38分18秒
******************************************************/
#include<cstdio>
#include<cstring>
#include<cmath>
#include<algorithm>
#include<iostream>
#include<queue>
using namespace std;
const int FFF = 33;
int first[FFF],e;
long long ans = -1,dis[FFF];
const int FF = 2005;
struct node
{
int u,v,next;
long long value;
}edge[FF];
bool flag;
bool vis[FFF];
void addedge(int x,int y,long long z)
{
edge[e].u=x;edge[e].v=y;edge[e].value=z;edge[e].next=first[x];first[x]=e++;
edge[e].v=x;edge[e].u=y;edge[e].value=z;edge[e].next=first[y];first[y]=e++;
}
struct PP
{
int u;long long dis;
bool operator < (const PP & a) const{
return dis > a.dis;
}
};
void findless(int s,int t,int del)
{
priority_queue<PP> q;
memset(vis,false,sizeof(vis));
memset(dis,-1,sizeof(dis));
dis[s] = 0;
PP now,tmp;
now.u = s;now.dis = 0;
q.push(now);
while(!q.empty())
{
now = q.top(); q.pop();
int u = now.u;
if(vis[u])
continue;
vis[u] = true;
for(int k = first[u]; ~k;k = edge[k].next)
{
int vv = edge[k].v;
// cout<<" vv = "<<vv<<endl;
if(vv == del) continue;
if(dis[vv] == -1 || dis[vv] > dis[u] + edge[k].value)
{
dis[vv] = dis[u] + edge[k].value;
tmp.u = vv;tmp.dis = dis[vv];
q.push(tmp);
}
}
}
//cout<<" del = "<<del<<" dis = "<<dis[t]<<" ans = "<<ans<<endl;
if(dis[t] == -1)
flag = true;
else
{
if(ans == -1)
ans = dis[t];
else
ans = max(ans,dis[t]);
}
}
int main()
{
int n,m;
while(scanf("%d%d",&n,&m),n+m)
{
long long x,y,z;
memset(first,-1,sizeof(first));
e = 0;
ans = -1;
for(int i = 0;i < m;i++)
{
//scanf("%d%d%d",&x,&y,&z);
cin>>x>>y>>z;
addedge(x,y,z);
}
flag = false;
for(int i = 2;i < n && !flag;i++)
{
findless(1,n,i);
}
if(flag)
printf("Inf\n");
else
{
cout<<ans<<endl;
}
}
return 0;
}