首先算出mp[i][j]:
以i为最高点，i到j的最短距离是多少。
然后对于每次询问，枚举最高点。
结果就为min（mp[i][st]+mp[i][ed]+val[i]）;
但是这个题我用SFL优化了一下。。
#include<stdio.h>
#include<string.h>
#include<stdlib.h>
#include<iostream>
#include<algorithm>
#include<vector>
#include<queue>
#include<map>
#include<math.h>
#include<deque>
using namespace std;
#define maxn 1010
#define maxm 20020
#define LL long long
#define INF (LL)99999999999999
struct listt
{int u,v;LL w;int next;
}edge[maxm*2];
int head[maxn];
int nums;
void init()
{memset(head,-1,sizeof(head));nums=1;
}
void add(int u,int v,LL w)
{// cout<<"add"<<u<<" "<<v<<" "<<w<<endl;edge[nums].u=u;edge[nums].v=v;edge[nums].w=w;edge[nums].next=head[u];head[u]=nums++;
}
LL mp[maxn][maxn];
LL val[maxn];
deque<int>que;
LL dist[maxn];
int vis[maxn];
int n;
void spfa(int st)
{for(int i=1;i<=n;i++)dist[i]=INF;memset(vis,0,sizeof(vis));dist[st]=0;que.push_back(st);vis[st]=1;while(!que.empty()){int x=que.front();que.pop_front();vis[x]=0;// cout<<x<<"-"<<endl;;for(int i=head[x];i!=-1;i=edge[i].next){int y=edge[i].v;// cout<<i<<" "<<y<<" "<<val[y]<<" "<<val[x]<<endl;if(val[y]>val[st])continue;if(dist[y]>dist[x]+edge[i].w){dist[y]=dist[x]+edge[i].w;if(!vis[y]){if(!que.empty()){if(dist[y]>dist[que.front()])que.push_back(y);else que.push_front(y);}else que.push_back(y);vis[y]=1;}}}}// printf("_________________%d\n",st);for(int i=1;i<=n;i++){mp[st][i]=dist[i];// printf("%lld ",mp[st][i]);}// cout<<endl;
}
struct lisq
{int st;int ed;LL minn;
}Q[maxm];
int main()
{int m,q;while(~scanf("%d%d",&n,&m)&&(n||m)){init();while(!que.empty())que.pop_back();for(int i=1;i<=n;i++){for(int j=1;j<=n;j++){mp[i][j]=INF;}mp[i][i]=0;}for(int i=1;i<=n;i++)scanf("%lld",&val[i]);int u,v;LL w;for(int i=1;i<=m;i++){scanf("%d%d%lld",&u,&v,&w);add(u,v,w);add(v,u,w);}scanf("%d",&q);int st,ed;for(int i=1;i<=q;i++){scanf("%d%d",&Q[i].st,&Q[i].ed);Q[i].minn=INF;}for(int i=1;i<=n;i++){spfa(i);for(int j=1;j<=q;j++){Q[j].minn=min(Q[j].minn,mp[i][Q[j].st]+mp[i][Q[j].ed]+val[i]);}}for(int i=1;i<=q;i++){if(Q[i].minn<INF)printf("%lld\n",Q[i].minn);else puts("-1");}cout<<endl;}return 0;
}