[Best training DAY16] KC's Can [Dynamic programming]

思路：

First converted into limited knapsack problem,But the card often card in the past

$code$

#include<iostream>
#include<cstdio>

#define re register

using namespace std;

int n, m;
int a[110][110], sum[110][110], p[110][110], f[110][10001], g[10001], s[10001];

int main() {
    
	scanf("%d%d", &n, &m);
	for(re int i = 1; i <= n; ++ i) {
    
		scanf("%d", &a[i][0]);
		for(re int j = 1; j <= a[i][0]; ++ j)
			scanf("%d", &a[i][j]), sum[i][j] = a[i][j] + sum[i][j - 1];
		for(re int j = 0; j <= a[i][0]; ++ j) {
    
			p[i][j] = 1e9;
			re int q = a[i][0] - j;
			for(re int k = 1; k + q - 1 <= a[i][0]; ++ k) {
    
				p[i][j] = min(p[i][j], sum[i][k + q - 1] - sum[i][k - 1]);
			}
			p[i][j] = sum[i][a[i][0]] - p[i][j];
		}
	}
	for(re int i = 1; i <= n; ++ i) {
    
		for(re int j = 0; j <= a[i][0]; ++ j) {
    
			for(re int k = m; k >= j; -- k) {
    
				if(g[k - j] + p[i][j] > f[j][k]) f[j][k] = g[k - j] + p[i][j];
				if(f[j][k] > s[k]) s[k] = f[j][k];
			}
		}
		for(re int k = m; k >= 0; -- k)
			g[k] = s[k];
	}
	printf("%d", g[m]);
	return 0;
}