POJ 3045 Cow Acrobats （贪心）

Cow Acrobats

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 3192		Accepted: 1260

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

这题 - - 没怎么懂 转载一发题解 贪心真是弱啊

按照w+s贪心叠，越大的越在下面。如果最优放置时，相邻两头牛属性分别为w1,s1,w2,s2,第一头牛在第二头上面，sum为第一头牛上面的牛的体重之和，那么

第一头牛风险：a=sum-s1;第二头牛风险:b=sum+w1-s2;

交换两头牛位置之后  a'=sum+w2-s1,b'=sum-s2，

由于是最优放置，所以w2-s1>=w1-s2，即w2+s2>=w1+s1，所以和最大的就该老实的在下面呆着= =！

AC代码如下：

//
//  POJ 3045 Cow Acrobats
//
//  Created by TaoSama on 2015-04-25
//  Copyright (c) 2015 TaoSama. All rights reserved.
//
#include <algorithm>
#include <cctype>
#include <cmath>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <iomanip>
#include <iostream>
#include <map>
#include <queue>
#include <string>
#include <set>
#include <vector>

using namespace std;
const int INF = 0x3f3f3f3f;
const int MOD = 1e9 + 7;
const int N = 1e5 + 10;

int n;
struct Cow {
    int w, s;
    bool operator < (const Cow& rhs) const {
        return w + s < rhs.w + rhs.s;
    }
} a[50005];

int main() {
#ifdef LOCAL
    freopen("in.txt", "r", stdin);
//  freopen("out.txt","w",stdout);
#endif
    ios_base::sync_with_stdio(0);

    scanf("%d", &n);
    for(int i = 1; i <= n; ++i)
        scanf("%d%d", &a[i].w, &a[i].s);
    sort(a + 1, a + 1 + n);
    int ans = -INF, sum = 0;
    for(int i = 1; i <= n; ++i){
		ans = max(ans, sum - a[i].s);
		sum += a[i].w;
    }
    printf("%d\n", ans);
    return 0;
}