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  • D. Table with Letters 2 夜

    http://codeforces.com/contest/253/problem/D

    挺麻烦的一道题

    代码:

    #include<iostream>
    #include<cstdio>
    #include<cstring>
    #include<string>
    #include<map>
    #include<vector>
    #include<stack>
    #include<queue>
    #include<algorithm>
    #define LL long long
    using namespace std;
    const int INF=0x3f3f3f3f;
    const int N=405;
    int sum[N][N];
    char s[N][N];
    int level[N];
    int n,m,k;
    LL ans;
    bool OK(int x1,int y1,int x2,int y2)
    {
        if(x1==x2)
        return true;
        int tmp=sum[x2][y2]-sum[x2][y1-1]-sum[x1-1][y2]+sum[x1-1][y1-1];
        if(tmp<=k)
        return true;
        return false;
    }
    void Fsearch(char w)
    {//cout<<w<<endl;
         for(int j1=1;j1<=m;++j1)
         for(int j2=j1+1;j2<=m;++j2)
         {
             int I=0;
             for(int i=1;i<=n;++i)
             {
                 if(s[i][j1]==w&&s[i][j2]==w)
                 level[I++]=i;
             }
             int l=0;
             int r=1;
             while(r<I)
             {
                 if(OK(level[l],j1,level[r],j2))
                 {
                     ans+=(r-l);
                     ++r;
                 }
                 else
                 {
                     ++l;
                 }
             }
         }
    }
    int main()
    {
        //freopen("data.in","r",stdin);
        freopen("input.txt","r",stdin);
        freopen("output.txt","w",stdout);
        while(scanf("%d %d %d",&n,&m,&k)!=EOF)
        {
            getchar();
            memset(sum,0,sizeof(sum));
            bool had[27];
            memset(had,false,sizeof(had));
            for(int i=1;i<=n;++i)
            gets(s[i]+1);
            for(int i=1;i<=n;++i)
            for(int j=1;j<=m;++j)
            {
                had[s[i][j]-'a']=true;
                if(s[i][j]=='a')
                sum[i][j]=1;
            }
            for(int i=1;i<=n;++i)
            for(int j=1;j<=m;++j)
            {
                sum[i][j]+=(sum[i-1][j]+sum[i][j-1]-sum[i-1][j-1]);
            }
            ans=0;
            for(char w='a';w<='z';++w)
            if(had[w-'a'])
            Fsearch(w);
            printf("%I64d\n",ans);
        }
        return 0;
    }
    

      

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  • 原文地址:https://www.cnblogs.com/liulangye/p/2810177.html
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