hdu 2363

枚举加最短路

#include <stdio.h>
#include <string.h>
#include <algorithm>
#include <queue>
using namespace std;
#define N 105
#define INF 0x7f7f7f7f
struct Height{
    int low,high,dif;
}H[N*N];
struct Edge{
    int u ,val,next;
}e[N*N];
int h[N],p[N],vis[N],d[N],n;
bool cmp(Height a,Height b)
{
    return a.dif<b.dif;
}
void add(int n,int m)
{
    int i,x,y,val,cout = 1;
    for(i = 1 ; i <= n ; i++) p[i] = -1;
    while(m--)
    {
        scanf("%d %d %d",&x,&y,&val);
        e[cout].u = y;
        e[cout].val = val;
        e[cout].next = p[x];
        p[x] = cout++;
        e[cout].u = x;
        e[cout].val = val;
        e[cout].next = p[y];
        p[y] = cout++;
    }
}
int spfa(Height temp)
{
    int i;
    memset(vis,0,sizeof(vis));
    for(i = 1 ; i <= n ; i++) d[i] = INF;
    queue<int>q;
    vis[1] = 1;
    d[1] = 0;
    q.push(1);
    while(!q.empty())
    {
        int cur = q.front();
        q.pop();
        if(h[cur]<temp.low||h[cur]>temp.high) continue;
        for(i = p[cur] ; i != -1 ; i = e[i].next)
        {

            if(d[e[i].u]>d[cur]+e[i].val && h[e[i].u]>=temp.low&&h[e[i].u]<=temp.high)
            {
                d[e[i].u] = d[cur] +e[i].val;
                if(!vis[e[i].u])
                {
                    vis[e[i].u] = 1;
                    q.push(e[i].u);
                }
            }
        }
        vis[cur] = 0;
    }
    return d[n];
}
int main()
{
    int t,m,x,y,val,i,j;
    scanf("%d",&t);
    while(t--)
    {
        scanf("%d %d",&n,&m);
        for(i = 1 ; i <= n ; i++)
            scanf("%d",&h[i]);
        add(n,m);
        int k = 0;
        for(i = 1 ; i <= n ; i++)
        {
            for(j = i ; j <= n ; j++)
            {
                H[k].high = max(h[i],h[j]);
                H[k].low = min(h[i],h[j]);
                H[k].dif = H[k].high - H[k].low;
                k++;
            }
        }
        sort(H,H+k,cmp);
        int ans;
        for(i = 0 ; i < k ; i++)
        {
            ans = spfa(H[i]);
            if(ans != INF) break;
        }
        printf("%d %d\n",H[i].dif,ans);
    }
}

hdu 2363