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layout : wiki
title : BOJ 16920. 확장 게임
summary :
date : 2019-04-14 23:01:01 +0900
updated : 2021-12-15 17:58:46 +0900
<https://www.acmicpc.net/problem/16920>
[문제 원 출처 링크](https://codeforces.com/contest/1105/problem/D )
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#include <cstdio>
#include <algorithm>
#include <vector>
#include <queue>
#include <stack>
#include <cstring>
#include <cstdlib>
#include <map>
using namespace std;
void printBoard();
void printVisit();
#define MAX_N 1010
#define MAX_P 12
#define EMPTY '.'
#define WALL '#'
int N, M, P;
char board[MAX_N][MAX_N];
int isVisited[MAX_N][MAX_N];
int range[MAX_P];
int dx[4] = { 0, -1, 1, 0 };
int dy[4] = { -1, 0, 0, 1 };
queue<pair<int, int> > q[MAX_P];
int bfs()
{
queue<int> playerQ;
for (int i = 1; i <= P; i++)
playerQ.push(i);
while (!playerQ.empty())
{
int i = playerQ.front();
playerQ.pop();
int rCnt = range[i];
while (!q[i].empty() && rCnt--)
{
int qSize = q[i].size();
for (int qs = 0; qs < qSize; qs++)
{
int r = q[i].front().first;
int c = q[i].front().second;
q[i].pop();
for (int j = 0; j < 4; j++)
{
int nr = r + dx[j];
int nc = c + dy[j];
if (nr < 0 || nr >= N || nc < 0 || nc >= M || isVisited[nr][nc] != 0)
continue;
isVisited[nr][nc] = i;
q[i].push({ nr, nc });
}
}
}
if (q[i].size() > 0)
playerQ.push(i);
}
return 0;
}
int main()
{
// freopen("in.txt", "r", stdin);
int testcase = 1;
// scanf("%d", &testcase);
for (int tc = 1; tc <= testcase; tc++)
{
memset(isVisited, 0, sizeof(isVisited));
scanf("%d%d%d", &N, &M, &P);
for (int i = 1; i <= P; i++)
scanf("%d", &range[i]);
for (int i = 0; i < N; i++)
{
scanf("%s", &board[i]);
for (int j = 0; j < M; j++)
{
if (board[i][j] != EMPTY && board[i][j] != WALL)
{
int idx = board[i][j] - '0';
q[idx].push({ i, j });
isVisited[i][j] = idx;
}
else if (board[i][j] == WALL)
isVisited[i][j] = -1;
}
}
bfs();
int res[MAX_P] = { 0, };
for (int i = 0; i < N; i++)
{
for (int j = 0; j < M; j++)
{
if (isVisited[i][j] <= 0)
continue;
res[isVisited[i][j]]++;
}
}
for (int i = 1; i <= P; i++)
printf("%d ", res[i]);
printf("\n");
}
}
void printBoard()
{
for (int i = 0; i < N; i++)
printf("%s\n", board[i]);
printf("\n");
}
void printVisit()
{
for (int i = 0; i < N; i++)
{
for (int j = 0; j < M; j++)
{
if (board[i][j] == WALL)
printf("%c", board[i][j]);
else
printf("%d", isVisited[i][j]);
}
printf("\n");
}
printf("\n");
}
```
## 2. 풀이
1. player 수 만큼의 queue가 필요
2. 각 player에 해당하는 queue들이 비어 있지 않으면 playerQ에 저장된다.
3. 'S칸 만큼 이동할 수 있는 모든 칸에 성을 동시에 만든다.' -> bfs에서 count 'S'만큼 확산되는 형태