hdu 5154 -- Harry and Magical Computer

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Harry and Magical ComputerTime Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 641    Accepted Submission(s): 295 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 Input3 23 12 13 33 22 11 3  Sample OutputYESNO

思路: 用拓撲排序求解.

 

 1 /*====================================================================== 2  *           Author :   kevin 3  *            Email :   [email protected] 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 0x7f7f7f7f16 #define M 1000517 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\n");70         }71         else{72             printf("NO\n");73         }74     }75     return 0;76 }

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hdu 5154 -- Harry and Magical Computer