【dp】 AreYouBusy

题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=3535

题意： 多组背包， 0类型为为至少去1样， 1为至多取1样， 2 为随意。

如果将2类型 再添加一组数据 （0， 0）， 则可转换为0类型， 即0,1 背包问题， 1类型为经典分组背包。

/***Good Luck***/
#define _CRT_SECURE_NO_WARNINGS
#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <string>
#include <algorithm>
#include <stack>
#include <map>
#include <queue>
#include <vector>
#include <set>
#include <functional>
#include <cmath>
#include <numeric>

#define Zero(a) memset(a, 0, sizeof(a))
#define Neg(a)  memset(a, -1, sizeof(a))
#define All(a) a.begin(), a.end()
#define PB push_back
#define inf 0x3f3f3f3f
#define inf2 0x7fffffffffffffff
#define ll long long
using namespace std;
//#pragma comment(linker, "/STACK:102400000,102400000")
void get_val(int &a) {
    int value = 0, s = 1;
    char c;
    while ((c = getchar()) == ‘ ‘ || c == ‘\n‘);
    if (c == ‘-‘) s = -s; else value = c - 48;
    while ((c = getchar()) >= ‘0‘ && c <= ‘9‘)
        value = value * 10 + c - 48;
    a = s * value;
}
const int maxn = 110;
int n, t;
int dp[maxn][maxn];
int C[maxn][maxn];
int G[maxn][maxn];
int TE[maxn];
int N1[maxn];

int main() {
    //freopen("data.out", "w", stdout);
    //freopen("data.in", "r", stdin);
    //cin.sync_with_stdio(false);
    while (cin >> n >> t){
        for (int i = 1; i < maxn; ++i)
        for (int j = 0; j < maxn; ++j)
            dp[i][j] = -inf;
        Zero(dp[0]);
        int n1, te;
        for (int i = 1; i <= n; ++i) {
            scanf("%d%d", &n1, &te);
            for (int j = 1; j <= n1; ++j) {
                scanf("%d%d", C[i] + j, G[i] + j);
            }
            if (te != 0) {// 情况一和情况三相似， 在情况三中添加一组0， 0 便可表示任意取
                n1++;     // 及可取可不取。
                C[i][n1] = 0;
                G[i][n1] = 0;
            }
            TE[i] = te;
            N1[i] = n1;
        }
        for (int i = 1; i <= n; ++i) {
            if (TE[i] == 0) {
                for (int ii = 1; ii <= N1[i]; ++ii) //dp[i][v] = max(dp[i][v], dp[i][v - C[i][ii]] + G[i][ii],dp[i - 1][v - C[i][ii]] + G[i][ii])
                for (int v = t; v >= C[i][ii]; --v) {
                    dp[i][v] = max(dp[i][v], dp[i][v - C[i][ii]] + G[i][ii]);
                    dp[i][v] = max(dp[i][v], dp[i - 1][v - C[i][ii]] + G[i][ii]);
                }
            }
            else if (TE[i] == 1) {
                for (int v = t; v >= 0; v--)
                for (int ii = 1; ii <= N1[i]; ++ii)
                if (v >= C[i][ii])
                    dp[i][v] = max(dp[i][v], dp[i - 1][v - C[i][ii]] + G[i][ii]);
            }
            else {
                for (int ii = 1; ii <= N1[i]; ++ii)
                for (int v = t; v >= C[i][ii]; --v) {
                    dp[i][v] = max(dp[i][v], dp[i][v - C[i][ii]] + G[i][ii]);
                    dp[i][v] = max(dp[i][v], dp[i - 1][v - C[i][ii]] + G[i][ii]);
                }
            }
        }
        if (dp[n][t] >= 0)
        printf("%d\n", dp[n][t]);
        else printf("-1\n");
    }
    return 0;
}