Problem 1538 - B - Stones II (动态规划)

Problem 1538 - B - Stones II
Time Limit: 1000MS Memory Limit: 65536KB
Total Submit: 371 Accepted: 54 Special Judge: No
Description

Xiaoming took the flight MH370 on March 8, 2014 to China to take the ACM contest in WHU. Unfortunately, when the airplane crossing the ocean, a beam of mystical light suddenly lit up the sky and all the passengers with the airplane were transferred to another desert planet.



When waking up, Xiaoming found himself lying on a planet with many precious stones. He found that:



There are n precious stones lying on the planet, each of them has 2 positive values ai and bi. Each time Xiaoming can take the ith of the stones ,after that, all of the stones’ aj (NOT including the stones Xiaoming has taken) will cut down bi units.



Xiaoming could choose arbitrary number (zero is permitted) of the stones in any order. Thus, he wanted to maximize the sum of all the stones he has been chosen. Please help him.
Input
The input consists of one or more test cases.

First line of each test case consists of one integer n with 1 <= n <= 1000.
Then each of the following n lines contains two values ai and bi.( 1<= ai<=1000, 1<= bi<=1000)
Input is terminated by a value of zero (0) for n.
Output
For each test case, output the maximum of the sum in one line.
Sample Input
1
100 100
3
2 1
3 1
4 1
0
Sample Output
100

6

dp[i][j]:前i项选择j项的最大值

dp[i][j]=max(dp[i-1][j],dp[i][j-1]+p[i].ai-p[i].bi*(j-1))

#include<bits/stdc++.h>
//#include<iostream>
//#include<cmath>
//#include<cstring>
//#include<string>
//#include<cstdlib>
//#include<cstdio>
//#include<map>
//#include<algorithm>
using namespace std;
template<class T>inline T read(T&x)
{
    char c;
    while((c=getchar())<=32);
    bool ok=false;
    if(c=='-')ok=true,c=getchar();
    for(x=0; c>32; c=getchar())
        x=x*10+c-'0';
    if(ok)x=-x;
    return x;
}
template<class T> inline void read_(T&x,T&y)
{
    read(x);
    read(y);
}
template<class T> inline void read__(T&x,T&y,T&z)
{
    read(x);
    read(y);
    read(z);
}
template<class T> inline void write(T x)
{
    if(x<0)putchar('-'),x=-x;
    if(x<10)putchar(x+'0');
    else write(x/10),putchar(x%10+'0');
}
template<class T>inline void writeln(T x)
{
    write(x);
    putchar('\n');
}
//-------ZCC IO template------
const int maxn=1001;
const double inf=999999999;
#define lson (rt<<1),L,M
#define rson (rt<<1|1),M+1,R
#define M ((L+R)>>1)
#define For(i,t,n) for(int i=(t);i<(n);i++)
typedef long long  LL;
typedef double DB;
#define bug printf("---\n");
struct node
{
    int ai,bi;
    bool operator<(node tmp)const
    {
        if(tmp.bi!=bi)return bi>tmp.bi;
        else return ai>tmp.ai;
    }
}p[maxn];
int dp[maxn][maxn];
int main()
{
    int n;
    while(read(n))
    {
        for(int i=1;i<=n;i++)
        {
            read_(p[i].ai,p[i].bi);
        }
        sort(p+1,p+n+1);
        For(i,0,n+1)dp[i][0]=0;
        For(i,1,n+1)For(j,1,i+1)
        {
            dp[i][j]=max(dp[i-1][j],dp[i-1][j-1]+p[i].ai-(j-1)*p[i].bi);
        }
        int ans=0;
        For(i,0,n+1)
           ans=max(ans,dp[n][i]);
        writeln(ans);
    }
    return 0;
}