POJ-1975 Median Weight Bead(Floyed)
2016-01-06 09:44
459 查看
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. Your task is to find the bead whose weight is median (the ((N+1)/2)th among all beads). The following comparison has been performed on some pairs of beads:
A scale is given to compare the weights of beads. We can determine which one is heavier than the other between two beads. As the result, we now know that some beads are heavier than others. We are going to remove some beads which cannot have the medium weight.
For example, the following results show which bead is heavier after M comparisons where M=4 and N=5.
1. Bead 2 is heavier than Bead 1.
Bead 4 is heavier than Bead 3.
Bead 5 is heavier than Bead 1.
Bead 4 is heavier than Bead 2.
From the above results, though we cannot determine exactly which is the median bead, we know that Bead 1 and Bead 4 can never have the median weight: Beads 2, 4, 5 are heavier than Bead 1, and Beads 1, 2, 3 are lighter than Bead 4. Therefore, we can remove these two beads.
Write a program to count the number of beads which cannot have the median weight.
Input
The first line of the input file contains a single integer t (1 <= t <= 11), the number of test cases, followed by the input data for each test case. The input for each test case will be as follows:
The first line of input data contains an integer N (1 <= N <= 99) denoting the number of beads, and M denoting the number of pairs of beads compared. In each of the next M lines, two numbers are given where the first bead is heavier than the second bead.
Output
There should be one line per test case. Print the number of beads which can never have the medium weight.
Sample Input
1
5 4
2 1
4 3
5 1
4 2
Sample Output
2
用两次Floyed就可以了
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. Your task is to find the bead whose weight is median (the ((N+1)/2)th among all beads). The following comparison has been performed on some pairs of beads:
A scale is given to compare the weights of beads. We can determine which one is heavier than the other between two beads. As the result, we now know that some beads are heavier than others. We are going to remove some beads which cannot have the medium weight.
For example, the following results show which bead is heavier after M comparisons where M=4 and N=5.
1. Bead 2 is heavier than Bead 1.
Bead 4 is heavier than Bead 3.
Bead 5 is heavier than Bead 1.
Bead 4 is heavier than Bead 2.
From the above results, though we cannot determine exactly which is the median bead, we know that Bead 1 and Bead 4 can never have the median weight: Beads 2, 4, 5 are heavier than Bead 1, and Beads 1, 2, 3 are lighter than Bead 4. Therefore, we can remove these two beads.
Write a program to count the number of beads which cannot have the median weight.
Input
The first line of the input file contains a single integer t (1 <= t <= 11), the number of test cases, followed by the input data for each test case. The input for each test case will be as follows:
The first line of input data contains an integer N (1 <= N <= 99) denoting the number of beads, and M denoting the number of pairs of beads compared. In each of the next M lines, two numbers are given where the first bead is heavier than the second bead.
Output
There should be one line per test case. Print the number of beads which can never have the medium weight.
Sample Input
1
5 4
2 1
4 3
5 1
4 2
Sample Output
2
用两次Floyed就可以了
#include <iostream> #include <string.h> #include <math.h> #include <algorithm> #include <stdlib.h> using namespace std; int a[105][105]; int b[105][105]; int n,m; void floyed1() { for(int k=1;k<=n;k++) { for(int i=1;i<=n;i++) { for(int j=1;j<=n;j++) { if(i!=j&&a[i][k]&&a[k][j]) a[i][j]=1; } } } } void floyed2() { for(int k=1;k<=n;k++) { for(int i=1;i<=n;i++) { for(int j=1;j<=n;j++) { if(i!=j&&b[i][k]&&b[k][j]) b[i][j]=1; } } } } int main() { int t; int x,y; scanf("%d",&t); while(t--) { memset(a,0,sizeof(a)); memset(b,0,sizeof(b)); scanf("%d%d",&n,&m); for(int i=1;i<=m;i++) { scanf("%d%d",&x,&y); a[x][y]=1; b[y][x]=1; } floyed1(); floyed2(); int res=0; for(int i=1;i<=n;i++) { int num1=0; for(int j=1;j<=n;j++) { if(a[i][j]) num1++; } int num2=0; for(int j=1;j<=n;j++) { if(b[i][j]) num2++; } if(num1>(n/2)||num2>(n/2)) res++; } printf("%d\n",res); } }
相关文章推荐
- 数据结构——bitmap
- Wifi小车资料
- [jstips]undefined和null的区别
- HTTP中Get与Post的区别
- *圣思园java se培训总结(102-)(synchronized相关)
- the 4th national media processing meeting
- SDK变量命名习惯小结
- 证书导入java的jre环境中
- Spark 1.6.0 (Scala 2.11)版本的编译与安装部署
- java151227ExceptionDemo2
- 交换机arp只能查询本身三层所配置的网段
- 2.2.1 创建一个 Path
- 如何解除改变phpmyadmin数据库导入文件大小限制?
- rk3288_USB_CAMERA+MIC_OK
- -bash: ulimit: pipe size: cannot modify limit: Invalid argument
- Micron 美光存储器件
- 分享php邮件管理器源码
- -bash: ulimit: pipe size: cannot modify limit: Invalid argument
- [从头学数学] 第32节 混合运算
- [Android]快递查询——API的简单应用