poj4022

模拟

利用pick公式area=on/2+in-1

计算in的时候从外围广搜所有不是in的点，再用总数减去。

#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <algorithm>
using namespace std;

#define maxn 150

struct Point
{
    int x, y;
} q[maxn * maxn];

char map[maxn][maxn];
int n, m;
bool vis[maxn][maxn];
int dir[4][2] =
{
{ 1, 1 },
{ 1, -1 },
{ -1, 1 },
{ -1, -1 } };

int dir1[4][2] =
{
{ 1, 0 },
{ 0, -1 },
{ 0, 1 },
{ -1, 0 } };

void input()
{
    scanf("%d%d", &n, &m);
    for (int i = 0; i < n; i++)
        scanf("%s", map[i]);
}

bool out_of_bound(Point &a)
{
    return a.x < 0 || a.y < 0 || a.x > n + 2 || a.y > m + 2;
}

bool ok(Point &a, Point &b)
{
    int x = min(a.x, b.x) - 1;
    int y = min(a.y, b.y) - 1;
    if (x < 0 || y < 0 || x >= n || y >= m)
        return true;
    return map[x][y] == '.';
}

bool empty(int x, int y, char ch)
{
    if (x < 0 || y < 0 || x >= n || y >= m)
        return true;
    return map[x][y] != ch;
}

bool on_the_bound(Point &a)
{
    return !empty(a.x - 2, a.y - 2, '\\') || !empty(a.x - 1, a.y - 1, '\\')
            || !empty(a.x - 1, a.y - 2, '/') || !empty(a.x - 2, a.y - 1, '/');
}

void work()
{
    memset(vis, 0, sizeof(vis));
    vis[0][0] = true;
    int front, rear;
    front = rear = 0;
    Point a;
    a.x = a.y = 0;
    q[rear++] = a;
    int cnt = 1;
    while (front != rear)
    {
        a = q[front++];
        for (int i = 0; i < 4; i++)
        {
            Point b;
            b.x = a.x + dir[i][0];
            b.y = a.y + dir[i][1];
            if (!out_of_bound(b) && !on_the_bound(b) && ok(a, b)
                    && !vis[b.x][b.y])
            {
                q[rear++] = b;
                vis[b.x][b.y] = true;
                cnt++;
            }
            b.x = a.x + dir1[i][0];
            b.y = a.y + dir1[i][1];
            if (!out_of_bound(b) && !on_the_bound(b) && !vis[b.x][b.y])
            {
                q[rear++] = b;
                vis[b.x][b.y] = true;
                cnt++;
            }
        }
    }
    int on = n * m;
    for (int i = 0; i < n; i++)
        on -= count(map[i], map[i] + m, '.');
    cnt += on;
    int in = (n + 3) * (m + 3) - cnt;
    int ans = on / 2 + in - 1;
    printf("%d\n", ans);
}

int main()
{
    //freopen("t.txt", "r", stdin);
    input();
    work();
}