poj3045Cow Acrobats(贪心， 二分）

本是一道练习二分的题目。。 但是发现贪心可用。。。二分反而没想到。。

题目大意是奶牛要叠罗汉了（杂技） 。。 求最小化最大危险值，  危险值等于 一头奶牛上面所有的奶牛体重之和减去这头的力量值。

证明略了。。看到网上写了很多了。。 结果就是按照w+s排序

这道题应该也可以用二分来最小化最大值。。但是A了之后就不太想了哎。。。。

题目：

Cow Acrobats

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 2245		Accepted: 888

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.

Source

USACO 2005 November Silver

 

代码：

#include <algorithm>
#include <cstdio>
#include <iostream>
using namespace std;
#define MAXN 50000+10
#define LL long long
#define min(x,y) x<y?x:y
#define max(x,y) x<y?y:x
int N;

struct P
{
    LL w,s;
    bool operator < (const P& x) const
    {
        return (w+s)<(x.w+x.s);
    }
};

P cow[MAXN];


int main()
{
    scanf("%d",&N);

    LL l =0 , r= 0;
    for(int i=0;i<N;i++)
    {
        scanf("%lld%lld",&cow[i].w,&cow[i].s);
    }
    sort( cow,cow+N);
    LL ans = -10000000000;
    LL sum =0;
    for(int i=0;i<N;i++)
    {
        ans = max( sum - cow[i].s, ans);
        sum+= cow[i].w;
    }
    printf("%lld
",ans);

    return 0;
}