AcWing 1127. 香甜的黄油 题解（最短路—spfa）

AcWing 1127. 香甜的黄油
还是用spfa算法求最短路，思路就是找到每个牧场为中心点放置糖，找到所有奶牛移动距离和最小的牧场作为结果，输出移动和的最小值

#include<bits/stdc++.h>

using namespace std;

const int N = 810, M = 4000, INF = 0x3f3f3f3f;

int h[N], e[M], ne[M], w[M], idx;
int d[N], q[N];
bool st[N];
int n, p, m;
int id[N];  //记录每个牧场的编号 

void add(int a, int b, int c){
    
	e[idx] = b;
	w[idx] = c;
	ne[idx] = h[a];
	h[a] = idx ++ ;
}

int spfa(int start){
    
	memset(d, 0x3f, sizeof d);
	d[start] = 0;
	
	int hh = 0, tt = 1;
	q[0] = start;
	st[start] = true;
	
	while(hh != tt){
    
		int t = q[hh ++ ];
		if(hh == N) hh = 0;
		st[t] = false;
		
		for(int i = h[t]; ~i; i = ne[i]){
    
			int j = e[i];
			if(d[j] > d[t] + w[i]){
    
				d[j] = d[t] + w[i];
				if(!st[j]){
    
					q[tt ++ ] = j;
					if(tt == N) tt = 0;
					st[j] = true;
				}
			}
		}
	}
	
	int res = 0;
	for(int i = 0; i < n; i ++ ){
    
		int j = id[i];
		if(d[j] == INF) return INF;
		res += d[j]; 
	}
	return res;
}

int main()
{
    
	cin>>n>>p>>m;
	for(int i = 0; i < n; i ++ ){
    
		cin>>id[i];
	}
	memset(h, -1, sizeof h);
	for(int i = 0; i < m; i ++ ){
    
		int a, b, c;
		cin>>a>>b>>c;
		add(a, b, c);
		add(b, a, c);
	}
	
	int minv = INF;
	for(int i = 1; i <= p; i ++ ){
    
		minv = min(minv, spfa(i));
	}
	cout<<minv<<endl;
	return 0;
}