To_Heart—题解——CF301B

思路：最短路

每两个点的距离为$(|x_1-x_2|+|y_1-y_2|)\ast d-a[j]$,其中j为走到第j个点的可恢复的单位时间的大小
然后跑一遍dijkstra即可

#include<bits/stdc++.h>
#include<queue>
#include<vector>
using namespace std;

struct zz{
    
    
	int v;
	int w;
	bool operator < (const zz x) const
      {
    
     return x.w<w; }
};

vector<zz> v[2505];
int n,d,INF;
int dp[2505];
int a[100005];
int x[100005];
int y[100005];
bool f[2505];

void Dijkstra(int s,int t){
    
    
	dp[s]=0;
	priority_queue<zz> q;
	zz Now;
	Now.v=s;
	Now.w=dp[s];
	q.push(Now); 
	while(q.size()){
    
    
		zz now=q.top();
		q.pop();
		if(f[now.v]){
    
    
			continue;
		}
		f[now.v]=1;
		int k=now.v;
		for(int j=0;j<v[k].size();j++){
    
    
			if(dp[k]+v[k][j].w<dp[v[k][j].v]){
    
    
				dp[v[k][j].v]=dp[k]+v[k][j].w;
				zz NOW;
				NOW.w=dp[v[k][j].v];
				NOW.v=v[k][j].v;
				q.push(NOW); 
			}
		}
	}
	printf("%d\n",dp[t]);
}

int main(){
    
    
	memset(dp,0x3f,sizeof dp);
	scanf("%d%d",&n,&d);
	for(int i=2;i<n;i++) scanf("%d",&a[i]); 
	for(int i=1;i<=n;i++) scanf("%d%d",&x[i],&y[i]);
	for(int i=1;i<=n;i++){
    
    
		for(int j=i+1;j<=n;j++){
    
    
			int dist=(abs(x[j]-x[i])+abs(y[j]-y[i]))*d;
			v[i].push_back(zz{
    
    j,dist-a[j]});
			v[j].push_back(zz{
    
    i,dist-a[i]});
		}
	}
		
	//cin>>s>>t;
	Dijkstra(1,n); 
	return 0;
}