P1194 买礼物

买礼物

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问题分析

• 都要买，问题在需要确定一个购买顺序
• 用了某个优惠关系就在两点间连一条边，最后出来是一棵树
• 那么用所有优惠关系建图，最后求最小生成树即可

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• 裸最小生成树
• dist 初始化为 A
• 如果边权大于A则边权赋值为A（坑点）
• 建议用prim，因为是针对点的，每个点都要买
• n = B, m = B * B;
• 总结：B个点B^2条边的最小生成树，记得针对点

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代码

#include <bits/stdc++.h>

#define pr pair<int, int>
#define mk make_pair

using namespace std;

const int N = 1e5 + 1;

struct Node{
	int v,w,nxt;
	Node(){}
	Node(int _v, int _w, int _nxt) : v(_v), w(_w), nxt(_nxt){}
}edge[N << 2]; 

int n,m,top,cnt,cost;
int vis[N],head[N],dist[N];
int A,B;

void addedge(int u, int v, int w){
	edge[++top].v = v;
	edge[top].w = w;
	edge[top].nxt = head[u];
	head[u] = top;
}

priority_queue<pr, vector<pr>, greater<pr> > q;
void prim(){
	for(int i = 1; i <= n; ++i) dist[i] = A;
	dist[1] = A;
	q.push(mk(dist[1], 1));
	while(!q.empty()){
		int u = q.top().second;
		int d = q.top().first; q.pop();
		if(vis[u]) continue;
		vis[u] = 1;
		cnt += 1;
		cost += d;
		for(int i = head[u]; i; i = edge[i].nxt){
			int v = edge[i].v;
			int w = edge[i].w;
			if(w <= dist[v] && !vis[v])
				dist[v] = w,
					q.push(mk(dist[v], v));
		}
		
	}
}

int main(){
	cin >> A >> B;
	n = B;
	m = B * B;
	for(int u = 1; u <= B; ++u){
		for(int v = 1; v <= B; ++v){
			int w;
				cin >> w;
					if(w == 0 && u != v || w > A) w = A;
						addedge(u, v, w);
		}
	}
	prim();
	printf("%d", cost);
	return 0;
}