HDU1401

1.题目描述：

Solitaire

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 4560    Accepted Submission(s): 1380

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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2.题意概述：

在8 * 8 的跳棋棋盘上，给你两个状态，每个状态4个点，判断一个状态8步内能否到达另一个状态。

3.解题思路：

给定了始末状态，很典型的双向广搜。从始末两个方向进行广度搜索，理想情况下可以减少二分之一的搜索量，从而提高搜索速度。问题的关键在于状态的保存，因为四个棋子是一样的，棋盘的0大小只有8*8，所以先按坐标排序之后，设定八个位置（四个坐标）的哈希，每次枚举当前的4个点和四个方向去判断，判断函数考虑用递归思路去判断那种有一个重棋子子的现象。而且限制条件为8步数，那么每次双向限制就是4步数。还有很多细节化的东西详见代码。

4.AC代码：

#include <bits/stdc++.h>#define INF 0x3f3f3f3f#define maxn 100100#define lson root << 1#define rson root << 1 | 1#define lent (t[root].r - t[root].l + 1)#define lenl (t[lson].r - t[lson].l + 1)#define lenr (t[rson].r - t[rson].l + 1)#define N 1111#define eps 1e-6#define pi acos(-1.0)#define e exp(1.0)using namespace std;const int mod = 1e9 + 7;typedef long long ll;typedef unsigned long long ull;char hs[8][8][8][8][8][8][8][8];int dir[4][2] = { {1, 0}, {0, 1}, {-1, 0}, {0, -1} };struct point{int x, y;friend bool operator< (point a, point b){if (a.x == b.x)return a.y < b.y;return a.x < b.x;}};struct node{point p[4];int step;};void sethash(node a, char k){hs[a.p[0].x][a.p[0].y][a.p[1].x][a.p[1].y][a.p[2].x][a.p[2].y][a.p[3].x][a.p[3].y] = k;}char gethash(node a){return hs[a.p[0].x][a.p[0].y][a.p[1].x][a.p[1].y][a.p[2].x][a.p[2].y][a.p[3].x][a.p[3].y];}bool judge(node &state, int i, int j, int m){if (m == 1){if (state.step >= 4)return 0;state.step++;}state.p[i].x += dir[j][0];state.p[i].y += dir[j][1];if (state.p[i].x >= 0 && state.p[i].x < 8 && state.p[i].y >= 0 && state.p[i].y < 8){for (int k = 0; k < 4; k++)if (i != k){if (state.p[i].x == state.p[k].x && state.p[i].y == state.p[k].y){if (m == 1)return judge(state, i, j, 2);return 0;}}sort(state.p, state.p + 4);return 1;}return 0;}bool bfs2(node a, node b){queue<node> q1, q2;q1.push(a);q2.push(b);node pre;while (!q1.empty() || !q2.empty()){if (!q1.empty()){pre = q1.front();q1.pop();for (int i = 0; i < 4; i++)for (int j = 0; j < 4; j++){node cur = pre;if (judge(cur, i, j, 1)){char k = gethash(cur);if (k == 2)return 1;if (k == 0){sort(cur.p, cur.p + 4);sethash(cur, 1);q1.push(cur);}}}}if (!q2.empty()){pre = q2.front();q2.pop();for (int i = 0; i < 4; i++)for (int j = 0; j < 4; j++){node cur = pre;if (judge(cur, i, j, 1)){char k = gethash(cur);if (k == 1)return 1;if (k == 0){sort(cur.p, cur.p + 4);sethash(cur, 2);q2.push(cur);}}}}}return 0;}int main(){#ifndef ONLINE_JUDGEfreopen("in.txt", "r", stdin);freopen("out.txt", "w", stdout);long _begin_time = clock();#endifint x, y;while (~scanf("%d%d", &x, &y)){memset(hs, 0, sizeof(hs));node a, b;a.p[0].x = --x;a.p[0].y = --y;for (int i = 1; i < 4; i++){scanf("%d%d", &a.p[i].x, &a.p[i].y);a.p[i].x--;a.p[i].y--;}sort(a.p, a.p + 4);sethash(a, 1);for (int i = 0; i < 4; i++){scanf("%d%d", &b.p[i].x, &b.p[i].y);b.p[i].x--;b.p[i].y--;}sort(b.p, b.p + 4);sethash(b, 2);a.step = b.step = 0;if (bfs2(a, b))puts("YES");elseputs("NO");}#ifndef ONLINE_JUDGElong _end_time = clock();printf("time = %ld ms.", _end_time - _begin_time);#endifreturn 0;}