技術標籤：圖論

思路：最短路

每兩個點的距離為 ( ∣ x 1 − x 2 ∣ + ∣ y 1 − y 2 ∣ ) ∗ d − a [ j ] (|x_1-x_2|+|y_1-y_2|)*d-a[j] (∣x1​−x2​∣+∣y1​−y2​∣)∗d−a[j],其中j為走到第j個點的可恢復的單位時間的大小
然後跑一遍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;
}