POJ 1198 Solitaire(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.

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.

Output

The output should contain one word 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

题意是四个棋子，给出起始位置和终止位置（无序），问是否在8步之内满足要求。

我用的是单向bfs

#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<cmath>
#include<iostream>
#include<algorithm>
#include<queue>

using namespace std;

#define MAXN 9

int fx[4][2] = {{1, 0}, {0, 1}, {-1, 0}, {0, -1}};
bool vis[MAXN][MAXN][MAXN][MAXN][MAXN][MAXN][MAXN][MAXN];
bool map[MAXN][MAXN];

struct node{
int x[5], y[5];
int dep;
};

node s, t;

bool check(node p){
for(int i = 1; i <= 4; i++)
if(!map[p.x[i]][p.y[i]]) return false;
return true;
}

bool ok(int X, int Y){
if(X <= 0 || Y <= 0 || X > 8 || Y > 8) return false;
return true;
}

bool ok_c(node t){
bool &tmp = vis[t.x[1]][t.x[2]][t.x[3]][t.x[4]][t.y[1]][t.y[2]][t.y[3]][t.y[4]];
if(tmp) return false;
return tmp = true;
}

bool empty(int X, int Y, node now){
for(int i = 1; i <= 4; i++)
if(X == now.x[i] && Y == now.y[i]) return false;
return true;
}

void bfs(){
queue<node> q;
q.push(s);
ok_c(s);

while(!q.empty()){
node now = q.front(); q.pop();

if(now.dep >= 8){
printf("NO\n");
return;
}

for(int i = 1; i <= 4; i++){
int X = now.x[i], Y = now.y[i];
for(int j = 0; j < 4; j++){
node next = now;
next.dep = now.dep+1;

int nx = X + fx[j][0], ny = Y + fx[j][1];

if(!ok(nx, ny)) continue;
next.x[i] = nx; next.y[i] = ny;

if(empty(nx, ny, now)){
next.x[i] = nx; next.y[i] = ny;
if(!ok_c(next)) continue;

if(check(next)){
printf("YES\n");
return;
}
q.push(next);
}else{
nx += fx[j][0]; ny += fx[j][1];
if(!ok(nx, ny)) continue;
next.x[i] = nx; next.y[i] = ny;
if(!ok_c(next) || !empty(nx, ny, now)) continue;
if(check(next)){
printf("YES\n");
return;
}

q.push(next);
}
}
}
}

printf("NO\n");
return;
}

int main(){
freopen("test.in", "r", stdin);
freopen("test.out", "w", stdout);

for(int i = 1; i <= 4; i++) scanf("%d%d", &s.x[i], &s.y[i]);
for(int i = 1; i <= 4; i++){
scanf("%d%d", &t.x[i], &t.y[i]);
map[t.x[i]][t.y[i]] = true;
}
s.dep = 0;

bfs();

return 0;
}

^-^come on!