Buildings

　　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 wi and strength si. When floors are stacked up, each floor has PDV(Potential Damage Value) equal to (Σwj)-si, where (Σwj) 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.

　　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 wi and si (0 <= wi, si <= 100000) separated by single spaces.
　　Please process until EOF (End Of File).

　　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.

3
10 6
2 3
5 4
2
2 2
2 2
3
10 3
2 5
3 3

1
0
2

#include <cstdio>
#include <iostream>
#include <algorithm>

using namespace std;
#define N 100000 + 10
#define INF 0x3f3f3f3f
#define LL long long

struct E
{
    int w, s;
}e[N];

int n;

int cmp(E a, E b)
{
    return a.s + a.w < b.s + b.w ;
}

int main()
{
    while(~scanf("%d", &n))
    {
        for(int i = 1; i <= n; i++)
        {
            scanf("%d%d", &e[i].w, &e[i].s);
        }
        sort(e + 1, e + 1 + n, cmp);
        LL tmp = e[1].w;
        LL ans = -INF;
        for(int i = 2; i <= n; i++)
        {
            LL t = tmp - e[i].s;
            ans = max(ans, t);
            tmp += e[i].w;
        }
        if(ans < 0)
        printf("0\n");
        else printf("%I64d\n", ans);
    }
    return 0;
}