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最新推荐文章于 2022-03-09 19:47:14 发布

原创最新推荐文章于 2022-03-09 19:47:14 发布 · 493 阅读

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http://poj.org/problem?id=1154

Description

A single-player game is played on a rectangular board divided in R rows and C columns. There is a single uppercase letter (A-Z) written in every position in the board.
Before the begging of the game there is a figure in the upper-left corner of the board (first row, first column). In every move, a player can move the figure to the one of the adjacent positions (up, down,left or right). Only constraint is that a figure cannot visit a position marked with the same letter twice.
The goal of the game is to play as many moves as possible.
Write a program that will calculate the maximal number of positions in the board the figure can visit in a single game.

Input

The first line of the input contains two integers R and C, separated by a single blank character, 1 <= R, S <= 20.
The following R lines contain S characters each. Each line represents one row in the board.

Output

The first and only line of the output should contain the maximal number of position in the board the figure can visit.

Sample Input

3 6
HFDFFB
AJHGDH
DGAGEH

Sample Output

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
这道题不是多组输入，只是单组输入，如果写成多组输入后就会出现Output Limit Exceeded，这应该算是一篇比较简单dfs，代码中的一大亮点是map标记数组的应用，很精奥。
#include<stdio.h>
#include<string.h>
#define NN 30
#define MAX(a,b) (a)>(b)?(a):(b)
int m,n,ans;
char huiyao[NN][NN];
int ope[4][2]={{0,1},{1,0},{0,-1},{-1,0}};
int map[27];
void dfs(int x,int y,int step)
{
int xl,yl,l;
ans=MAX(ans,step);
map[huiyao[x][y]-'A']=0;
for(l=0;l<4;l++)
{
xl=x+ope[l][0];
yl=y+ope[l][1];
if(xl<0||yl<0||xl>=m||yl>=n)
continue;
if(map[huiyao[xl][yl]-'A']==0)
continue;
map[huiyao[xl][yl]-'A']=0;
dfs(xl,yl,step+1);
map[huiyao[xl][yl]-'A']=1;
}

}
int main()
{
int i,j,k;
scanf("%d%d",&m,&n);

getchar();
memset(map,0,sizeof(map));
for(i=0;i<m;i++)
scanf("%s",huiyao[i]);
for(i=0;i<m;i++)
{
for(j=0;j<n;j++)
{
map[huiyao[i][j]-'A']=1;
}
}
ans=1;
dfs(0,0,1);
printf("%d\n",ans);
/*for(i=0;i<m;i++)
{
for(j=0;j<n;j++)
{
printf("%c",huiyao[i][j]);
}
printf("\n");
}*/
return 0;
}