HDOJ 题目1401 Solitaire（双向BFS）

Solitaire

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 3850    Accepted Submission(s): 1165

Problem Description

Solitaire is a game played on a chessboard 8x8. The rows and columns of the chessboard are numbered from 1 to 8, from the top to the bottom and from left to right respectively.

There are four identical pieces on the board. In one move it is allowed to:

> move a piece to an empty neighboring field (up, down, left or right),

> jump over one neighboring piece to an empty field (up, down, left or right).



There are 4 moves allowed for each piece in the configuration shown above. As an example let's consider a piece placed in the row 4, column 4. It can be moved one row up, two rows down, one column left or two columns right.

Write a program that:

> reads two chessboard configurations from the standard input,

> verifies whether the second one is reachable from the first one in at most 8 moves,

> writes the result to the standard output.

 

Input

Each of two input lines contains 8 integers a1, a2, ..., a8 separated by single spaces and describes one configuration of pieces on the chessboard. Integers a2j-1 and a2j (1 <= j <= 4) describe the position of one piece - the row number and the column number respectively. Process to the end of file.

 

Output

The output should contain one word for each test case - YES if a configuration described in the second input line is reachable from the configuration described in the first input line in at most 8 moves, or one word NO otherwise.

 

Sample Input

4 4 4 5 5 4 6 52 4 3 3 3 6 4 6

 

Sample Output

YES

 

Source

Southwestern Europe 2002

 

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 题目大意：就是给了初始4个棋子的坐标，又给了目标4个旗子的坐标，可以通过走上下左右4个方向，或隔着自己的一个子跳一步，问能不能再8步之内完成

ac代码

#include<stdio.h>#include<string.h>#include<stdlib.h>#include<queue>#include<algorithm>#include<iostream>using namespace std;char Hash[8][8][8][8][8][8][8][8];struct p{int x,y;};struct s{p node[4];int step;}st,ed,a,temp;int cmp(p a,p b){if(a.x==b.x)return a.y<b.y;return a.x<b.x;}void biaoji(struct s a,int w){Hash[a.node[0].x][a.node[0].y][a.node[1].x][a.node[1].y][a.node[2].x][a.node[2].y][a.node[3].x][a.node[3].y]=w;}char getnum(struct s a){return Hash[a.node[0].x][a.node[0].y][a.node[1].x][a.node[1].y][a.node[2].x][a.node[2].y][a.node[3].x][a.node[3].y];}int dirx[4]={0,-1,0,1};int diry[4]={1,0,-1,0};int jud(s &a,int i,int j,int m){if(m==1){if(a.step>=4)return 0;a.step++;}a.node[i].x+=dirx[j];a.node[i].y+=diry[j];if(a.node[i].x<0||a.node[i].x>=8||a.node[i].y<0||a.node[i].y>=8)return 0;int k;for(k=0;k<4;k++){if(i!=k){if(a.node[i].x==a.node[k].x&&a.node[i].y==a.node[k].y){if(m==1){return jud(a,i,j,2);}elsereturn 0;}}}if(k>=4){sort(a.node,a.node+4,cmp);return 1;}}int bfs(){queue<struct s>q1;queue<struct s>q2;memset(Hash,0,sizeof(Hash));sort(st.node,st.node+4,cmp);sort(ed.node,ed.node+4,cmp);st.step=0;ed.step=0;q1.push(st);biaoji(st,1);q2.push(ed);biaoji(ed,2);int tmp1,tmp2;tmp1=tmp2=0;while(!q1.empty()||!q2.empty()){if(!q1.empty()){int i,j;for(i=0;i<4;i++)for(j=0;j<4;j++){a=q1.front();if(jud(a,i,j,1)){char c=getnum(a);if(c==2)return 1;if(c==0){biaoji(a,1);q1.push(a);}}}q1.pop();}//tmp1=q1.front().step;if(!q2.empty()){int i,j;for(i=0;i<4;i++)for(j=0;j<4;j++){a=q2.front();if(jud(a,i,j,1)){char c=getnum(a);if(c==1)return 1;if(c==0){biaoji(a,2);q2.push(a);}}}q2.pop();}}return 0;}int main(){while(scanf("%d%d",&st.node[0].x,&st.node[0].y)!=EOF){st.node[0].x--;st.node[0].y--;int i;for(i=1;i<4;i++){scanf("%d%d",&st.node[i].x,&st.node[i].y);st.node[i].x--;st.node[i].y--;}for(i=0;i<4;i++){scanf("%d%d",&ed.node[i].x,&ed.node[i].y);ed.node[i].x--;ed.node[i].y--;}if(bfs())printf("YES\n");elseprintf("NO\n");}}