BZOJ 1085 SCOI 2005 骑士精神 IDA*

标签：bzoj   ida   搜索   剪枝   scoi2005   

题目大意：有一张5*5的棋盘，上面有12和黑棋还有12个白棋。问最少多步可以到达目标状态。

思路：搜索+剪枝。至于剪枝我就用ID+A*的组合了，因为都不难想，估价函数就是当前图和目标图有多少个方块不一样。如果当前步数+估价大于当前迭代加深的层数就退出。

CODE：

#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
using namespace std;
const int dx[] = {0,1,1,2,2,-1,-1,-2,-2};
const int dy[] = {0,2,-2,1,-1,-2,2,-1,1};

int cases;
char src[10][10],aim[10][10];

void Pretreatment()
{
	for(int i = 1; i <= 5; ++i)
		aim[1][i] = '1',aim[5][i] = '0';
	for(int i = 2; i <= 5; ++i)
		aim[2][i] = '1';
	aim[3][4] = aim[3][5] = aim[4][5] = '1';
	for(int i = 1; i <= 4; ++i)
		aim[4][i] = '0';
	aim[3][1] = aim[3][2] = aim[2][1] = '0'; 
	aim[3][3] = '*';
}

bool IDA_(int deep,int max_deep)
{
	int difference = 0;
	int x,y;
	for(int i = 1; i <= 5; ++i)
		for(int j = 1; j <= 5; ++j) {
			if(src[i][j] == '*')
				x = i,y = j;
			difference += (src[i][j] != aim[i][j]);
		}
	if(!difference)	return true;
	if(difference - 1 + deep > max_deep)	return false;
	for(int i = 1; i <= 8; ++i) {
		int fx = x + dx[i];
		int fy = y + dy[i];
		if(fx <= 0 || fy <= 0 || fx > 5 || fy > 5)	continue;
		swap(src[x][y],src[fx][fy]);
		if(IDA_(deep + 1,max_deep))
			return true;
		swap(src[x][y],src[fx][fy]);
	}
	return false;
}

int main()
{
	Pretreatment();
	for(cin >> cases; cases; --cases) {
		for(int i = 1; i <= 5; ++i)
			scanf("%s",src[i] + 1);
		bool flag = false;
		for(int i = 0; i <= 15 && !flag; ++i) {
			if(IDA_(0,i)) {
				flag = true;
				printf("%d\n",i);
			}
		}
		if(!flag)	puts("-1");
	}
	return 0;
}

BZOJ 1085 SCOI 2005 骑士精神 IDA*

标签：bzoj   ida   搜索   剪枝   scoi2005   

原文地址：http://blog.csdn.net/jiangyuze831/article/details/40587487