【最短路】洛谷P1119 灾后重建

Link

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解

细看题目，发现其数据的时间线都给人安排好了，我们只用考虑加入点之后如何跑最短路。
点，是按顺序加的，我们使用Floyed。
以新加入的点为转移点跑即可。
注意询问时要判断两点是否都已建好。

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代码

#include<cstdio>
#include<iostream>
#include<cmath>
using namespace std;
int timee = -1, s[300][300], u, v, ww, n, q, m, t[300], qt, cz = -1;
int read(){
    
      //快读
	int x = 0; char c = getchar();
	while(c > '9' || c < '0') c = getchar();
	while(c <= '9' && c >= '0'){
    
    
		x = x * 10 + c - 48;
		c = getchar(); 
	}
	return x;
}
void work(int d){
    
      //以d点为转移点跑Floyed
	int k = d; 
	for(int i = 0; i < n; ++i)
		if(i!=k) for(int j = 0; j < n; ++j)
		if(j!=k && j!=i && s[i][k] != -1 && s[k][j] != -1) {
    
    
			if(s[i][j] < 0) s[i][j] = s[j][i] = s[i][k] + s[k][j];
			else s[i][j] = s[j][i] = min(s[i][j], s[i][k] + s[k][j]);
		}
}
int main(){
    
    
	n = read(); m = read();
	for(int i = 0; i < n; ++i){
    
    
		t[i] =  read();
		for(int j = 0; j < i; ++j)
			s[i][j] = s[j][i] = -1;
	}
		
	for(int i = 1; i <= m; ++i){
    
    
		u = read(); v = read(); ww = read();
		s[u][v] = s[v][u] = ww; 
	}
	q = read();
	while(q--){
    
    
		u = read(); v = read(); qt = read();
		for(int j = cz + 1; j < n; ++j){
    
    
			if(t[j] <= qt){
    
     //时间内建好的点
				work(j); //以d点为转移点跑Floyed
				cz = j;
			}
			else break;
		}
		if(u == v) printf("0\n");
		else if(t[u] > qt || t[v] > qt) printf("-1\n");  //判断
		else printf("%d\n", s[u][v]);
	}
}