HDU-1198-Farm Irrigation

来源：互联网发布：js 中文编码编辑：程序博客网时间：2024/05/01 01:21

ACM模版

描述

题解

使用二进制标记匹配状态，并查集搞搞，思路倒是不难，并查集的改造需要注意细节~~~

还可以DFS搞。

代码

#include <iostream>#include <cstdio>#include <cstring>#define mem(a, b) memset(a, b, sizeof(a))using namespace std;//  二进制表示状态const int map[] = {3, 6, 9, 12, 10, 5, 7, 11, 13, 14, 15};//  只用考虑两个方向衔接const int dir[2][2] = {{0, 1}, {1, 0}};const int MAXN = 555;int flag;int maxRoom;int M, N;int pre[MAXN * MAXN];char farm[MAXN][MAXN];int root;int find(int x){    int r = x;    while (pre[r] != r)    {        r = pre[r];    }    int i = x, j;    while (i != r)    {        j = pre[i];        pre[i] = r;        i = j;    }    return r;}void join(int x, int y){    int flag = 1;   //  表示衔接方向，0为上下衔接，1为左右衔接    for (int i = 0; i < 2; i++)    {        int x_ = x + dir[i][0];        int y_ = y + dir[i][1];        if (y_ > M)        {            flag = 0;            continue;        }        if (x_ > N)        {            break;        }        //  转成状态        int posA = farm[x][y] - 'A';        int posB = farm[x_][y_] - 'A';        int state = 0;  //  标记是否连通        if (flag)        {            if (((map[posA] >> 2) & 1) && (map[posB] & 1))            {                state = 1;            }            flag = 0;        }        else        {            if ((map[posA] >> 3) & 1 && ((map[posB] >> 1) & 1))            {                state = 1;            }        }        if (state)        {            int fx = find((x - 1) * M + y);            int fy = find((x_ - 1) * M + y_);            if (fx != fy)            {                pre[fx] = fy;                root--;            }        }    }    return ;}int main(int argc, const char * argv[]){    while (cin >> N >> M, N != -1 || M != -1)    {        root = N * M;        for (int i = 0; i <= root; i++)        {            pre[i] = i;        }        for (int i = 1; i <= N; i++)        {            scanf("%s", farm[i] + 1);        }        for (int i = 1; i <= N; i++)        {            for (int j = 1; j <= M; j++)            {                join(i, j);            }        }        cout << root << '\n';    }    return 0;}

0 0