hdu 1401 Solitaire (char数组判重节约内存+伪dbfs)
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Solitaire
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 2611 Accepted Submission(s): 850
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.
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
感想:
哎。。。!这题又坑了我一下午。后来发现是一个小错误导致WA了--输入的数据应该减1后再处理
题目还是蛮卡内存的,dbfs第一次被逼到只开一个char数组判重。
题意:
在一个8×8的棋盘中,给定你4个棋子A,再给你4个棋子B,问你在8步之内能不能够从A位置移动到B位置;
规则:棋子能能上下左右移动,同时能跳过相邻的棋子到达相邻棋子的空地方。
规则:棋子能能上下左右移动,同时能跳过相邻的棋子到达相邻棋子的空地方。
思路:
还是蛮明显的,8维数组判重,直接模拟棋子的移动就够了。状态有点多,最好用dbfs,因为只要判断YES、NO,我用了一个伪dbfs--先正着搜4步,然后反着搜,如果已经标记了的话就输出YES,否则就NO。
陷阱:
4个棋子是没有区别的,所以最好将棋子排序,避免状态是一样的,但是在程序中状态却不一样。
代码:
#include <iostream>#include <cstdio>#include <cstring>#include <algorithm>#include <queue>using namespace std;int n,m,ans;char vis[8][8][8][8][8][8][8][8];int dx[]= {-1,1,0,0};int dy[]= {0,0,-1,1};bool mp[8][8]; // 用来判断每一时刻棋盘上某一位置有无棋子struct Node{ int x,y;} p1[4],p2[4];struct node{ int step; Node pp[4];} cur,now;queue<node>q1,q2;bool cmp(const Node&xx1,const Node&xx2) // 从小到大排序{ if(xx1.x!=xx2.x) return xx1.x<xx2.x; return xx1.y<xx2.y;}bool bfs(){ int i,j,nst,tst,nstep,nx,ny,tx,ty; memset(vis,0,sizeof(vis)); memset(mp,0,sizeof(mp)); while(!q1.empty()) q1.pop(); sort(p1,p1+4,cmp); for(i=0; i<4; i++) { cur.pp[i]=p1[i]; } cur.step=0; vis[p1[0].x][p1[0].y][p1[1].x][p1[1].y][p1[2].x][p1[2].y][p1[3].x][p1[3].y]='1'; q1.push(cur); while(!q1.empty()) // 初始状态向前走4步 走过的标记为‘1’ { now=q1.front(); q1.pop(); nstep=now.step; if(nstep>=4) break ; for(i=0; i<4; i++) // 放棋子 { mp[now.pp[i].x][now.pp[i].y]=1; } for(i=0; i<4; i++) // 第几个piece { nx=now.pp[i].x; ny=now.pp[i].y; for(j=0; j<4; j++) // u d l r { cur=now; // 这行代码需加在这里 而不是上面 cur.step=nstep+1; tx=nx+dx[j]; ty=ny+dy[j]; if(tx<0||tx>=8||ty<0||ty>=8) continue ; if(mp[tx][ty]) // 如果前面是棋子 则跳过棋子 { tx+=dx[j]; ty+=dy[j]; if(tx<0||tx>=8||ty<0||ty>=8||mp[tx][ty]) continue ; } cur.pp[i].x=tx; cur.pp[i].y=ty; sort(cur.pp,cur.pp+4,cmp); if(!vis[cur.pp[0].x][cur.pp[0].y][cur.pp[1].x][cur.pp[1].y][cur.pp[2].x][cur.pp[2].y][cur.pp[3].x][cur.pp[3].y]) { vis[cur.pp[0].x][cur.pp[0].y][cur.pp[1].x][cur.pp[1].y][cur.pp[2].x][cur.pp[2].y][cur.pp[3].x][cur.pp[3].y]='1'; q1.push(cur); } } } for(i=0; i<4; i++) // 拿走棋子 { mp[now.pp[i].x][now.pp[i].y]=0; } } memset(mp,0,sizeof(mp)); while(!q2.empty()) q2.pop(); sort(p2,p2+4,cmp); for(i=0; i<4; i++) { cur.pp[i]=p2[i]; } cur.step=0; if(vis[p2[0].x][p2[0].y][p2[1].x][p2[1].y][p2[2].x][p2[2].y][p2[3].x][p2[3].y]) return true; else vis[p2[0].x][p2[0].y][p2[1].x][p2[1].y][p2[2].x][p2[2].y][p2[3].x][p2[3].y]='2'; q2.push(cur); while(!q2.empty()) // 最终状态向后走4步 { now=q2.front(); q2.pop(); nstep=now.step; if(nstep>=4) break ; for(i=0; i<4; i++) { mp[now.pp[i].x][now.pp[i].y]=1; } for(i=0; i<4; i++) // 第几个piece { nx=now.pp[i].x; ny=now.pp[i].y; for(j=0; j<4; j++) // u d l r { cur=now; cur.step=nstep+1; tx=nx+dx[j]; ty=ny+dy[j]; if(tx<0||tx>=8||ty<0||ty>=8) continue ; if(mp[tx][ty]) { tx+=dx[j]; ty+=dy[j]; if(tx<0||tx>=8||ty<0||ty>=8||mp[tx][ty]) continue ; } cur.pp[i].x=tx; cur.pp[i].y=ty; sort(cur.pp,cur.pp+4,cmp); if(!vis[cur.pp[0].x][cur.pp[0].y][cur.pp[1].x][cur.pp[1].y][cur.pp[2].x][cur.pp[2].y][cur.pp[3].x][cur.pp[3].y]) { // 没有标记过就标记为‘2’ vis[cur.pp[0].x][cur.pp[0].y][cur.pp[1].x][cur.pp[1].y][cur.pp[2].x][cur.pp[2].y][cur.pp[3].x][cur.pp[3].y]='2'; q2.push(cur); } // 已经标记为‘1’ 说明在能走到 else if(vis[cur.pp[0].x][cur.pp[0].y][cur.pp[1].x][cur.pp[1].y][cur.pp[2].x][cur.pp[2].y][cur.pp[3].x][cur.pp[3].y]=='1') return true ; } } for(i=0; i<4; i++) { mp[now.pp[i].x][now.pp[i].y]=0; } } return false ;}int main(){ int i,j,xx,yy; while(~scanf("%d%d",&xx,&yy)) { p1[0].x=xx-1; p1[0].y=yy-1; for(i=1; i<4; i++) { scanf("%d%d",&xx,&yy); p1[i].x=xx-1; p1[i].y=yy-1; } for(i=0; i<4; i++) { scanf("%d%d",&xx,&yy); p2[i].x=xx-1; p2[i].y=yy-1; } if(bfs()) printf("YES\n"); else printf("NO\n"); } return 0;}
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