[计算几何]poj 2308 Toys

#include <cstdio>
#include <cstring>
#include <algorithm>
#include <iostream>
#include <queue>
using namespace std;
struct Node
{
    int x, y, se, d, c;
};
queue < Node >  Q;

int move[4][2] = {-1, 0, 1, 0, 0, -1, 0, 1};
int vis[11][11], mat[11][11], card[4], flag, N, M;
char Map[11][11];

inline int off(int x, int y, int c)
{
    if (!(x >= 0 && x < N && y >= 0 && y < M)) return 0;
    if (mat[x][y] != -1 && mat[x][y] != c) return 0;
    return 1;
}

inline int isAB(int a, int b)
{
    for (int i = 0; i < N; i++)
    {
        for (int j = 0; j < M; j++)
        {
            if (mat[i][j] == a)
            {
                if (mat[i + 1][j + 1] == a && mat[i][j + 1] == b && mat[i + 1][j] == b) return 1;
                else return 0;
            }
            if (mat[i][j] == b)
            {
                if (mat[i + 1][j + 1] == b && mat[i][j + 1] == a && mat[i + 1][j] == a) return 1;
                else return 0;
            }
        }
    }
}

inline int cut(void)
{
    for (int i = 0; i < 4; i ++ )
    {
        for (int j = i + 1; j < 4; j ++ )
        {
            if (card[i] == card[j] && card[i] == 2)
            {
                if (isAB(i, j)) return 1;
                else return 0;
            }
        }
    }
}

int dfs(int count)
{
    if (!count) return flag = 1;
    if (cut()) return 0;

    for (int i = 0; i < N; i ++ )
        for (int j = 0; j < M; j ++ )
        {
            if (flag) return 1;

            if (mat[i][j] != -1)
            {
                int c = mat[i][j];
                int pos[10][2], cnt = 0;

                memset(vis, 0, sizeof(vis));
                Node nt, nt1;
                nt.x = i;
                nt.y = j;
                nt.c = -1;nt.se = 0;
                nt.d = -1;
                vis[i][j] = 1;
                Q.push(nt);

                while(!Q.empty())
                {
                    nt = Q.front();Q.pop();
                    if (nt.c == c)
                    {
                        pos[cnt][0] = nt.x;
                        pos[cnt][1] = nt.y;
                        cnt ++ ; continue;
                    }
                    for (int k = 0; k < 4; k ++ )
                    {
                        int x = nt.x + move[k][0], y = nt.y + move[k][1];

                        if (off(x, y, c) && !vis[x][y])
                        {
                            if (nt.d == k || nt.d == -1) nt1.se = nt.se;
                            else nt1.se = nt.se + 1;

                            if (nt1.se <= 2)
                            {
                                nt1.x = x;
                                nt1.y = y;
                                nt1.c = mat[x][y];
                                nt1.d = k;
                                vis[x][y] = 1;
                                Q.push(nt1);
                            }
                        }
                    }
                }
                mat[i][j] = -1;
                card[c] -= 2;

                for (int k = 0; k < cnt; k ++ )
                {
                    if (flag) return 1;
                    int x = pos[k][0], y = pos[k][1];
                    mat[x][y] = -1;
                    dfs(count - 2);
                    mat[x][y] = c;
                }
                mat[i][j] = c;
                card[c] += 2;
            }
        }
    return 0;
}

int main(void)
{
// freopen("test.txt", "r", stdin);
    while (scanf("%d%d", &N, &M) && N && M)
    {
        memset(card, 0, sizeof(card));
        memset(mat, -1, sizeof(mat));
        int count = 0; flag = 1;
        for (int i = 0; i < N; i++ ) scanf("%s", Map[i]);
        {
            for (int i = 0;i < N;i ++ )
            {
                for (int j = 0;j < M;j ++ )
                {
                    if (Map[i][j] == 'A') { card[0] ++ ;mat[i][j] = 0;}
                    else if (Map[i][j] == 'B') { card[1] ++ ;mat[i][j] = 1;}
                    else if (Map[i][j] == 'C') { card[2] ++ ;mat[i][j] = 2;}
                    else if (Map[i][j] == 'D') { card[3] ++ ;mat[i][j] = 3;}
                    else mat[i][j] = -1;
                }
                for (int i = 0;i < 4;i ++ )
                {
                    count += card[i];
                    if (card[i]%2) flag = 0;
                }
                if (!flag) printf("no\n");
                else
                {
                    flag = 0;
                    dfs(count);
                    if (flag) printf("yes\n");
                    else printf("no\n");
                }
            }
        }
    }
    return 0;
} 

1.设直线的方程ax+bx+c=0(a>0);若点(x0,y0)代入方程：

　　若ax0+by0+c<0，则点在直线的左侧；

　　若ax0+by0+c=0，则点在直线上；

　　若ax0+by0+c<0，则点在直线的右侧;

　　但是前提是a>0,否则结果左右对调(WA了一次).

2.(x1,y1)和(x2,y2)为某直线的两个点，直线的方程就可以表示为(y2-y1)(x-x1)-(x2-x1)(y-y1)=0；因为n较大,而直线与直线对点具有序的关系，即点如果在某条直线的左侧，那么肯定在那条直线右侧所有的直线的左侧，

所以可用二分搜索把时间复杂度降为O(nlogn);