ZOJ 3956 Course Selection System 01背包

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Course Selection System

Time Limit: 1 Second Memory Limit: 65536 KB

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:

$(\sum_{i=1}^{m} H_{x_i})^2-(\sum_{i=1}^{m} H_{x_i})\times(\sum_{i=1}^{m} C_{x_i})-(\sum_{i=1}^{m} C_{x_i})^2$

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

Hint

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

For the second case, Edward should select no courses.

Author: WANG, Yucheng
Source: The 17th Zhejiang University Programming Contest Sponsored by TuSimple

因为c的值很小，所以以c看做重量，把h当做价值

01背包，取尽可能大的h

因为当c固定时，这个式子只需要看前面 h*h-h*c=h*（h-c）

h大于c时，肯定h越大越好

h小于c时，表达式的值必定小于零，肯定是不考虑的

#include<bits/stdc++.h>using namespace std;long long int dp[50005],h[505],c[505];int main(){int n,t,i,j;while(cin>>t){while(t--){scanf("%d",&n);long long maxx=0,now;memset(dp,-1,sizeof(dp));for(i=0;i<n;i++){scanf("%lld %lld",&h[i],&c[i]);}dp[0]=0;for(i=0;i<n;i++){for(j=50000;j>=c[i];j--){if(dp[j-c[i]]==-1)continue;if(dp[j-c[i]]+h[i]>dp[j]){dp[j]=dp[j-c[i]]+h[i];now=dp[j]*dp[j]-j*dp[j]-j*j;if(now>maxx)                      maxx=now;}}}printf("%lld\n",maxx);}}return 0;}

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