Cow Acrobats(贪心)
Time Limit: 1000MS | Memory Limit: 65536K | |
Total Submissions: 3686 | Accepted: 1428 |
Description
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
* Lines 2..N+1: Line i+1 describes cow i with two space-separated integers, W_i and S_i.
Output
Sample Input
3
10 3
2 5
3 3
Sample Output
2
题解:错的心碎啊,还是代码太弱,会写而wa。。。。重的而且抗压的显然在最下面;
代码:
#include<cstdio>
#include<cstring>
#include<cmath>
#include<algorithm>
using namespace std;
const int MAXN=100010;
typedef long long LL;
const int inf=1<<29;
struct Node{
LL w,s;
friend bool operator < (Node a,Node b){
return a.w+a.s>b.w+b.s;
}
};
Node dt[MAXN];
int main(){
int N;
while(~scanf("%d",&N)){
for(int i=0;i<N;i++){
scanf("%lld%lld",&dt[i].w,&dt[i].s);
}
sort(dt,dt+N);
LL ans=-inf,temp=0;
if(N==1)ans=-dt[0].s;
for(int i=N-1;i>=0;i--){
if(temp-dt[i].s>ans)ans=temp-dt[i].s;
temp+=dt[i].w;
}
printf("%I64d\n",ans);
} return 0;
}
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