ZOJ

题目链接：http://acm.zju.edu.cn/onlinejudge/showProblem.do?problemCode=3956点击打开链接

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. The i-th course is described by two values: happiness H_i and credit C_i. If a student selects m 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 and C_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.

h越大值越大 c越大值越小

这样h就是value c就是weight

01背包 最后带入公式判断最大情况

#include <iostream>#include <algorithm>#include <cstring>#include <vector>#include <stdio.h>#define maxn 2005using namespace std;struct xjy{    long long int h;    long long int c;};vector<xjy > s;int main(){    int t;    scanf("%d",&t);    while(t--)    {        long long int m;long long int sum=0;        long long int dp[55555];        memset(dp,0,sizeof(dp));        s.clear();        scanf("%lld",&m);        for(int i=1;i<=m;i++)        {            xjy mid;            scanf("%lld%lld",&mid.h,&mid.c);            s.push_back(mid);            sum+=mid.c;        }        for(int i=0;i<m;i++)        {           for(int j=sum;j>=s[i].c;j--)                dp[j]=max(dp[j],dp[j-s[i].c]+s[i].h);        }        long long int ans=0;        for(int i=1;i<=sum;i++)            ans=max(ans,dp[i]*dp[i]-dp[i]*i-i*i);        printf("%lld\n",ans);    }}