B. Checkout Assistant 01背包变形

http://codeforces.com/problemset/problem/19/B

对于每个物品，能偷多ti个，那么先让ti + 1， 表示选了这个东西后，其实就是选了ti + 1个了。那么只需要选出>=n个即可。

一开始的时候想不到ti + 1，一直不知道能多选ti个后，本来是选了多少个。

#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <cmath>
#include <algorithm>
#include <assert.h>
#define IOS ios::sync_with_stdio(false)
using namespace std;
#define inf (0x3f3f3f3f)
typedef long long int LL;
 
#include <iostream>
#include <sstream>
#include <vector>
#include <set>
#include <map>
#include <queue>
#include <string>
const int maxn = 2e3 + ;
int w[maxn];
int val[maxn];
LL dp[ *  + ];
void work() {
    int n;
    cin >> n;
    int sum = ;
    for (int i = ; i <= n; ++i) {
        cin >> w[i] >> val[i];
        w[i]++;
        w[i] = min(w[i], );
        sum += w[i];
    }
    for (int i = ; i <=  *  - ; ++i) {
        dp[i] = 1e17L;
    }
    dp[] = ;
    sum = min(, sum);
    for (int i = ; i <= n; ++i) {
        for (int j = sum; j >= w[i]; --j) {
            dp[j] = min(dp[j], dp[j - w[i]] + val[i]);
        }
    }
    LL ans = 1e17L;
    for (int i = n; i <= sum; ++i) {
        ans = min(ans, dp[i]);
    }
    cout << ans << endl;
}
 
int main() {
#ifdef local
    freopen("data.txt", "r", stdin);
//    freopen("data.txt", "w", stdout);
#endif
    work();
    return ;
}

写了一个超时的二维费用背包

dp[i][j]表示前n个数选出i个，有j个可以免费拿的情况。

直接超时。

#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <cmath>
#include <algorithm>
#include <assert.h>
#define IOS ios::sync_with_stdio(false)
using namespace std;
#define inf (0x3f3f3f3f)
typedef long long int LL;
 
#include <iostream>
#include <sstream>
#include <vector>
#include <set>
#include <map>
#include <queue>
#include <string>
const int maxn = 2e3 + ;
LL dp[maxn][maxn];
int ti[maxn];
int cost[maxn];
 
void work() {
    int n;
    cin >> n;
    int sum = ;
    for (int i = ; i <= n; ++i) {
        cin >> ti[i] >> cost[i];
        sum += ti[i];
    }
    for (int i = ; i <=  + ; ++i) {
        for (int j = ; j <=  + ; ++j) {
            dp[i][j] = 1e17L;
        }
    }
    dp[][] = ;
    sum = min(, sum);
    for (int i = ; i <= n; ++i) {
        for (int j = i; j >= ; --j) {
            for (int k = sum; k >= ti[i]; --k) {
                dp[j][k] = min(dp[j][k], dp[j - ][k - ti[i]] + cost[i]);
            }
        }
    }
    LL ans = 1e17L;
    for (int i = ; i <= n; ++i) {
        for (int j = sum; j >= ; --j) {
            dp[i][j] = min(dp[i][j], dp[i][j + ]);
        }
    }
    for (int i = ; i <= n; ++i) {
        ans = min(ans, dp[i][n - i]);
    }
    cout << ans << endl;
//    cout << dp[3][5] << endl;
}
 
int main() {
#ifdef local
    freopen("data.txt", "r", stdin);
//    freopen("data.txt", "w", stdout);
#endif
    IOS;
    work();
    return ;
}