SDOI2005 位图

题目链接

题解

从每一个白块周围扩展即可。

#include <cstdio>
#include <cstdlib>
#include <algorithm>
#include <cstring>
#include <cctype>
#define INF 2000000000
using namespace std;
typedef long long ll;
int read(){
    int f = 1, x = 0;
    char c = getchar();
    while(c < '0' || c > '9'){if(c == '-') f = -f; c = getchar(); }
    while(c >= '0' && c <= '9')x = x * 10 + c - '0', c = getchar();
    return f * x; 
}
int n, m, mp[155][155], ans[155][155];
int que[22505][2], r, l;
int dx[] = {0, 0, -1, 1}, dy[] = {-1, 1, 0, 0};
void init(){
    n = read(), m = read();
    memset(ans, -1, sizeof(ans));
    for(int i = 0; i < n; ++i)
        for(int j = 0; j < m; ++j)
            mp[i][j] = read();
    r = l = 0;
    for(int i = 0; i < n; ++i)
        for(int j = 0; j < m; ++j){
            if(mp[i][j]) ans[i][j] = 0;
            else{
                for(int k = 0; k < 4; ++k){
                    int ex = i + dx[k], ey = j + dy[k];
                    if(ex >= 0 && ex < n && ey >= 0 && ey < m && mp[ex][ey]){
                        que[r][0] = i, que[r++][1] = j;
                        ans[i][j] = 1;
                        break;
                    }
                }
            }
        }
}
void solve(){
    while(r - l){
        int cx = que[l][0], cy = que[l++][1];
        for(int i = 0; i < 4; ++i){
            int ex = cx + dx[i], ey = cy + dy[i];
            if(ex >= 0 && ex < n && ey >= 0 && ey < m && ans[ex][ey] < 0){
                ans[ex][ey] = ans[cx][cy] + 1;
                que[r][0] = ex, que[r++][1] = ey;
            }
        }
    }
    for(int i = 0; i < n; ++i){
        for(int j = 0; j < m - 1; ++j)
            printf("%d ", ans[i][j]);
        printf("%d\n", ans[i][m - 1]);
    }
}
int main(){
    init();
    solve();
    return 0;
}