hdu 4328

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题目

求颜色全为红的矩形周长和颜色全为蓝的矩形周长和颜色交替的矩形周长,输出周长最大值.

类似于hdu1505,求满足条件的最大子矩阵,在处理颜色交替时,判断条件加一个颜色比较就行了.

#include<stdio.h>#include<string.h>#include<algorithm>using namespace std;int t,n,m;char s[1002][1002];int rr[1002][1002],b[1002][1002],rb[1002][1002];//rr[i][j],以第i行,第j列为顶边的红色矩形的高,(i,j从1开始)int r[1002],l[1002];int main(){    int ca=1;    scanf("%d",&t);    while(t--)    {        memset(rr,0,sizeof(rr));        memset(b,0,sizeof(b));        memset(br,0,sizeof(br));        memset(rb,0,sizeof(rb));        memset(s,0,sizeof(s));        scanf("%d%d",&n,&m);        for(int i=0;i<n;i++)            scanf("%s",s[i]);        for(int i=1;i<=n;i++)        {            for(int j=1;j<=m;j++)            {                if(s[i-1][j-1]=='R')                    rr[i][j]=rr[i][j-1]+1;                else                    rr[i][j]=0;                if(s[i-1][j-1]=='B')                    b[i][j]=b[i][j-1]+1;                else                    b[i][j]=0;                if(j==1) rb[i][j]=1;                else if(j>1)                {                    if(s[i-1][j-1]!=s[i-1][j-2]) rb[i][j]=rb[i][j-1]+1;                    else rb[i][j]=1;                }            }        }        int ans=0;        for(int j=1;j<=m;j++)        {            for(int i=1;i<=n;i++)            {                l[i]=r[i]=i;            }            for(int i=1;i<=n;i++)            {                while(b[i][j]<=b[l[i]-1][j]&&l[i]>=1&&b[i][j]>0) l[i]=l[l[i]-1];            }            for(int i=n;i>=1;i--)            {                while(b[i][j]<=b[r[i]+1][j]&&r[i]<=n&&b[i][j]>0) r[i]=r[r[i]+1];            }            for(int i=1;i<=n;i++)                ans=max(ans,b[i][j]*2+(r[i]-l[i]+1)*2);        }        for(int j=1;j<=m;j++)        {            for(int i=1;i<=n;i++)            {                l[i]=r[i]=i;            }            for(int i=1;i<=n;i++)            {                while(rr[i][j]<=rr[l[i]-1][j]&&l[i]>=1&&rr[i][j]>0) l[i]=l[l[i]-1];            }            for(int i=n;i>=1;i--)            {                while(rr[i][j]<=rr[r[i]+1][j]&&r[i]<=n&&rr[i][j]>0) r[i]=r[r[i]+1];            }            for(int i=1;i<=n;i++)                ans=max(ans,rr[i][j]*2+(r[i]-l[i]+1)*2);        }        for(int j=1;j<=m;j++)        {            for(int i=1;i<=n;i++)            {                l[i]=r[i]=i;            }            for(int i=1;i<=n;i++)            {                while(rb[i][j]<=rb[l[i]-1][j]&&l[i]>=1&&s[l[i]-1-1][j-1]!=s[l[i]-1][j-1]) l[i]=l[l[i]-1];            }            for(int i=n;i>=1;i--)            {                while(rb[i][j]<=rb[r[i]+1][j]&&r[i]<=n&&s[r[i]+1-1][j-1]!=s[r[i]-1][j-1]) r[i]=r[r[i]+1];            }            for(int i=1;i<=n;i++)                ans=max(ans,rb[i][j]*2+(r[i]-l[i]+1)*2);        }        printf("Case #%d: %d\n",ca++,ans);    }}