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https://www.acmicpc.net/problem/13460
13460번: 구슬 탈출 2
첫 번째 줄에는 보드의 세로, 가로 크기를 의미하는 두 정수 N, M (3 ≤ N, M ≤ 10)이 주어진다. 다음 N개의 줄에 보드의 모양을 나타내는 길이 M의 문자열이 주어진다. 이 문자열은 '.', '#', 'O', 'R', 'B'
www.acmicpc.net
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#include <iostream>
#include <cstdio>
#include <queue>
#include <algorithm>
struct INFO {
int ry, rx, by, bx, count;
};
INFO start;
char map[10][11];
using namespace std;
int bfs(){
const int dy[] = {-1, 1, 0, 0};
const int dx[] = {0, 0, -1, 1};\
int visited[10][10][10][10] = {0, };
queue <INFO> q;
q.push(start);
visited[start.ry][start.rx][start.by][start.bx] = 1;
int ret = -1;
while (!q.empty())
{
INFO cur = q.front();
q.pop();
if(cur.count>10) break;
if(map[cur.ry][cur.rx] == 'O' && map[cur.by][cur.bx] != 'O'){
ret = cur.count;
break;
}
for(int dir = 0; dir < 4; ++dir){
int next_ry = cur.ry;
int next_rx = cur.rx;
int next_by = cur.by;
int next_bx = cur.bx;
while(1){
if(map[next_ry][next_rx] != '#' && map[next_ry][next_rx] !='O'){
next_ry += dy[dir], next_rx += dx[dir];
}
else {
if(map[next_ry][next_rx] == '#' ){
next_ry -= dy[dir], next_rx -= dx[dir];
}
break;
}
}
while(1){
if(map[next_by][next_bx] != '#' && map[next_by][next_bx] !='O'){
next_by += dy[dir], next_bx += dx[dir];
}
else {
if(map[next_by][next_bx] == '#' ){
next_by -= dy[dir], next_bx -= dx[dir];
}
break;
}
}
if(next_ry == next_by && next_rx == next_bx){
if(map[next_ry][next_rx] != 'O'){
int red_dist = abs(next_ry - cur.ry) + abs(next_rx - cur.rx);
int blue_dist = abs(next_by - cur.by) + abs(next_bx - cur.bx);
if(red_dist > blue_dist){
next_ry -= dy[dir], next_rx -= dx[dir];
}
else{
next_by -= dy[dir], next_bx -= dx[dir];
}
}
}
if(visited[next_ry][next_rx][next_by][next_bx] == 0){
visited[next_ry][next_rx][next_by][next_bx] = 1;
INFO next;
next.ry = next_ry;
next.rx = next_rx;
next.by = next_by;
next.bx = next_bx;
next.count = cur.count + 1;
q.push(next);
}
}
}
return ret;
}
int main(){
int n, m;
scanf("%d %d", &n, &m);
for(int i = 0; i < n; ++i){
scanf("%s", map[i]);
}
for(int y = 0; y < n; ++y){
for(int x = 0; x < m; ++x){
if(map[y][x] == 'R'){
start.ry = y, start.rx = x;
}
if(map[y][x] == 'B'){
start.by = y, start.bx = x;
}
}
}
start.count = 0;
int ret = bfs();
cout<<ret<<endl;
return 0;
}
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