pat 1055 The World's Richest

来源：互联网发布：微信恶搞踢出群软件编辑：程序博客网时间：2024/05/22 14:19

题意其实是求数组中的前k小的数。简单的方法大家都想得到，先排序，再取前k个，第二个不出意料的超时。要想以线性期望时间通过，应该得用随机选择算法来做。暂时先放着，以后再来解决。

代码：

#include<stdio.h>#include<stdlib.h>#include<string.h>const int MAXN = 100005;struct human{char name[10];int age;int net_worth;}rich[MAXN];int cmp(const void *a,const void *b){human *x=(human *)a;human *y = (human *)b;if(x->net_worth == y->net_worth && x->age == y->age)return strcmp(x->name, y->name);if(x->net_worth == y->net_worth)return x->age - y->age;return y->net_worth - x->net_worth;}int main(){int nPeople, nQuery, nOutput, aMin, aMax;int i,j;freopen("C:\\Documents and Settings\\Administrator\\桌面\\input.txt","r",stdin);scanf("%d%d",&nPeople, &nQuery);for(i = 0; i < nPeople; i++){scanf(" %s%d%d",rich[i].name, &rich[i].age, &rich[i].net_worth);}qsort(rich, nPeople, sizeof(rich[0]), cmp);// for(i = 0; i < nPeople; i++){// printf("%s %d %d\n",rich[i].name, rich[i].age, rich[i].net_worth);// // }int res;for(j = 0; j < nQuery; j++){scanf("%d%d%d",&nOutput, &aMin, &aMax);printf("Case \#%d:\n", j+1); res = 0;for(i = 0; i < nPeople && res < nOutput; i++){if(rich[i].age >= aMin && rich[i].age <= aMax){printf("%s %d %d\n",rich[i].name, rich[i].age, rich[i].net_worth);res++;}}if(res == 0)printf("None\n");}return 0;}