最短路练习 8/poj/3660/Cow Contest
2017-04-24 14:04
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题目链接:http://poj.org/problem?id=3660
Cow Contest
Description
N (1 ≤ N ≤ 100) cows, conveniently numbered 1..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 among the competitors.
The contest is conducted in several head-to-head rounds, each between two cows. If cow A has a greater skill level than cow B (1 ≤ A ≤ N; 1 ≤ B ≤ N; A ≠ B), then cow A will
always beat cow B.
Farmer John is trying to rank the cows by skill level. Given a list the results of M (1 ≤ M ≤ 4,500) two-cow rounds, determine the number of cows whose ranks can be precisely determined from the results. It is guaranteed that the results
of the rounds will not be contradictory.
Input
* Line 1: Two space-separated integers: N and M
* Lines 2..M+1: Each line contains two space-separated integers that describe the competitors and results (the first integer, A, is the winner) of a single round of competition: A and B
Output
* Line 1: A single integer representing the number of cows whose ranks can be determined
Sample Input
Sample Output
题意:有n头牛,有m组关系,这m组关系不一定能得出所有的牛的排名,可能会有一些关系不确定,最多可以确定几头牛的排名。
思路:用Floyd:a>b,b>c 能得到a>c; a>b,c>b得不出a和c; 所以关系是单向的;
有关系的为1,没关系的为0;最后确定一头牛的排名确定不确定,看 他若与其它牛的关系都确定,排名则确定。
最后统计数量。
可惜了,我到最后统计数量的时候没想到这个方法。
AC代码:
#include<iostream>
#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<algorithm>
#include<cmath>
#include<queue>
#include<vector>
#include<map>
#include<string>
#define LL long long
#define eps 1e-8
using namespace std;
const int mod = 1e7+7;
const int inf = 0x3f3f3f3f;
const int maxn = 1e6 +10;
int n,m,a,b;
int ma[105][105];
int main()
{
while(~scanf("%d%d",&n,&m))
{
memset(ma,0,sizeof(ma));
for(int i=0;i<m;i++)
{
scanf("%d%d",&a,&b);
ma[a][b]=1;
}
for(int i=1;i<=n;i++)
for(int j=1;j<=n;j++)
for(int k=1;k<=n;k++)
if(ma[j][i]&&ma[i][k])
ma[j][k]=1;
int sum,ans=0;
for(int i=1;i<=n;i++)
{
sum=0;
for(int j=1;j<=n;j++)
if(ma[i][j]||ma[j][i])
sum++;
if(sum==n-1)
ans++;
}
printf("%d\n",ans);
}
}
Cow Contest
Time Limit: 1000MS | Memory Limit: 65536K | |
Total Submissions: 11308 | Accepted: 6280 |
N (1 ≤ N ≤ 100) cows, conveniently numbered 1..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 among the competitors.
The contest is conducted in several head-to-head rounds, each between two cows. If cow A has a greater skill level than cow B (1 ≤ A ≤ N; 1 ≤ B ≤ N; A ≠ B), then cow A will
always beat cow B.
Farmer John is trying to rank the cows by skill level. Given a list the results of M (1 ≤ M ≤ 4,500) two-cow rounds, determine the number of cows whose ranks can be precisely determined from the results. It is guaranteed that the results
of the rounds will not be contradictory.
Input
* Line 1: Two space-separated integers: N and M
* Lines 2..M+1: Each line contains two space-separated integers that describe the competitors and results (the first integer, A, is the winner) of a single round of competition: A and B
Output
* Line 1: A single integer representing the number of cows whose ranks can be determined
Sample Input
5 5 4 3 4 2 3 2 1 2 2 5
Sample Output
2
题意:有n头牛,有m组关系,这m组关系不一定能得出所有的牛的排名,可能会有一些关系不确定,最多可以确定几头牛的排名。
思路:用Floyd:a>b,b>c 能得到a>c; a>b,c>b得不出a和c; 所以关系是单向的;
有关系的为1,没关系的为0;最后确定一头牛的排名确定不确定,看 他若与其它牛的关系都确定,排名则确定。
最后统计数量。
可惜了,我到最后统计数量的时候没想到这个方法。
AC代码:
#include<iostream>
#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<algorithm>
#include<cmath>
#include<queue>
#include<vector>
#include<map>
#include<string>
#define LL long long
#define eps 1e-8
using namespace std;
const int mod = 1e7+7;
const int inf = 0x3f3f3f3f;
const int maxn = 1e6 +10;
int n,m,a,b;
int ma[105][105];
int main()
{
while(~scanf("%d%d",&n,&m))
{
memset(ma,0,sizeof(ma));
for(int i=0;i<m;i++)
{
scanf("%d%d",&a,&b);
ma[a][b]=1;
}
for(int i=1;i<=n;i++)
for(int j=1;j<=n;j++)
for(int k=1;k<=n;k++)
if(ma[j][i]&&ma[i][k])
ma[j][k]=1;
int sum,ans=0;
for(int i=1;i<=n;i++)
{
sum=0;
for(int j=1;j<=n;j++)
if(ma[i][j]||ma[j][i])
sum++;
if(sum==n-1)
ans++;
}
printf("%d\n",ans);
}
}
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