Color a Tree POJ - 2054

一、内容

题意：给定一棵树，求染完所有节点的最少代价。

二、思路

三、代码

#include <cstdio>
#include <iostream>
using namespace std;
const int N = 1e3 + 5;
struct Node{
	int size, sum, fa;
	double avg;
} nodes[N];
int n, root, a, b;
int find() {
	int v = 0;
	double mavg = -1;
	//查找出最大的avg
	for (int i = 1; i <= n; i++) {
		if (i != root && mavg < nodes[i].avg) {
			v = i;
			mavg = nodes[i].avg;
		}
	} 
	return v;
}
int main() {
	while (scanf("%d%d", &n, &root), n) {
		int ans = 0; 
		for (int i = 1; i <= n; i++) {
			Node &t = nodes[i];
			scanf("%d", &t.sum);
			t.size = 1;
			t.avg = t.sum;
			//加上每个节点原始的一次
			ans += t.sum; 
		}
		for (int i = 1; i < n; i++) {
			scanf("%d%d", &a, &b);
			nodes[b].fa = a;
		}
		//每次找节点中等效权值avg最大的
		for (int k = 1; k < n; k++) {
			int v = find();
			//加上父亲的层次
			int fa = nodes[v].fa;
			ans += nodes[v].sum * nodes[fa].size;  
			//进行合并
			nodes[fa].sum += nodes[v].sum;
			nodes[fa].size += nodes[v].size;		
			nodes[fa].avg = (double)nodes[fa].sum / nodes[fa].size;		
			for (int i = 1; i <= n; i++) {
				if (nodes[i].fa == v) {
					nodes[i].fa = fa;
				} 
			}	
			nodes[v].avg = -1;//等于删除操作 
		}
		printf("%d\n", ans);	
	}
 
	return 0;
}