Codeforces Round #290 (Div. 2) _B找矩形环的三种写法

http://codeforces.com/contest/510/status/B

题目大意 给一个n*m  找有没有相同字母连起来的矩形串

第一种并查集 瞎搞一下

第一次的时候把val开成字符串了 所以wa

改了AC

#include<cstdio>
#include<map>
//#include<bits/stdc++.h>
#include<vector>
#include<stack>
#include<iostream>
#include<algorithm>
#include<cstring>
#include<cmath>
#include<set>
#include<queue>
#include<cstdlib>
#include<climits>
#define PI acos(-1.0)
#define INF 0x3fffffff
using namespace std;
typedef long long ll;
typedef __int64 int64;
const ll mood=1e9+7;
const int64 Mod=998244353;
const double eps=1e-9;
const int N=100;
const int MAXN=1e4+5;
typedef int rl;
inline void r(rl&num){
    num=0;rl f=1;char ch=getchar();
    while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}
    while(ch>='0'&&ch<='9')num=num*10+ch-'0',ch=getchar();
    num*=f;
}
char ch[N][N];
int  val[N][N];
int par[MAXN],sum[MAXN];
int n,m;
char tem;
int dx[]={1,0},dy[]={0,1};
void init()
{
    int s=0;
    for(int i=0;i<n;i++)
    {
        for(int j=0;j<m;j++)
        {

            val[i][j]=s;//cout<<val[i][j]<<endl;
            par[s]=s;
            sum[s]=1;
            s++;
        }
    }
}
int find(int x)
{
    if(x==par[x]) return x;
    return par[x]=find(par[x]);
}
bool unite(int x,int y)
{
    x=find(x);y=find(y);
    if(x==y&&sum[x]>=4) return true;
    if(x==y) return false;
    par[x]=y;
    sum[y]+=sum[x];
    return false;
}
bool check(int x,int y)
{
    if(x>=0&&x<n&&y<m&&y>=0&&ch[x][y]==tem) return true;
    else return false;
}
int main()
{
    r(n);r(m);
    for(int i=0;i<n;i++)
    {
        scanf("%s",ch[i]);
    }
    init();
    bool mk=false;
    for(int i=0;i<n;i++)
    {
        for(int j=0;j<m;j++)
        {
            tem=ch[i][j];
            for(int k=0;k<2;k++)
            {
                int x=dx[k]+i,y=dy[k]+j;
                if(check(x,y))
                {
                    if(unite(val[i][j],val[x][y]))
                    {
                        mk=true;
                        break;
                    }
                }
            }
            if(mk) break;
        }
        if(mk) break;
    }
    if(mk) puts("Yes");
    else puts("No");
    return 0;
}
/*
50 50
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*/
并查集

这个类似联通块 dfs做 首尾相交就输出YES 做一下

 

#include<cstdio>
char ch[52][52];
int vis[52][52];
int flag;
int n,m;
int dx[]={0,0,-1,1},dy[]={1,-1,0,0};
void dfs(int atx,int aty,int x,int y,char z)
{
    if(vis[x][y])
    {
        flag=1;
        return ;
    }
    vis[x][y]=1;
    int nx,ny;
    for(int i=0;i<4;i++)
    {
        nx=x+dx[i];ny=y+dy[i];
        if(nx<0||nx>=n||ny<0||ny>=m||z!=ch[nx][ny]||(atx==nx&&ny==aty)) continue;
        dfs(x,y,nx,ny,z);
    }
    return ;
}
int main()
{
    scanf("%d%d",&n,&m);
    for(int i=0;i<n;i++)
    {
        scanf("%s",ch[i]);
    }
    for(int i=0;i<n;i++)
        for(int j=0;j<m;j++)
        {
            if(!vis[i][j])
            {
                dfs(-1,-1,i,j,ch[i][j]);
            }
            if(flag){printf("Yes\n");return 0;}
        }
    printf("No\n");
    return 0;
}
其实dfs做的很不顺利

 

还有一个方法 构成矩形块的充要条件 自己的上下左右要有和自己相同的块至少两个 小于两个的块必然不是 可以标记扔掉 

这里感叹下高手对细节的处理真的好强 等自己写的时候才知道问题在哪里 仔细看下人家的才恍然大悟 这样才不会越界

就是在n*m的周围搞上一圈空

大致意思就是把不符合的点全部标记并回到1，2点并不是回到起点 1，1点如果满足条件没被标记的话 就会留下1，1点 所以判断ans>1

#include<cstdio>
char ch[52][52];
int main()
{
    int n,m;
    scanf("%d%d",&n,&m);
    for(int i=1;i<=n;i++)
    {
        getchar();
        for(int j=1;j<=m;j++)
        {
            ch[i][j]=getchar();
        }
    }
    for(int i=1;i<=n;i++)
        for(int j=1;j<=m;j++)
        {
            if(ch[i][j]!='%')
            {
                 int tem=0;
                if(ch[i][j]==ch[i-1][j]) tem++;
                if(ch[i][j]==ch[i+1][j]) tem++;
                if(ch[i][j]==ch[i][j-1]) tem++;
                if(ch[i][j]==ch[i][j+1]) tem++;
                if(tem<2) {ch[i][j]='%';i=j=1;} //回到起点
            }
        }
        int ans=0;
      for(int i=1;i<=n;i++)
        for(int j=1;j<=m;j++)
        {
            if(ch[i][j]!='%') ans++;
            if(ans>1)
            {
                puts("Yes");
                return 0;
            }
        }
        puts("No");
        return 0;
}
神奇的姿势

现在搞一下回到起点的 只要判断有点留下 那么就输出Yes；

#include<cstdio>
char ch[52][52];
int main()
{
    int n,m;
    scanf("%d%d",&n,&m);
    for(int i=1;i<=n;i++)
    {
        getchar();
        for(int j=1;j<=m;j++)
        {
            ch[i][j]=getchar();
        }
    }
    for(int i=1;i<=n;i++)
        for(int j=1;j<=m;j++)
        {
            if(ch[i][j]!='%')
            {
                int tem=0;
                if(ch[i][j]==ch[i-1][j]) tem++;
                if(ch[i][j]==ch[i+1][j]) tem++;
                if(ch[i][j]==ch[i][j-1]) tem++;
                if(ch[i][j]==ch[i][j+1]) tem++;
                if(tem<2) {ch[i][j]='%';i=1;j=0;}

            }
        }
        int ans=0;
      for(int i=1;i<=n;i++)
        for(int j=1;j<=m;j++)
        {
            if(ch[i][j]!='%') 
            {
                puts("Yes");
                return 0;
            }
        }
        puts("No");
        return 0;
}
理解版神奇的姿势

 

posted @ 2016-04-04 10:37  zxMrlc  阅读(218)  评论(0)  编辑  收藏  举报