poj 2585 拓扑排序

这题主要在于建图。对9个2*2的小块,第i块如果出现了不等于i的数字,那么一定是在i之后被brought的。可以从i到该数字建一条边。

图建好后,进行一次拓扑排序,判段是否存在环。若存在环,那么就是BROKEN,否则是CLEAN。

#include<iostream>
#include<cstdio>
#include<cstring>
#define Maxn 102
#define Maxm 10010
using namespace std;
int graphic[Maxn][Maxn],indegree[Maxn],n,m,map[Maxn][Maxn],vi[10];
int Topsort()
{
    int i,j,k,num=0;
    for(i=1;i<=9;i++)
    {
        for(j=1;j<=9;j++)
        {
            if(indegree[j]==0)
            {
                indegree[j]--;
                for(k=1;k<=9;k++)
                    if(graphic[j][k])
                        indegree[k]--;
                num++;
                break;
            }
        }
    }
    if(num<9)
        return 0;
    return 1;
}
int main()
{
    char str[20];
    int i,j;
    while(scanf("%s",&str),strlen(str)<9)
    {
        n=0;
        memset(vi,0,sizeof(vi));
        memset(graphic,0,sizeof(graphic));
        memset(indegree,0,sizeof(indegree));
        for(i=1;i<=4;i++)
        {
            for(j=1;j<=4;j++)
            {
                scanf("%d",&map[i][j]);
            }
        }
        int k=0;
        for(i=1;i<=3;i++)
            for(j=1;j<=3;j++)
            {    
                k++;
                if(map[i][j]!=k)
                {
                    if(!graphic[k][map[i][j]])
                    {
                        graphic[k][map[i][j]]=1;
                        indegree[map[i][j]]++;
                    }
                }
                if(map[i][j+1]!=k)
                {
                    if(!graphic[k][map[i][j+1]])
                    {
                        graphic[k][map[i][j+1]]=1;
                        indegree[map[i][j+1]]++;
                    }
                }
                if(map[i+1][j]!=k)
                {
                    if(!graphic[k][map[i+1][j]])
                    {
                        graphic[k][map[i+1][j]]=1;
                        indegree[map[i+1][j]]++;
                    }
                }
                if(map[i+1][j+1]!=k)
                {
                    if(!graphic[k][map[i+1][j+1]])
                    {
                        graphic[k][map[i+1][j+1]]=1;
                        indegree[map[i+1][j+1]]++;
                    }
                }
            }
        scanf("%s",&str);
        if(Topsort())
            printf("THESE WINDOWS ARE CLEAN
");
        else
            printf("THESE WINDOWS ARE BROKEN
");
    }
    return 0;
}
原文地址:https://www.cnblogs.com/wangfang20/p/3191047.html