POJ - 3045 Cow Acrobats (贪心)

Cow Acrobats

Time Limit: 1000MS Memory Limit: 65536KTotal Submissions: 4243 Accepted: 1630

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

310 32 53 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:有n头奶牛堆积木，其中第i头奶牛的重量为wi，力量为si，每头奶牛有一个难受值di，等于他上面所有奶牛的重量和减去他的力量，现在找出一种方案，使得这n头奶牛中最大的难受值尽量小。

假设某头牛的上方的包括牛本身的重量为W,那么W - wi - si即为承受重量，那么W - (wi + si),我们要尽可能让wi + si在面对更多的W时更大，所以对它进行排序处理，取最大值即可

#include <cstdio>#include <cstring>#include <algorithm>using namespace std;typedef long long LL;const int MAXN = 5e4 + 5;const int INF = 0x3f3f3f3f;int N;struct o {int w, s;o() {}o(int w, int s) : w(w), s(s) {}bool operator < (const o &p) const {return w + s < p.w + p.s;}} A[MAXN];int main() {while(~scanf("%d", &N)) {for(int i = 0; i < N; i ++) {scanf("%lld%lld", &A[i].w, &A[i].s);}sort(A, A + N);LL w = 0, me = -INF;for(int i = 0; i <= N - 1; i ++) {me = max(me, w - A[i].s);w += A[i].w;}printf("%lld\n", me);}return 0;}