Bzoj 1682: [usaco2005 Mar] out of hay crisis

Description

The hay is used up! Bessie planned to explore the disaster.

There are N (2 ≤ n ≤ 2000) farms, M (≤ m ≤ 10000) two-way road connecting them, the length cannot exceed 10 ^ 9. every farm is connected to farm 1. bessie is going to visit every farm. every time she takes a long journey, it consumes a unit of water. from one farm to another, she needs to carry water that equals the length of the road. please help her determine the minimum tank capacity. that is to say, determine a scheme to minimize the length of the longest road that passes through all farms. If necessary, she can go back. input

Input two integers n and m in the second row. Input three integers in each row in the next m row, indicating the start and end points and length of a road. Output

Returns an integer that represents the minimum value of the longest path on the route.

Question:

The DJK calculates the minimum value of the maximum edge in the path from 1 to each vertex, and takes the max value.

Code:

#include<iostream>#include<cstdio>#include<algorithm>#include<cstring>#include<queue>//by zrt//problem:using namespace std;typedef long long LL;const int inf(0x3f3f3f3f);const double eps(1e-9);int H[2005],X[20005],P[20005];LL E[20005];LL d[2005];int tot;inline void add(int x,int y,LL z){    P[++tot]=y;X[tot]=H[x];H[x]=tot;E[tot]=z;}int n,m;int x,y;LL z;struct N{    int x;    LL w;    N(int a=0,int b=0){        x=a,w=b;    }    friend bool operator < (N a,N b){        return a.w>b.w;    }};priority_queue<N> q;bool vis[2005];int main(){    #ifdef LOCAL    freopen("in.txt","r",stdin);    freopen("out.txt","w",stdout);    #endif    scanf("%d%d",&n,&m);    memset(d,0x3f,sizeof d);    for(int i=0;i<m;i++){        scanf("%d%d%lld",&x,&y,&z);        add(x,y,z);        add(y,x,z);    }    d[1]=0;    q.push(N(1,0));    while(!q.empty()){        x=q.top().x;q.pop();        if(vis[x]) continue;        vis[x]=1;        for(int i=H[x];i;i=X[i]){            if(d[P[i]]>max(d[x],E[i])){                d[P[i]]=max(d[x],E[i]);                q.push(N(P[i],d[P[i]]));            }        }    }    LL ans=0;    for(int i=2;i<=n;i++) ans=max(ans,d[i]);    printf("%lld\n",ans);    return 0;}

Bzoj 1682: [usaco2005 Mar] out of hay crisis