POJ3045-Cow Acrobats

Cow Acrobats

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 5013		Accepted: 1898

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

题意：n头牛叠罗汉，每头牛都有自己的重量wi和力量si，每头牛都有风险值为自己上面的牛的总重减去自己的力量，问这n头牛叠起来面临的危险值最小是多少

解题思路：假设目前排放的的序列是最优排列。任意位置有第一头牛和第二头牛分别有w1，s1；w2，s2。第一头牛上面的牛的重量总和为sum。且第一头牛在第二头牛上面，可以知道危险指数分别为a=sum-s1，b=sum+w1-s2。现在调换两头牛的位置，可得a'=sum+w2-s1，b'=sum-s2。  因为之前是最优排列，所以得知w2-s1>=w1-s2，   移项可得：w2+s2>=w1+s1，所以体重和力量之和越大越在底下

#include <iostream>
#include <cstdio>
#include <cstring>
#include <string>
#include <algorithm>
#include <cmath>
#include <map>
#include <cmath>
#include <set>
#include <stack>
#include <queue>
#include <vector>
#include <bitset>
#include <functional>

using namespace std;

#define LL long long
const int INF = 0x3f3f3f3f;

int n;
struct node
{
	int w, s;
}x[50009];

bool cmp(node a, node b)
{
	return a.w + a.s > b.w + b.s;
}

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