POJ 1182：食物链

并查集

想法很简单

开两个数组

吃 与 被吃

然而 炸了我一下午

mlgb

贴个错误代码

吃饭去了

烦

#include<stdio.h>
#include<string.h>
#include<algorithm>
using namespace std;

const int maxn=55555;
int fa[maxn];
int eat[maxn];
int ate[maxn];


int find(int x)
{
 return fa[x]==x?x:fa[x]=find(fa[x]);
}

void merge(int x,int y)
{
 int fx,fy;
 fx=find(x);
 fy=find(y);
 if(fx!=fy)
 fa[fy]=fx;
}

int main()
{
    int n,k;
    int com,n1,n2;
    scanf("%d%d",&n,&k);
    for(int i=1;i<=n;i++)
    {
            eat[i]=0;
            ate[i]=0;
            fa[i]=i;
    }
    int num=0;
    for(int i=0;i<k;i++)
    {
            scanf("%d%d%d",&com,&n1,&n2);
            if(com==1)
            {
                if(n1>n||n2>n)
                        {num++;continue;}
                if(find(n1)==find(eat[n2])||find(n2)==find(eat[n1]))
                        {num++;continue;}
                if(!(eat[n1]==0&&ate[n2]==0))
                {
                   if(find(eat[n1])==find(ate[n2]))
                        {num++;continue;}
                }

//                if(find(n2)==find(ate[n1])||find(n1)==find(ate[n2]))
//                        {num++;continue;}

                        merge(n1,n2);

            }

            else if(com==2)
            {
                if(n1>n||n2>n)
                        {num++;continue;}
//                else
//                {
                        if(n1==n2)
                                {num++;continue;}
                        if(find(n1)==find(n2))
                                {num++;continue;}
//                        if(find(ate[n1])==find(n2))
//                                {num++;continue;}
                        if(find(n1)==find(eat[n2]))
                                {num++;continue;}

                          if(!(eat[n1]==0&&ate[n2]==0))
                {
                   if(find(eat[n1])==find(ate[n2]))
                        {num++;continue;}
                }
                        if(!(eat[n1]==0&&eat[n2]==0))
                                //if(eat[n1]==find(eat[n2])||eat[n2]==find(eat[n1]))
                        {
                                if(find(eat[n1])==find(eat[n2]))
                                        {num++;continue;}
                        }


                        if(eat[n1]==0)
                                eat[n1]=n2;
                        else
                                merge(eat[n1],n2);

                        if(ate[n2]==0)
                                ate[n2]=n1;
                        else
                                merge(ate[n2],n1);

               // }
            }
    }
    printf("%d\n",num);

    return 0;
}

POJ 1182：食物链

原文：http://blog.csdn.net/whereyousink/article/details/47342079