hdu 5154 Harry and Magical Computer(BestCoder Round #25)

標籤：bestcoder   編程   algorithm   acm   

Harry and Magical Computer                                                       Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
                                                                             Total Submission(s): 472    Accepted Submission(s): 222


Problem DescriptionIn reward of being yearly outstanding magic student, Harry gets a magical computer. When the computer begins to deal with a process, it will work until the ending of the processes. One day the computer got n processes to deal with. We number the processes from 1 to n. However there are some dependencies between some processes. When there exists a dependencies (a, b), it means process b must be finished before process a. By knowing all the m dependencies, Harry wants to know if the computer can finish all the n processes. 
InputThere are several test cases, you should process to the end of file.
For each test case, there are two numbers n m on the first line, indicates the number processes and the number of dependencies. 1≤n≤100,1≤m≤10000
The next following m lines, each line contains two numbers a b, indicates a dependencies (a, b). 1≤a,b≤n 
OutputOutput one line for each test case. 
If the computer can finish all the process print "YES" (Without quotes).
Else print "NO" (Without quotes). 
Sample Input

3 23 12 13 33 22 11 3

 
Sample Output

YESNO

 
SourceBestCoder Round #25 
     
       判斷拓撲排序是否能夠完成，拓撲排序不能完成的條件是沒有環，只要判斷圖裡是否有環是否就可以了，用flody和深搜應該都可以吧，我用的flody做的。

#include <iostream>#include <cstdio>#include <algorithm>#include <cstring>using namespace std;int map[150][150];int n,m;void flody(){    for(int k=1;k<=n;k++)       for(int i=1;i<=n;i++)           for(int j=1;j<=n;j++)               if(map[i][k]&&map[k][j])                   map[i][j]=1;}int main(){    while(~scanf("%d%d",&n,&m))    {        memset(map,0,sizeof(map));        int sign=0;        int a,b;        for(int i=0;i<m;i++)        {            scanf("%d%d",&a,&b);            if(a==b)            {                sign=1;                continue;            }            map[a][b]=1;        }        if(sign)        printf("NO\n");        else        {            flody();            for(int i=1;i<=n;i++)            if(map[i][i])            sign=1;            if(sign)            printf("NO\n");            else            printf("YES\n");        }    }    return 0;}

hdu 5154 Harry and Magical Computer(BestCoder Round #25)