題目鏈接http://bailian.openjudge.cn/practice/1154/總時間限制:?1000ms?內存限制:?65536kB描述

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.

輸入

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.

輸出

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

樣例輸入

3 6 HFDFFB AJHGDH DGAGEH

樣例輸出

6

來源

Croatia OI 2002 Regional Competition - Juniors

算法：深搜

代碼一：

1 #include<iostream> 2 using namespace std; 3 int bb[26]={0},s,r,sum=1,s1=1; 4 char aa[25][25]; 5 int dir[4][2]={-1,0,1,0,0,-1,0,1}; 6 void dfs(int a,int b) 7 { 8 int a1,b1; 9 if(s1>sum) sum=s1; //更新最大數值 10 for(int i=0;i<4;i++) 11 { 12 a1=a+dir[i][0]; //用bb數組記錄訪問過的字母 13 b1=b+dir[i][1]; 14 if(a1>=0&&a1<s&&b1>=0&&b1<r&&!bb[aa[a1][b1]-'A']) 15 { 16 s1++; 17 bb[aa[a1][b1]-'A']=1; //如果在這條單線上沒有記錄改字母被訪問過，則總數++； 18 dfs(a1,b1); //第一個字母總要被訪問，所以不用回溯； 19 bb[aa[a1][b1]-'A']=0; //回溯反標記 20 s1--; //臨時記錄恢復 21 } 22 } 23 } 24 int main() 25 { 26 cin>>s>>r; 27 for(int i=0;i<s;i++) 28 for(int j=0;j<r;j++) 29 cin>>aa[i][j]; 30 bb[aa[0][0]-'A']=1; 31 dfs(0,0); 32 cout<<sum<<endl; 33 return 0; 34 } View Code

代碼二：

1 #include <stdio.h> 2 #include<iostream> 3 using namespace std; 4 int qq[25][25]; 5 int fx[4]={1,0,-1,0},fy[4]={0,-1,0,1},pd[30],sum,ans;//右下左上 6 void fun(int x,int y) 7 { 8 if(ans<sum)ans=sum; 9 if(qq[x][y]==0) return; 10 for(int i=0;i<4;i++) 11 { 12 if(qq[x+fx[i]][y+fy[i]]!=0&&pd[qq[x+fx[i]][y+fy[i]]]==0) 13 { 14 sum++; 15 pd[qq[x+fx[i]][y+fy[i]]]=1; 16 fun(x+fx[i],y+fy[i]); 17 pd[qq[x+fx[i]][y+fy[i]]]=0; 18 sum--; 19 } 20 } 21 } 22 int main(int argc, char *argv[]) 23 { 24 int r,s; 25 scanf("%d%d",&r,&s); 26 for(int i=1;i<=r;i++) 27 for(int j=1;j<=s;j++) 28 { 29 char t; 30 cin>>t; 31 qq[i][j]=t-'A'+1; 32 } 33 pd[qq[1][1]]=1; 34 sum=ans=1; 35 fun(1,1); 36 printf("%d",ans); 37 return 0; 38 } View Code

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轉載于:https://www.cnblogs.com/huashanqingzhu/p/10630870.html

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