HDU5137How Many Maos Does the Guanxi Worth（最短路）

题意：从2~n-1这几个点中任意去掉一个点，使得从1到n的最短路径最大，如果任意去掉一个点1~n无通路输出Inf。

思路：枚举去掉2~n-1个点，依次用spfa算法求取最短路，去最大即可

#include<bits/stdc++.h>using namespace std;#define pii pair<int, int>const int maxn = 35;int n, m;struct Edge{    int to, w;    Edge(){}    Edge(int to, int w):to(to), w(w){}};vector<Edge> a[maxn];int dis[maxn], fib;bool vis[maxn];void dij(int s){    memset(dis, 63, sizeof(dis));    memset(vis, false, sizeof(vis));    dis[s] = 0;    vis[fib] = true;    priority_queue<pii, vector<pii>, greater<pii> > q;    q.push(pii(dis[s], s));    while(!q.empty())    {        pii now = q.top();        q.pop();        int u = now.second, v;        if(vis[u])            continue;        for(int i = 0; i < a[u].size(); i++)        {            v = a[u][i].to;            if(dis[v] > dis[u] + a[u][i].w)            {                dis[v] = dis[u] + a[u][i].w;                q.push(pii(dis[v], v));            }        }    }}int main(){    int u, v, w;    while(scanf("%d%d", &n, &m) != EOF)    {        if(n == 0 && m == 0) break;        for(int i = 1; i <= n; i++)            a[i].clear();        for(int i = 1; i <= m; i++)        {            scanf("%d%d%d", &u, &v, &w);            a[u].push_back(Edge(v, w));            a[v].push_back(Edge(u, w));        }        int ans = 0;        for(fib = 2; fib < n; fib++)        {            dij(1);            ans = max(ans, dis[n]);        }        if(ans == 0x3f3f3f3f)            puts("Inf");        else            printf("%d\n", ans);    }    return 0;}