POJ 3050

来源：互联网发布：张宇唱功知乎编辑：程序博客网时间：2024/04/29 03:40

题意：从一个起点，走5步，能组成几个六位数
题解:
用set直接排除重复项最easy，爆搜每一个起点

dfs(int x,int y,int step,int s)if(step==5){   if(ansset.find(s)==ansset.end()){         ansset.insert(s);    }    return;}

实际上并不需要判断，因为set不会装重复元素

#include<iostream>#include<stdio.h>#include<stdlib.h>#include<string.h>#include<queue>#define pr(x) cout<<#x<<" "<<x;#define pl(x) cout<<#x<<" "<<x<<endl;#include<math.h>#include<algorithm>#include<set>#include<map>#include<bitset>#include<string>#include<cstring>#include<sstream>using namespace std;int maze[5][5];set<int> ansset;int dx[] = {-1,1,0,0};int dy[] = {0,0,-1,1};bool isin(int x,int y){    if(x>=0&&x<5&&y>=0&&y<5){return true;}    else{return false;}}void dfs(int x,int y,int step,int s){        if(step==5){            if(ansset.find(s)==ansset.end()){                ansset.insert(s);            }            return;        }        for(int i =0;i<4;i++){            int nx = x + dx[i];            int ny = y + dy[i];            if(isin(nx,ny)){            dfs(nx,ny,step+1,s*10+ maze[nx][ny]);            }        }}int main(){    for(int i = 0;i<5;i++){        for(int j = 0;j<5;j++){            cin>>maze[i][j];        }    }    for(int i = 0;i<5;i++){        for(int j = 0;j<5;j++){            dfs(i,j,0,maze[i][j]);        }    }    cout<<ansset.size()<<endl;}

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