E - Souvenir(图论典型例题)

思路：对于有很多询问的题，一般都是先初始化。我们求出每个点到其他点的最短路径以及相同路径下最大的价值和即可。

代码：

#include <bits/stdc++.h>
#define pb push_back
#define a first
#define b second
using namespace std;
typedef long long int ll;
typedef pair <int,int> P;
typedef unsigned long long ull;char str[1005];
int main(){int n;scanf("%d",&n);vector <ll> A(n);for(int i=0;i<n;i++) scanf("%lld",&A[i]);vector < vector<int> > mp(n);for(int i=0;i<n;i++){scanf("%s",&str);mp[i].resize(n);for(int j=0;j<n;j++){mp[i][j]=str[j]=='Y';}mp[i][i]=1;}ll inf=1000000000000000;vector < vector<ll> > ans(n);vector < vector<int> > vd(n);for(int i=0;i<n;i++){ans[i].resize(n);for(int j=0;j<n;j++) ans[i][j]=-inf;vector <int> dist(n,n+1);queue <int> que;dist[i]=0;ans[i][i]=A[i];que.push(i);while(!que.empty()){int v=que.front();que.pop();for(int j=0;j<n;j++){if(mp[v][j]){if(dist[j]==n+1){dist[j]=dist[v]+1;que.push(j);}if(dist[j]==dist[v]+1){ans[i][j]=max(ans[i][j],ans[i][v]+A[j]);}}}}vd[i]=dist;}int Q;scanf("%d",&Q);for(int i=0;i<Q;i++){int u,v;scanf("%d%d",&u,&v);u--,v--;if(ans[u][v]==-inf) puts("Impossible");else printf("%d %lld\n",vd[u][v],ans[u][v]);}
}