CF815C Karen and Supermarket

题目链接

CF815C Karen and Supermarket

题解

只要在最大化数量的前提下，最小化花费就好了
这个数量枚举ok，
dp[i][j][1/0]表示节点i的子树中买了j件商品 i 优惠了 / 没优惠
复杂度是n^2的
因为每次是新儿子节点的siz * 之前儿子几点的siz，
就相当于树上的节点两两匹配，这个匹配只会在lca处计算一次

代码

#include<cstdio> 
#include<cstring> 
#include<algorithm> 
#define LL long long 
inline int read() { 
	int x = 0,f = 1; 
	char c = getchar(); 
	while(c < '0' || c  > '9') { if(c == '-')f = -1; c = getchar(); } 
	while(c <= '9' && c >= '0') x = x * 10 + c - '0',c = getchar(); 
	return x * f; 
} 
const int maxn = 5007; 
int n; LL b;int c[maxn], d[maxn]; 
struct node {
	int v,nxt; 
} edge[maxn]; 
int head[maxn],num = 0 ;  
inline void add_edge(int u,int v) { 
	edge[++ num].v = v; edge[num].nxt = head[u];head[u] = num; 
}  
LL dp[maxn][maxn][2]; 
int siz[maxn]; 
void dfs(int x) { 
	siz[x] = 1;
	dp[x][0][0] = 0; 
	dp[x][1][0] = c[x] ; 
	dp[x][1][1] = c[x] - d[x]; 
	//for(int i = head[x];i;i = edge[i].nxt)  dfs(edge[i].v),  siz[x] += siz[edge[i].v]; 
	for(int i = head[x];i;i = edge[i].nxt) { 
		int v = edge[i].v; 
		dfs(v); 
		for(int j = siz[x];j >= 0;-- j) { 
			for(int k = 0;k <= siz[v];++ k) {  
				dp[x][j + k][0] = std::min(dp[x][j + k][0],dp[x][j][0] + dp[v][k][0]); 
				dp[x][j + k][1] = std::min(dp[x][j + k][1],dp[x][j][1] + std::min(dp[v][k][1],dp[v][k][0])); 
			} 
		} siz[x] += siz[v]; 
	} 
} 
int main() {  
	memset(dp,0x3f,sizeof dp); 
	n = read(), b = read(); 
	c[1] = read(); 	d[1] = read(); 
	for(int pre, i = 2;i <= n;++ i) { 
		c[i] = read(),d[i] = read(); pre = read(); 
		add_edge(pre,i); 
	} 
	dfs(1); 
	int ans = 0; 
	for(int i = 1;i <= n;++ i) 
		if(std::min(dp[1][i][0],dp[1][i][1]) <= b)ans = i;   
	printf("%d\n",ans); 
	return 0; 
}