POJ3045 Cow Acrobats —— 思维证明

题目链接：http://poj.org/problem?id=3045

Cow Acrobats

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

题解：

1.证明过程（来自cqbzwja）：http://blog.csdn.net/cqbzwja/article/details/47451687

2.自己的思考：根据承受力来排序，体重可能会走向极端；根据体重来排序，承受力也可能会走向极端。所以片面的考虑是得不到结果的（做题都能映射出人生，还能说些什么），既然体重和承受力共同影响这结果，所以就需要综合两者来考虑，即两者之和。

代码如下：

 1 #include <iostream>
 2 #include <cstdio>
 3 #include <cstring>
 4 #include <cmath>
 5 #include <algorithm>
 6 #include <vector>
 7 #include <queue>
 8 #include <stack>
 9 #include <map>
10 #include <string>
11 #include <set>
12 #define ms(a,b) memset((a),(b),sizeof((a)))
13 using namespace std;
14 typedef long long LL;
15 const double EPS = 1e-8;
16 const int INF = 2e9;
17 const LL LNF = 2e18;
18 const int MAXN = 1e5+10;
19
20 struct node
21 {
22     int w, s;
23     bool operator<(const node &x)const{
24         return (w+s)<(x.w+x.s);
25     }
26 }a[MAXN];
27
28
29 int main()
30 {
31     int n;
32     while(scanf("%d", &n)!=EOF)
33     {
34         for(int i = 1; i<=n; i++)
35             scanf("%d%d", &a[i].w, &a[i].s);
36         sort(a+1, a+1+n);
37         LL ans = -INF, tot = 0;
38         for(int i = 1; i<=n; i++)
39         {
40             ans = max(ans, tot-a[i].s);
41             tot += a[i].w;
42         }
43         printf("%lld\n", ans);
44     }
45 }