Cow Acrobats
| Time Limit: 1000MS | Memory Limit: 65536K | |
| Total Submissions: 4889 | Accepted: 1850 |
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.
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.
* 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.
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
代码实现
1 #include<iostream> 2 #include <stdio.h> 3 #include <string.h> 4 #include <algorithm> 5 using namespace std; 6 const int MAXN = 50017; 7 #define INF 0x3f3f3f3f 8 typedef struct 9 { 10 int w, s; 11 int ss; 12 } COW; 13 COW cow[MAXN]; 14 bool cmp(COW a, COW b) 15 { 16 return a.ss < b.ss; 17 } 18 int main() 19 { 20 int n; 21 int sum[MAXN]; 22 while(~scanf("%d",&n)) 23 { 24 memset(sum,0,sizeof(sum)); 25 //将sum中当前位置后面的sizeof(sum)个字节用0替换并返回 s 26 //作用是在一段内存块中填充某个给定的值，它是对较大的结构体或数组进行清零操作的一种最快方法 27 for(int i = 1; i <= n; i++) 28 { 29 scanf("%d%d",&cow[i].w,&cow[i].s); 30 cow[i].ss = cow[i].w+cow[i].s; 31 } 32 sort(cow+1,cow+n+1,cmp); 33 int ans = -INF ; 34 for(int i = 1; i <= n; i++) 35 { 36 sum[i] = sum[i-1]+cow[i].w; 37 ans = max(ans,sum[i-1]-cow[i].s); 38 } 39 printf("%d ",ans); 40 } 41 return 0; 42 }
思路分析：
题意：这道题居然和今年成都赛区的倒数第二题一模一样。。。或者说该反过来说、、给你n头牛叠罗汉，
每头都有自己的重量w和力量s，承受的风险数就是该牛上面牛的总重量减去它的力量，题目要求设计一个
方案使得所有牛里面风险最大的要最小。
题解：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= =！