RQNOJ 30 愚蠢的矿工 解题报告

　　树形动态规划，第一次接触这样的树形动态规划，应该说是彻底的，以前的那个什么没有上司的舞会都是小儿科，嗯，思路见网上同类报告（偷懒）
　　代码：

#include <stdio.h>
#include <string.h>
#include <stdlib.h>
int num[1001];
struct node{
	int left, right;
	int sum;
}tree[1001];
int f[1002][102];
int m, n;

int srch(int root, int count)
{
	int i;
	int t;
	if(f[root][count] != 0){
		return f[root][count];
	}
	if(tree[root].right != n + 1){
		f[root][count] = srch(tree[root].right, count);
	}
	for(i = 0; i < count; i++){
		t = srch(tree[root].left, i) + srch(tree[root].right, count - i - 1) + num[root];
		if(f[root][count] < t){
			f[root][count] = t;
		}
	}
	return f[root][count];
}

int main(int argc, char **argv)
{
	int i, j;
	int a, b;
	scanf("%d%d", &n, &m);
	tree[0].left = tree[0].right = n + 1;
	for(i = 1; i <= n; i++){
		tree[i].left = tree[i].right = n + 1;
		scanf("%d", &num[i]);
	}
	for(i = 1; i <= n; i++){
		scanf("%d%d", &a, &b);
		if(tree[a].left == n + 1){
			tree[a].left = b;
		}else{
			j = tree[a].left;
			while(tree[j].right != n + 1){
				j = tree[j].right;
			}
			tree[j].right = b;
		}
	}
	for(i = 1; i <= m + 1; i++){
		f[n + 1][i] = -1000000;
	}
	printf("%d\n", srch(0, m + 1));
	return 0;
}

　　

 

转载于:https://www.cnblogs.com/yylogo/archive/2011/08/27/RQNOJ-30.html