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。

View Code

#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();

}