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最新推荐文章于 2020-12-20 11:03:31 发布

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最新推荐文章于 2020-12-20 11:03:31 发布

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分类专栏： BestCoder

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本文探讨了在处理多个任务时，如何通过依赖关系确保正确的执行顺序，从而实现高效的任务处理流程。通过实例分析了两种解决方法：Floyd算法和拓扑排序，帮助读者理解和实践任务间的依赖关系。

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Harry and Magical Computer

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 230 Accepted Submission(s): 108

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.

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

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

Sample Output

YES
NO

昨天bestcoder比赛打得好伤心，这道题想错了，题意是存在依存关系的就必须按照依存关系的顺序处理，如果不存在就无所谓了，我一直以为不存在就不能process了。

第一种方法：即bestcoder题解

用floyd判断，如果a->b的路径存在且b->a的路径存在，就NO

#include <iostream>
#include <cstring>
#include <algorithm>
#include <cassert>
#include <cstdio>
using namespace std ;

const int N = 100 + 11 ;

bool arr[N][N] ;
int n , m ;

void floyd() {
    for(int i = 1 ; i <= n ; ++i) {
        for(int j = 1 ; j <= n ; ++j) {
            if(i == j || arr[i][j]) continue ;
            for(int k = 1 ; k <= n ; ++k) {
                if(k == i ) continue ;
                arr[i][j] = arr[i][k] & arr[k][j] ;
            }
        }
    }
    for(int i = 1 ; i <= n ; ++i) {
        for(int j = i ; j <= n ; ++j) {
            if(arr[i][j] && arr[j][i]) {
                printf("NO\n") ; return ;
            }
        }
    }
    printf("YES\n") ;
}

int main() {//freopen("data.in" , "r" ,stdin) ;
    int a , b ;
    while(scanf("%d%d" ,&n ,&m)==2) {
        memset(arr , 0 , sizeof(arr)) ;
        while(m--) {
            scanf("%d%d" ,&a ,&b) ;
            arr[a][b] = true ;
        }
        floyd() ;
    }
}

第二种方法：拓扑排序

如果能把所有的点排序则YES

#include <iostream>
#include <cstring>
#include <algorithm>
#include <cstdio>
#include <list>
#include <queue>
#include <vector>
#include <cassert>
using namespace std ;

const int N = 100 + 11 ;

vector<int> err[N] ;
int degree[N] ;
int n , m ;

void topsort() {
    queue<int> que ;
    for(int i = 1 ; i <= n ; ++i) {
        if(degree[i] == 0) {
            que.push(i) ;
        }
    }
    int k = 0 ;
    while(!que.empty()) {
        ++k ;
        int f = que.front() ;
        que.pop() ;
        for(int i = 0 ; i < err[f].size() ; ++i) {
            int tmp = err[f][i] ;
            --degree[tmp] ;
            if(degree[tmp] == 0) {
                que.push(tmp) ;
            }
        }
    }
    if(k == n) printf("YES\n") ;
    else printf("NO\n") ;
}

int main() {//freopen("data.in" , "r" , stdin) ;
    int a , b ;
    while(scanf("%d%d" ,&n ,&m)==2) {
        memset(degree , 0 ,sizeof(degree)) ;for(int i = 1 ; i <= n ; ++i) err[i].clear() ;
        while(m--) {
            scanf("%d%d" ,&a ,&b) ;
            err[b].push_back(a) ;
            ++degree[a] ;
        }
        topsort() ;
    }
}