Cow Acrobats（贪心）

Cow Acrobats

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 2336		Accepted: 929

Description

Farmer John's N (1 <= N <= 50,000) cows (numbered 1..N) are planning to run away and join the circus. Their hoofed feet prevent them from tightrope walking and swinging from the trapeze (and their last attempt at firing a cow out of a cannon met with a dismal failure). Thus, they have decided to practice performing acrobatic stunts. 

The cows aren't terribly creative and have only come up with one acrobatic stunt: standing on top of each other to form a vertical stack of some height. The cows are trying to figure out the order in which they should arrange themselves ithin this stack. 

Each of the N cows has an associated weight (1 <= W_i <= 10,000) and strength (1 <= S_i <= 1,000,000,000). The risk of a cow collapsing is equal to the combined weight of all cows on top of her (not including her own weight, of course) minus her strength (so that a stronger cow has a lower risk). Your task is to determine an ordering of the cows that minimizes the greatest risk of collapse for any of the cows.

Input

* Line 1: A single line with the integer N. 

* Lines 2..N+1: Line i+1 describes cow i with two space-separated integers, W_i and S_i. 

Output

* Line 1: A single integer, giving the largest risk of all the cows in any optimal ordering that minimizes the risk.

Sample Input

3
10 3
2 5
3 3

Sample Output

2

Hint

OUTPUT DETAILS: 

Put the cow with weight 10 on the bottom. She will carry the other two cows, so the risk of her collapsing is 2+3-3=2. The other cows have lower risk of collapsing.

 

       题意：

       给出 N （1 ~ 50000），代表有 N 头牛，每头牛都有一个 W （1 ~ 1000）和 S （1 ~ 1000000000）。现有个马戏团表演，要求把牛搭牛，所以对应每头牛都会有一个风险值。这个风险值等于这头牛以上所有牛的 W 之和 - 这头牛的 S。现要求你设计叠放的顺序，使最大值最小化，输出这个最小化最大值。

 

       思路：

       贪心。先对总和（S + W）由小到大排序，后算出最大值即为最小化的最大值。风险值 = sum（W） - （Wi + Si ），故要求最大值最小，那么要求 ( Wi + Si ) 最大。

 

       AC：

#include <cstdio>
#include <algorithm>

using namespace std;

const int INF = 1000000005;

typedef struct { int w, s, n;} node;
node no[50005];

int cmp(node a, node b) { return a.n < b.n; }

int main () {
        int n;
        scanf("%d", &n);
        for (int i = 0; i < n; ++i) {
                scanf("%d%d", &no[i].w, &no[i].s);
                no[i].n = no[i].w + no[i].s;
        }

        sort(no, no + n, cmp);

        int sum = 0, max_num = -INF;
        for (int i = 0; i < n; ++i) {
                sum += no[i].w;
                if (max_num < sum - no[i].n)
                    max_num = sum - no[i].n;
        }

        printf("%d\n", max_num);
        return 0;
}