# Bzoj 1085: [SCOI2005]骑士精神

dfs + 剪枝.

1.每次交换只能改变一个位置.若发现之间相差的步数加上以前走的步数大于15的话,直接舍弃这一状态.
2.初始时,$ans$设为$16$

照这节奏,SCOI2005就刷完了???

#include <iostream>
#include <cstdio>
#define X 6
using namespace std;

const int gx[] = {0,-2,-2,-1,-1,1,1,2,2};
const int gy[] = {0,-1,1,-2,2,-2,2,-1,1};

char map[X][X];
char c[X][X] = {
'0','0','0','0','0','0',
'0','1','1','1','1','1',
'0','0','1','1','1','1',
'0','0','0','*','1','1',
'0','0','0','0','0','1',
'0','0','0','0','0','0'
};

int ans;

int inint(){
int num = 0;
for(int i = 1;i <= 5;++ i){
for(int j = 1;j <= 5;++ j){
if(c[i][j] != map[i][j])num ++;
}
}
return num;
}

void dfs(int x,int y,int d,int tmp){
int l = inint();
if(d + l > 16)return;
if(d > ans)return;
if(l == 0) ans = d;

for(int i = 1;i <= 8;++ i){
if(x + gx[i] < 1 || x + gx[i] > 5)continue;
if(y + gy[i] < 1 || y + gy[i] > 5)continue;
if(tmp + i == 9)continue;
swap(map[x][y],map[x + gx[i]][y + gy[i]]);
dfs(x + gx[i],y + gy[i],d + 1,i);
swap(map[x][y],map[x + gx[i]][y + gy[i]]);
}
}

void work(){
int x,y;
for(int i = 1;i <= 5;++ i)cin >> map[i] + 1;
for(int i = 1;i <= 5;++ i){
for(int j = 1;j <= 5;++ j){
if(map[i][j] == '*')
x = i,y = j;
}
}
ans = 16;
dfs(x,y,0,0);
printf("%d\n",ans == 16 ? -1 : ans);
return;
}

int main(){
int t;
scanf("%d",&t);
while(t --){
work();
}
return 0;
}

posted @ 2018-09-29 20:32  Rlif  阅读(89)  评论(0编辑  收藏