BZOJ4247 挂饰（动态规划）

　　相当于一个有负体积的背包。显然如果确定了选哪些，应该先把体积小的挂上去。于是按体积从小到大排序，就是一个裸的背包了。

#include<iostream>
#include<cstdio>
#include<cmath>
#include<cstdlib>
#include<cstring>
#include<algorithm>
using namespace std;
int read()
{
    int x=,f=;char c=getchar();
    while (c<''||c>'') {if (c=='-') f=-;c=getchar();}
    while (c>=''&&c<='') x=(x<<)+(x<<)+(c^),c=getchar();
    return x*f;
}
#define N 2010
#define inf 2000000000
int n,f[N][N];
struct data{int x,y;
}a[N];
bool cmp(const data&a,const data&b)
{
    return a.x>b.x;
}
int main()
{
#ifndef ONLINE_JUDGE
    freopen("bzoj4247.in","r",stdin);
    freopen("bzoj4247.out","w",stdout);
    const char LL[]="%I64d\n";
#else
    const char LL[]="%lld\n";
#endif
    n=read();
    for (int i=;i<=n;i++) a[i].x=read(),a[i].y=read();
    sort(a+,a+n+,cmp);
    for (int i=;i<=n;i++)
        for (int j=;j<=n+;j++) f[i][j]=-inf;
    f[][]=;
    for (int i=;i<=n;i++)
        for (int j=;j<=n;j++)
        f[i][j]=max(f[i-][j],f[i-][max(j-a[i].x,)+]+a[i].y);
    for (int i=;i<=n;i++) f[n][]=max(f[n][],f[n][i]);
    cout<<f[n][];
    return ;
}