hdu4296 Buildings

Buildings

Time Limit: 5000/2000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 5317    Accepted Submission(s): 1902

Problem Description

　　Have you ever heard the story of Blue.Mary, the great civil engineer? Unlike Mr. Wolowitz, Dr. Blue.Mary has accomplished many great projects, one of which is the Guanghua Building.
　　The public opinion is that Guanghua Building is nothing more than one of hundreds of modern skyscrapers recently built in Shanghai, and sadly, they are all wrong. Blue.Mary the great civil engineer had try a completely new evolutionary building method in project of Guanghua Building. That is, to build all the floors at first, then stack them up forming a complete building.
　　Believe it or not, he did it (in secret manner). Now you are face the same problem Blue.Mary once stuck in: Place floors in a good way.
　　Each floor has its own weight w_i and strength s_i. When floors are stacked up, each floor has PDV(Potential Damage Value) equal to (Σw_j)-s_i, where (Σw_j) stands for sum of weight of all floors above.
　　Blue.Mary, the great civil engineer, would like to minimize PDV of the whole building, denoted as the largest PDV of all floors.
　　Now, it’s up to you to calculate this value.

 

Input

　　There’re several test cases.
　　In each test case, in the first line is a single integer N (1 <= N <= 10⁵) denoting the number of building’s floors. The following N lines specify the floors. Each of them contains two integers w_i and s_i (0 <= w_i, s_i <= 100000) separated by single spaces.
　　Please process until EOF (End Of File).

 

Output

　　For each test case, your program should output a single integer in a single line - the minimal PDV of the whole building.
　　If no floor would be damaged in a optimal configuration (that is, minimal PDV is non-positive) you should output 0.

 

Sample Input

310 62 35 422 22 2310 32 53 3

 

Sample Output

102

 

Source

/*
有n个板，每个板有重量和强度两个属性，把板叠在一起，对于每个板有个PDV值，计算方式为这个板上面的板的重量和减去这个板的强度， 
对于每种叠放方式,取这个叠放方式中，所有板的PDV值最大的值为代表值，求所有叠放方式中最小的代表值
设sum为所有楼层的w之和，那么，对于最下面的楼层b，其PDV = sum - f[b].w - f[b].s = sum - (f[b].w + f[b].s)；要使这层最小，
那么 f[b].w + f[b].s应尽量大，对于从下面数上去的第二层，策略也是一样的。 
*/

#include<cstdio>  
#include<cstring>  
#include<algorithm>  
using namespace std;  

#define LL long long  //数据过大，使用longlong 
struct point  
{  
    int w,s;  
}node[100005];  
int n;  

bool cmp(point a,point b)  
{  
    return a.w+a.s<b.w+b.s;  //按两个值之和从小到大排列 
}  

int main()  
{  
    while(scanf("%d",&n)!=EOF) 
    {  
        for(int i=0;i<n;++i)  
        scanf("%d%d",&node[i].w,&node[i].s);  
        
        sort(node,node+n,cmp);  
        LL summ=0,maxx=0;  
        for(int i=0;i<n;++i)  //从上往下设计 
        {  
            maxx=max(maxx,summ-node[i].s);  
            summ+=node[i].w;  
        }  
        printf("%I64d\n",maxx);  
    }  
    return 0;  
}