HDU 4328 Contest 3

悬线法可解,稍有点烦琐。

#include <iostream>
#include <cstdio>
#include <cstring>
#include <string.h>
using namespace std;

char map[1010][1010];
int lefts[1010][1010],heights[1010][1010],rights[1010][1010];

int CountSingle(int n,int m,char sing){
	int ret=0;
	for(int i=1;i<=n;i++){
		if(i==1){
			int l=0;
			for(int j=1;j<=m;j++){
				if(map[i][j]!=sing){
					heights[i][j]=0;
					lefts[i][j]=m+1;
					l=j;
				}
				else{
					heights[i][j]=1;
					lefts[i][j]=l+1;
				}
			}
			int r=m+1;
			for(int j=m;j>=1;j--){
				if(map[i][j]!=sing){
					rights[i][j]=0;
					r=j;
				}
				else{
					rights[i][j]=r-1;
					ret=max(ret,heights[i][j]*2+2*(rights[i][j]-lefts[i][j]+1));
				}
			}
		}
		else{
			int l=0;
			for(int j=1;j<=m;j++){
				if(map[i][j]!=sing){
					heights[i][j]=0;
					lefts[i][j]=m+1;
					l=j;
				}
				else{
					if(map[i-1][j]==sing){
						heights[i][j]=heights[i-1][j]+1;
						lefts[i][j]=max(lefts[i-1][j],l+1);
					}
					else{
						heights[i][j]=1;
						lefts[i][j]=l+1;
					}
				}
			}
			int r=m+1;
			for(int j=m;j>=1;j--){
				if(map[i][j]!=sing){
					rights[i][j]=0;
					r=j;
				}
				else{
					if(map[i-1][j]==sing){
						rights[i][j]=min(rights[i-1][j],r-1);
						ret=max(ret,heights[i][j]*2+2*(rights[i][j]-lefts[i][j]+1));
					}
					else{
						rights[i][j]=r-1;
						ret=max(ret,heights[i][j]*2+2*(rights[i][j]-lefts[i][j]+1));
					}
				}
			}
		}
	}
	return ret;
}

int CountCross(int n,int m){
	int ret=0;
	for(int i=1;i<=n;i++){
		if(i==1){
			int l=0;
			for(int j=1;j<=m;j++){
				if(j==1){
					heights[i][j]=1;
					lefts[i][j]=1;
				}
				else{
					if(map[i][j-1]==map[i][j]){
						lefts[i][j]=j;
						l=j-1;
						heights[i][j]=1;
					}
					else{
						lefts[i][j]=l+1;
						heights[i][j]=1;
					}
				}
			}
			int r=m+1;
			for(int j=m;j>=1;j--){
				if(j==m){
					heights[i][j]=1;
					rights[i][j]=m;
				}
				else{
					if(map[i][j]==map[i][j+1]){
						heights[i][j]=1;
						rights[i][j]=j;
						r=j+1;
					}
					else{
						rights[i][j]=r-1;
						heights[i][j]=1;
					}
				}
				ret=max(ret,heights[i][j]*2+2*(rights[i][j]-lefts[i][j]+1));
			}
		}
		else{
			int l=0;
			for(int j=1;j<=m;j++){
				if(j==1){
					if(map[i][j]!=map[i-1][j]){
						heights[i][j]=heights[i-1][j]+1;
						lefts[i][j]=1;
					}
					else{
						heights[i][j]=1;
						lefts[i][j]=1;
					}
				}
				else{
					if(map[i][j]!=map[i-1][j]){
						heights[i][j]=heights[i-1][j]+1;
						if(map[i][j]==map[i][j-1]){
							lefts[i][j]=j;
							l=j-1;
						}
						else{
							lefts[i][j]=max(lefts[i-1][j],l+1);
						}
					}
					else{
						heights[i][j]=1;
						if(map[i][j]==map[i][j-1]){
							lefts[i][j]=j;
							l=j-1;
						}
						else{
							lefts[i][j]=l+1;
						}
					}
				}
			}
			int r=m+1;
			for(int j=m;j>=1;j--){
				if(j==m){
					if(map[i][j]!=map[i-1][j]){
						heights[i][j]=heights[i-1][j]+1;
						rights[i][j]=m;
					}
					else{
						heights[i][j]=1;
						rights[i][j]=m;
					}
				}
				else{
					if(map[i][j]!=map[i-1][j]){
						heights[i][j]=heights[i-1][j]+1;
						if(map[i][j]==map[i][j+1]){
							rights[i][j]=j;
							r=j+1;
						}
						else{
							rights[i][j]=min(rights[i-1][j],r-1);
						}
					}
					else{
						heights[i][j]=1;
						if(map[i][j]==map[i][j+1]){
							rights[i][j]=j;
							r=j+1;
						}
						else{
							rights[i][j]=r-1;
						}
					}
				}
				ret=max(ret,heights[i][j]*2+2*(rights[i][j]-lefts[i][j]+1));
			}
		}
	}
	return ret;
}

int main(){
	int ansR,ansB,ansC,kase=0;
	int T,n,m;
	scanf("%d",&T);
	while(T--){
		scanf("%d%d",&n,&m);
		getchar();
		for(int i=1;i<=n;i++)
			gets(map[i]+1);
		ansR=CountSingle(n,m,'R');
		ansB=CountSingle(n,m,'B');
		ansC=CountCross(n,m);
/*		for(int i=1;i<=3;i++){
		for(int j=1;j<=3;j++)
		cout<<heights[i][j]<<' ';
		cout<<endl;
		}
		cout<<endl;
		for(int i=1;i<=3;i++){
		for(int j=1;j<=3;j++)
		cout<<lefts[i][j]<<' ';
		cout<<endl;
		}
		cout<<endl;
		for(int i=1;i<=3;i++){
		for(int j=1;j<=3;j++)
		cout<<rights[i][j]<<' ';
		cout<<endl;
		}
		cout<<ansC<<endl;*/
		cout<<"Case #"<<++kase<<": "<<max(ansR,max(ansB,ansC))<<endl;
	}
	return 0;
}

  

原文地址:https://www.cnblogs.com/jie-dcai/p/4093434.html