Solitaire(双向BFS)
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.
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 5
2 4 3 3 3 6 4 6
Sample Output
YES
题意：给你四个点的起始坐标，还有四个点的终点坐标；
在每一步中你可以让一个点往上下左右移动一格，如果移动一格后撞到其他的点上，你可以跳过这一格，既移动两格。
问是否能在最多八步完成。
//思路：我一开始想到是用一个九维数组来存它的转态，不过一直在爆内存； //后来看了大佬们的博客后发现要用hash值来存，把四个点的坐标转换为一个8位数的数字就可以了； //注意，转换数字时一定要先排序，否则会重复； //再因为一些细节WA了几发后，终于A了 #define _CRT_SECURE_NO_WARNINGS #include<iostream> #include<cstring> #include<stdio.h> #include<algorithm> #include<map> #include<queue> #include<set> #include <sstream> #include<vector> #include<cmath> #include<stack> #include<time.h> #include<ctime> using namespace std; #define inf 1<<30 #define eps 1e-7 #define LD long double #define LL long long #define maxn 100000005 int step = 0; struct node { int x[5] = {}; int y[5] = {}; }start, last, t, v; set<int>vis[2]; int dir[5][3] = { {1,0},{-1,0},{0,1},{0,-1} }; queue<node>q[2]; int flag = 0;//旗帜 int number(node p)//转换为hash值 { int sum = 0; for (int i = 1; i <= 4; i++) { sum = sum * 100 + p.x[i] * 10 + p.y[i]; } return sum; } void paixu(node& t)//对四个点排序 { int x; for (int i = 1; i <= 4; i++) { x = i; for (int j = i + 1; j <= 4; j++) if (t.x[x] > t.x[j]) x = j; else if (t.x[x] == t.x[j] && t.y[x] > t.y[j]) x = j; swap(t.x[i], t.x[x]); swap(t.y[i], t.y[x]); } } int judge(node& w, int i)//判断一下是否合法，是否重复 { if (w.x[i] >= 1 && w.x[i] <= 8 && w.y[i] >= 1 && w.y[i] <= 8) { for (int j = 1; j <= 4; j++) { if (j == i) { continue; } if (w.x[i] == w.x[j] && w.y[i] == w.y[j]) { return 1; } } return 0; } return 1; } void BFS(int k)//进行一次搜索 { int sum = q[k].size();//获得当前队列中要搜索的个数 while (sum--) { t = q[k].front(); q[k].pop(); for (int i = 1; i <= 4; i++)//四个点嘛 { for (int j = 0; j < 4; j++)//四个方向 { v = t; v.x[i] = t.x[i] + dir[j][0]; v.y[i] = t.y[i] + dir[j][1]; if (judge(v, i) == 1)//撞到了格子，再走一步 { v.x[i] = v.x[i] + dir[j][0]; v.y[i] = v.y[i] + dir[j][1]; if (judge(v, i) == 1) { continue; } } paixu(v); int num; num = number(v); if (vis[k].count(num)==1)continue; if (vis[1 - k].count(num)==1) { flag = 1; return; } vis[k].insert(num); q[k].push(v); } } } } void BBFS() { int k; paixu(start); paixu(last); k = number(start); vis[0].insert(k); k = number(last); vis[1].insert(k); q[0].push(start); q[1].push(last); while (1) { if (q[0].size() < q[1].size()) { BFS(0); } else { BFS(1); } step++;//记录步数 if (step == 8)break; if (flag == 1)break; } } int main() { while (scanf("%d%d", &start.x[1], &start.y[1]) != EOF) { vis[0].clear(); vis[1].clear(); flag = 0, step = 0;//初始化啊，WA了两发。。。。 while (!q[0].empty()) { q[0].pop(); } while (!q[1].empty()) { q[1].pop(); } for (int i = 2; i <= 4; i++) { scanf("%d%d", &start.x[i], &start.y[i]); } for (int i = 1; i <= 4; i++) { scanf("%d%d", &last.x[i], &last.y[i]); } BBFS(); if (flag) printf("YES\n"); else printf("NO\n"); } return 0; }
