ZOJ 3956 Course Selection System(01背包变形)

Course Selection System

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Time Limit: 1 Second      Memory Limit: 65536 KB

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There are n courses in the course selection system of Marjar University. Thei-th course is described by two values: happiness H_i and creditC_i. If a student selectsm courses x₁,x₂, ...,x_m, then his comfort level of the semester can be defined as follows:

Edward, a student in Marjar University, wants to select some courses (also he can select no courses, then his comfort level is 0) to maximize his comfort level. Can you help him?

Input

There are multiple test cases. The first line of input contains an integer T, indicating the number of test cases. For each test case:

The first line contains a integer n (1 ≤ n ≤ 500) -- the number of cources.

Each of the next n lines contains two integers H_i andC_i (1 ≤H_i ≤ 10000, 1 ≤C_i ≤ 100).

It is guaranteed that the sum of all n does not exceed 5000.

We kindly remind you that this problem contains large I/O file, so it's recommended to use a faster I/O method. For example, you can use scanf/printf instead of cin/cout in C++.

Output

For each case, you should output one integer denoting the maximum comfort.

Sample Input

2310 15 12 1021 102 10

Sample Output

1910

Hint

For the first case, Edward should select the first and second courses.

For the second case, Edward should select no courses.

【题意】每门课有一个H值和C值，学生选课的满意值=ΣH^2-ΣH*ΣC-ΣC^2。求学生选课的最大满意值。

【分析】满意值=ΣH(ΣH-ΣC)-ΣC^2，可见当C一定时，H越大，满意值越高。因此转化为C作为容量的01背包来求解。

#include<cstdio>#include<iostream>#include<algorithm>#include<cstring>using namespace std;#define maxn 505int N,sum;int h[maxn],c[maxn];long long ans,dp[50005];int main(){    int T;    scanf("%d",&T);    while(T--){        scanf("%d",&N);        sum=0;ans=0;        memset(dp,0,sizeof(dp));        for(int i=0;i<N;i++){            scanf("%d%d",&h[i],&c[i]);            sum+=c[i];        }        for(int i=0;i<N;i++){            for(int j=sum;j>=c[i];j--){                if(dp[j]<=dp[j-c[i]]+h[i]){                    dp[j]=dp[j-c[i]]+h[i];                    ans=max(ans,dp[j]*dp[j]-dp[j]*j-j*j);                }            }        }        printf("%lld\n",ans);    }    return 0;}