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Re:复杂度只有o(n*lgn*m) 怎么会超时呢?求大牛指点迷津In Reply To:复杂度只有o(n*lgn*m) 怎么会超时呢?求大牛指点迷津 Posted by:snowbirds at 2009-04-23 21:16:09 > #include<stdio.h>
> #include<string.h>
> #include<stdlib.h>
> struct banji
> {
> int score;
> char name[31];
> };
> int cmp1(const void *a,const void *b)
> {
> banji *p=(banji *)a,*q=(banji *)b;
> return strcmp(p->name,q->name);
> }
> int cmp2(const void *a,const void *b)
> {
> banji *p=(banji *)a,*q=(banji *)b;
> return q->score-p->score;
> }
> banji a[20001],b[20001];
> int main()
> {
> int i,j,n,m,k;
> scanf("%d",&n);
>
> for(i=0;i<n;i++)
> scanf("%s",a[i].name);
> scanf("%d",&m);
> for(i=0;i<m;i++)
> {
> qsort(a,n,sizeof(banji),cmp1);
> for(j=0;j<n;j++)
> scanf("%d %s",&b[j].score,b[j].name);
> qsort(b,n,sizeof(banji),cmp1);
> for(j=0;j<n;j++)
> a[j].score=a[j].score+b[j].score;
> qsort(a,n,sizeof(banji),cmp2);
> for(j=0;j<n;j++)
> {
> if(strcmp(a[j].name,"JLY")==0)
> {k=a[j].score;
> break;
> }
> }
> for(j=0;j<n;j++)
> {if(a[j].score>k)continue;else break;}
> printf("%d\n",j+1);
>
> }
> return 0;
> }
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