Median Weight Bead Time Limit: 1000ms Memory Limit: 30000KB This problem will be judged on PKU. Original ID: 197564-bit integer IO format: %lld      Java class name: Main   There are N beads which of the same shape and size, but with different weight…
链接:poj 1975 题意:n个珠子,给定它们之间的重量关系.按重量排序.求确定肯定不排在中间的珠子的个数 分析:由于n为奇数.中间为(n+1)/2,对于某个珠子.若有至少有(n+1)/2个珠子比它重或轻,则它肯定不排在中间 能够将能不能确定的权值初始化为0,能确定重量关系的权值设为1 #include<stdio.h> #include<string.h> int a[110][110]; int main() { int T,n,m,i,j,k,d,x,sum; scanf(…
Median Weight Bead Time Limit: 1000MS Memory Limit: 30000K Total Submissions: 3162 Accepted: 1630 Description There are N beads which of the same shape and size, but with different weights. N is an odd number and the beads are labeled as 1, 2, -, N.…
Description There are N beads which of the same shape and size, but with different weights. N is an odd number and the beads are labeled as 1, 2, ..., N. Your task is to find the bead whose weight is median (the ((N+1)/2)th among all beads). The foll…
Description There are N beads which of the same shape and size, but with different weights. N is an odd number and the beads are labeled as 1, 2, ..., N. Your task is to find the bead whose weight is median (the ((N+1)/2)th among all beads). The foll…
描述 There are N beads which of the same shape and size, but with different weights. N is an odd number and the beads are labeled as 1, 2, ..., N. Your task is to find the bead whose weight is median (the ((N+1)/2)th among all beads). The following com…
链接:poj 3660 题意:给定n头牛,以及某些牛之间的强弱关系.按强弱排序.求能确定名次的牛的数量 思路:对于某头牛,若比它强和比它弱的牛的数量为 n-1,则他的名次能够确定 #include<stdio.h> #include<limits.h> int a[110][110]; int main() { int n,m,i,j,k,s,sum; while(scanf("%d%d",&n,&m)!=EOF){ for(i=1;i<=…
Description There are N beads which of the same shape and size, but with different weights. N is an odd number and the beads are labeled as 1, 2, ..., N. Your task is to find the bead whose weight is median (the ((N+1)/2)th among all beads). The foll…
题目链接:http://poj.org/problem?id=3660 Description N ( ≤ N ≤ ) cows, conveniently numbered ..N, are participating in a programming contest. As we all know, some cows code better than others. Each cow has a certain constant skill rating that is unique am…
<题目链接> 题目大意:FJ想按照奶牛产奶的能力给她们排序.现在已知有N头奶牛$(1 ≤ N ≤ 1,000)$.FJ通过比较,已经知道了M$1 ≤ M ≤ 10,000$对相对关系.每一对关系表示为“X Y”,意指X的产奶能力强于Y.现在FJ想要知道,他至少还要调查多少对关系才能完成整个排序. 解题分析: 因为完全图只需要$n*(n-1)/2$对关系就能确定所有的所有节点的顺序,所以这里我们只需要求出传递闭包之后,能够确定的关系数,然后相减即可.因为本题 $n\leq10^3$,$O(n^3…