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  • 搜索专题： HDU1372Knight Moves

    Knight Moves

    Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
    Total Submission(s): 11832    Accepted Submission(s): 6969


    Problem Description
    A friend of you is doing research on the Traveling Knight Problem (TKP) where you are to find the shortest closed tour of knight moves that visits each square of a given set of n squares on a chessboard exactly once. He thinks that the most difficult part of the problem is determining the smallest number of knight moves between two given squares and that, once you have accomplished this, finding the tour would be easy.
    Of course you know that it is vice versa. So you offer him to write a program that solves the "difficult" part.

    Your job is to write a program that takes two squares a and b as input and then determines the number of knight moves on a shortest route from a to b.
     

    Input
    The input file will contain one or more test cases. Each test case consists of one line containing two squares separated by one space. A square is a string consisting of a letter (a-h) representing the column and a digit (1-8) representing the row on the chessboard.
     

    Output
    For each test case, print one line saying "To get from xx to yy takes n knight moves.".
     

    Sample Input
    e2 e4
a1 b2
b2 c3
a1 h8
a1 h7
h8 a1
b1 c3
f6 f6
     

    Sample Output
    To get from e2 to e4 takes 2 knight moves.
To get from a1 to b2 takes 4 knight moves.
To get from b2 to c3 takes 2 knight moves.
To get from a1 to h8 takes 6 knight moves.
To get from a1 to h7 takes 5 knight moves.
To get from h8 to a1 takes 6 knight moves.
To get from b1 to c3 takes 1 knight moves.
To get from f6 to f6 takes 0 knight moves.
     

    Source
    University of Ulm Local Contest 1996 
    Problem : 1372 ( Knight Moves )     Judge Status : Accepted
    RunId : 21142000    Language : G++    Author : hnustwanghe
    Code Render Status : Rendered By HDOJ G++ Code Render Version 0.01 Beta
    #include<iostream>
#include<cstdio>
#include<queue>
#include<cstring>

using namespace std;

const int N = 8+5;
int visit[N][N];
typedef struct node{
    int x,y,step;
}Node;
const int dir[8][2]={{2,1},{-2,1},{-2,-1},{2,-1},{1,2},{1,-2},{-1,-2},{-1,2}};
int print(){
    for(int i=1;i<=8;i++)
    for(int j=1;j<=8;j++){
        printf("%d",visit[i][j]);
        if(j==7)
            printf("
");
    }
}
int DBFS(int x,int y,int goalx,int goaly){
    queue<Node> Q1,Q2;
    Node t,s;
    int step1=0,step2=0,flag =1,cnt,newx,newy,ans;
    memset(visit,0,sizeof(visit));
    t.x = x,t.y = y,t.step = 0;
    s.x = goalx,s.y = goaly,s.step = 0;
    visit[t.x][t.y] = 1;
    visit[s.x][s.y] = 2;
    Q1.push(t);
    Q2.push(s);
    while(flag){
        cnt = Q1.size();
        while(cnt--){
                t = Q1.front();
                if(t.x == goalx && t.y == goaly) return t.step;
            for(int d=0;d<8;d++){
                newx = t.x + dir[d][0];
                newy = t.y + dir[d][1];
                step1 = t.step + 1;
                if(newx>0 && newx <= 8 && newy>0 && newy<=8){
                    if(visit[newx][newy]==2){
                        ans = step1 + step2;
                        return ans;
                    }
                    if(!visit[newx][newy]){
                        s.x = newx,s.y= newy,s.step = step1;
                        Q1.push(s);
                        visit[newx][newy] = 1;
                    }
                }

            }
            Q1.pop();
        }
        cnt = Q2.size();
        while(cnt--){
            t = Q2.front();
            if(t.x==x && t.y==y) return t.step;
            for(int d=0;d<8;d++){
                newx = t.x + dir[d][0];
                newy = t.y + dir[d][1];
                step2 = t.step + 1;
                if(newx>0 && newx<=8 && newy>0 && newy <=8){
                    if(visit[newx][newy]==1){
                        ans = step1 + step2;
                        return ans;
                    }
                    if(!visit[newx][newy]){
                        s.x = newx,s.y = newy,s.step = step2;
                        Q2.push(s);
                        visit[newx][newy] = 2;
                    }
                }
            }
            Q2.pop();
        }
    }

    return -1;
}
void Input_and_solve(){
    char ch[10];
    int x,y,goalx,goaly;
    while(gets(ch)!=NULL){
        x = ch[0]-'a'+1;
        y = ch[1]-'0';
        goalx = ch[3]-'a'+1;
        goaly = ch[4]-'0';
        printf("To get from %c%c to %c%c takes %d knight moves.
",ch[0],ch[1],ch[3],ch[4],DBFS(x,y,goalx,goaly));
        //print();
    }
}
int main(){
    Input_and_solve();

    }
    Problem : 1372 ( Knight Moves )     Judge Status : Accepted
RunId : 21135178    Language : G++    Author : hnustwanghe
Code Render Status : Rendered By HDOJ G++ Code Render Version 0.01 Beta
    #include<iostream>
#include<cstdio>
#include<queue>
#include<cstring>
using namespace std;

typedef struct node{
    int x,y,step;
}Node;
bool visit[10][10];
const int dir[8][2]={{-2,1},{-2,-1},{2,-1},{2,1},{1,2},{1,-2},{-1,2},{-1,-2}};
int BFS(int x,int y,int goalx,int goaly){
    queue<Node>Q;
    memset(visit,0,sizeof(visit));
    int newx,newy;
    Node t,s;
    t.x = x,t.y = y,t.step = 0;
    visit[t.x][t.y] = true;
    Q.push(t);

    while(!Q.empty()){
        t = Q.front();
        if(t.x==goalx && t.y==goaly) return t.step;
        for(int d=0;d<8;d++){
            newx = t.x+dir[d][0];
            newy = t.y+dir[d][1];
            if(newx>=1 && newx<=8 && newy>=1 && newy<=8 && !visit[newx][newy]){
                s.x = newx;
                s.y = newy;
                s.step = t.step+1;
                Q.push(s);
            }
        }
        Q.pop();
    }
    return -1;
}
void Input_data_and_solve(){
    char a,c,ch[10];
    int b,d,x,y,goalx,goaly;
    while(gets(ch)!=NULL){
    a = ch[0];
    b = ch[1]-'0';
    c = ch[3];
    d = ch[4]-'0';
    x = a-'a'+1;
    y = b;
    goalx = c-'a'+1;
    goaly = d;
    BFS(x,y,goalx,goaly);
    printf("To get from %c%d to %c%d takes %d knight moves.
",a,b,c,d,BFS(x,y,goalx,goaly));
    }
}
int main(){
    Input_data_and_solve();
}


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  • 原文地址：https://www.cnblogs.com/Pretty9/p/7347698.html
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