luogu P1119 灾后重建

传送门

突然想起第一次讲课的时候说错了floyd的事...

当时LZY还在然后被直接打脸就很爽

也不知道这一次自己能不能剩下来 又会有多少人走掉

 

这题利用了floyd的滚动数组优化掉的那一维 也就是f[k][i][j]表示中间点仅有前k个时i->j的最短路

所以按时间排序直接做floyd 然后如果有询问就做了 所以要离线

(貌似输入按时间排好序了不用离线)

Code:

#include<cstdio>
#include<cstring>
#include<algorithm>
#include<cmath>
#include<queue>
#include<vector>
#include<iostream>
#include<iomanip>
#define itn int
#define ms(a,b) memset(a,b,sizeof a)
#define rep(i,a,n) for(int i = a;i <= n;i++)
#define per(i,n,a) for(int i = n;i >= a;i--)
#define inf 2147483647
using namespace std;
typedef long long ll;
ll read() {
    ll as = 0,fu = 1;
    char c = getchar();
    while(c < '0' || c > '9') {
        if(c == '-') fu = -1;
        c = getchar();
    }
    while(c >= '0' && c <= '9') {
        as = as * 10 + c - '0';
        c = getchar();
    }
    return as * fu;
}
//head
const int N = 806;
int n,m,Q;
int t[N];
int dp[N][N];
void add(int x,int y) {
    dp[x][y] = dp[y][x] = read();
}

void Floyd(int k) {
    rep(i,1,n) rep(j,1,n) {
        dp[i][j] = min(dp[i][j],dp[i][k] + dp[k][j]);
    }
}

int main() {
    n = read(),m = read();
    ms(dp,63);
    rep(i,1,n) t[i] = read(),dp[i][i] = 0;
    rep(i,1,m) add(read()+1,read()+1);
    Q = read();
    int cur = 1;
    //[1,cur)
    while(Q--) {
        int x = read()+1,y = read()+1,tim = read();
        while(cur <= n && t[cur] <= tim) Floyd(cur++);
        if(t[x] > tim || t[y] > tim || dp[x][y] == dp[0][0]) puts("-1");
        else printf("%d\n",dp[x][y]);
    }
    return 0;
}