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  • hdu 5154 -- Harry and Magical Computer

     

    Harry and Magical Computer

    Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
    Total Submission(s): 641    Accepted Submission(s): 295


    Problem Description
    In 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.
     

    Input
    There 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. 1n100,1m10000
    The next following m lines, each line contains two numbers a b, indicates a dependencies (a, b). 1a,bn
     

    Output
    Output 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 2 3 1 2 1 3 3 3 2 2 1 1 3
     

    Sample Output
    YES NO
     

    思路: 用拓扑排序求解.

     1 /*======================================================================
     2  *           Author :   kevin
     3  *            Email :   zerocode.kevin@gmail.com
     4  *         Filename :   HarryAndMagicalComputer.cpp
     5  *       Creat time :   2015-01-23 09:17
     6  *      Description :
     7 ========================================================================*/
     8 #include <iostream>
     9 #include <algorithm>
    10 #include <cstdio>
    11 #include <cstring>
    12 #include <queue>
    13 #include <cmath>
    14 #define clr(a,b) memset(a,b,sizeof(a))
    15 #define INF 0x7f7f7f7f
    16 #define M 10005
    17 using namespace std;
    18 inline int min_32(int (a),int (b)){return (a)<(b)?(a):(b);}
    19 inline int max_32(int (a),int (b)){return (a)>(b)?(a):(b);}
    20 inline long long min_64(long long (a),long long (b)){return (a)<(b)?(a):(b);}
    21 inline long long max_64(long long (a),long long (b)){return (a)>(b)?(a):(b);}
    22 int head[M],in[M];
    23 struct Node{
    24     int to,next;
    25 }node[M];
    26 void addEdges(int i,int j,int k)
    27 {
    28     node[k].to = j;
    29     node[k].next = head[i];
    30     head[i] = k;
    31 }
    32 int slove(int n)
    33 {
    34     queue<int>que;
    35     for(int i = 1; i <= n; i++){
    36         if(in[i] == 0){
    37             que.push(i);
    38         }
    39     }
    40     int k,cnt = 0;
    41     while(!que.empty()){
    42         int t = que.front();
    43         que.pop();
    44         cnt++;
    45         for(k = head[t]; k != -1; k = node[k].next){
    46             if(--in[node[k].to] == 0){
    47                 que.push(node[k].to);
    48             }
    49         }
    50     }
    51     if(cnt == n){
    52         return 1;
    53     }
    54     return 0;
    55 }
    56 int main(int argc,char *argv[])
    57 {
    58     int n,m;
    59     while(scanf("%d%d",&n,&m)!=EOF){
    60         clr(head,-1);
    61         clr(in,0);
    62         int a,b;
    63         for(int i = 0; i < m; i++){
    64             scanf("%d%d",&a,&b);
    65             addEdges(b,a,i);
    66             in[a]++;
    67         }
    68         if(slove(n)){
    69             printf("YES
    ");
    70         }
    71         else{
    72             printf("NO
    ");
    73         }
    74     }
    75     return 0;
    76 }
    View Code
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  • 原文地址:https://www.cnblogs.com/ubuntu-kevin/p/4244117.html
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