hdu 3496 Watch The Movie(二维费用背包)

二维费用背包问题

有两个限制条件：电影时长和看电影的部数。需要注意一下初始化dp[i][0] = 0,其余未知，因为看电影是一部一部来，所以当前dp值只能衔接前一部的dp值。

/*
Author:Zhaofa Fang
Lang:C++
*/
#include <cstdio>
#include <cstdlib>
#include <sstream>
#include <iostream>
#include <cmath>
#include <cstring>
#include <algorithm>
#include <string>
#include <utility>
#include <vector>
#include <queue>
#include <stack>
#include <map>
#include <set>
using namespace std;

typedef long long ll;
#define DEBUG(x) cout<< #x << ':' << x << endl
#define REP(i,n) for(int i=0;i < (n);i++)
#define FOR(i,s,t) for(int i = (s);i <= (t);i++)
#define FORD(i,s,t) for(int i = (s);i >= (t);i--)
#define PII pair<int,int>
#define PB push_back
#define MP make_pair
#define ft first
#define sd second
#define lowbit(x) (x&(-x))
#define INF (1<<30)

int dp[1005][105],ti[105],v[105];
int main()
{
    //freopen("in","r",stdin);
    //freopen("out","w",stdout);
    int T;
    scanf("%d",&T);
    while(T--)
    {
        int n,m,L;
        scanf("%d%d%d",&n,&m,&L);
        REP(i,n)scanf("%d%d",&ti[i],&v[i]);
        memset(dp,-1,sizeof(dp));
        REP(i,L+1)dp[i][0] = 0;
        REP(i,n)
        {
            FORD(j,L,ti[i])
            {
                FORD(k,m,1)
                {
                    if(dp[j-ti[i]][k-1]!=-1)dp[j][k] = max(dp[j][k],dp[j-ti[i]][k-1]+v[i]);
                }
            }
        }
        if(dp[L][m]!=-1)printf("%d\n",dp[L][m]);
        else puts("0");
    }
    return 0;
}