[ $\frak{Link}$]

三種物品集合，第一種至少選一個，第二種至多選一個，第三種隨便選。
第三種不帶約束的很簡單，01

#include<cstdio>
#include<algorithm>
#include<iostream>
#include<cstring>
using namespace std;
int n,T;
int m[105], s[105];
int c[105][105], g[105][105];
int F[105][105];
int
 
 main() {
	while (~scanf("%d%d", &n, &T)) {
		for (int i = 1; i <= n; ++i) {
			scanf("%d%d", &m[i], &s[i]);
			for (int j = 1; j <= m[i]; ++j) {
				scanf("%d%d", &c[i][j], &g[i][j]);
			}
		}
		memset(F, 0, sizeof(F));
		for (int i = 1; i <= n; ++i) {
			if (s[i] ==
 
 0) memset(F[i], 0xcf, sizeof(F[i]));
			else memcpy(F[i], F[i-1], sizeof(F[i-1]));
			for (int k = 1; k <= m[i]; ++k) {
				for (int j = T; j >= 0; --j) {
					if (j < c[i][k]) continue;
					if (s[i] == 0) {
						F[i][j] = max(F[i][j], F[i][j-c[i][k]] + g[i][k]);
						F[i][j] = max(F[i][j], F[i-1][j-c[i][k]] + g[i][k]);
					}
					else {
						if (s[i] == 1) F[i][j] = max(F[i][j], F[i-1][j-c[i][k]] + g[i][k]);
						if (s[i] == 2) F[i][j] = max(F[i][j], F[i][j-c[i][k]] + g[i][k]);
					}
				}
			}
		}
		printf( "%d\n", max(F[n][T],-1) );
	}
	return 0;
}			}
			}
		}
		printf( "%d\n", max(F[n][T],-1) );
	}
	return 0;
}