poj1062 Expensive gift (dijkstra+enum)

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The main idea of ​​the title: When buying things, there are substitutes for each thing. After having substitutes, you can get preferential prices. The owner of each item has its own level. Those whose level exceeds m cannot be directly or indirectly traded. Ask the minimum purchase price for item No. 1 what is the price.

Idea: I thought of dfs at first, but it is more troublesome to have a level of no more than m. After reading other people's practices, I found that it is too clever to convert this problem into the shortest path (I am too weak). The starting point is 0, which means There is nothing , the price of each item is the weight from 0 to i, and then the preferential price is the weight of u and i, which is converted into the shortest path, but the starting point is 0, and the end point is 1. For the level problem , enumerate the levels of each node in turn , assuming the lowest level, then traverse it, run Dij Tesla, and calculate the minimum value. (Enumerate with this method, dfs should also work).

See the code comments for details.

#include<iostream>
#include<stdio.h>
#include<stdlib.h>
#include<string.h>
#include<string>
#include<math.h>
#include<cmath>
#include<time.h>
#include<map>
#include<set>
#include<vector>
#include<queue>
#include<algorithm>
#include<numeric>
using namespace std;
int m,n;
const int maxn=110;
int dis[maxn],v[maxn],g[maxn][maxn],vis[maxn],vv[maxn];
int djks(){
	for(int i=1;i<=n;i++){
		dis[i]=g[0][i];//Indicates starting from 0 (when there is nothing)
	}
	dis[0]=0;
	vis[0]=1;
	for(int i=1;i<=n;i++){
		int p,minn=0x3f3f3f3f;
		for(int j=1;j<=n;j++){
			if(!vis[j]&&dis[j]<minn){
				from = dis [j];
				p=j;
			}
		}
		force[p]=1;
		for(int j=1;j<=n;j++){
			if(!vis[j]&&dis[j]>dis[p]+g[p][j]){
				dis[j]=dis[p]+g[p][j];
			}
		}
	}
	return dis[1];//(return to 1 point)
}
int main(){
	scanf("%d%d",&m,&n);
	memset(g,0x3f3f3f3f,sizeof(g));//图
	for(int i=1;i<=n;i++){
		int k;
		scanf("%d%d%d",&g[0][i],&v[i],&k);//The cost of u item to i item (the weight of reaching u node to i node)
		while(k--){
			int u, w;
			scanf("%d%d",&u,&w);
			g[u][i]=w;
		}
	}
	int minn=0x3f3f3f3f;//The minimum answer
	for(int i=1;i<=n;i++){//Enumerate the level of each node and set the current node to the lowest level
		int va = v [i];
		if(vv[va])continue;
		vv [va] = 1;
		memset(vis,0,sizeof(vis));
		for(int j=1;j<=n;j++){
			if(v[j]-va>m)vis[j]=1;//Exclude m above yourself
			if(v[j]<va)vis[j]=1;//Exclude those lower than yourself (because the enumeration is the lowest level)
		}
		minn = min (minn, djks ());
	}
	printf("%d\n",minn);
}