HDU_1198 Farm Irrigation(并查集)

/*
开始思路想错了，直接用并查集把输入的square并到并查集里边了，后来想到可能一种
square连续出现多次，这样直接合并肯定会出问题的。然后就一直再想其他思路，昨晚睡觉时突
然想到可以给这些square重新编号，把对应的square的pipes type继承过来，然后再用
并查集合并。最终以6A告终。不容易啊，大牛鄙视我吧。
*/

#include <iostream>
#include <cstdio>
#include <cstring>

using namespace std;

const int N = 3000;

class pipes
{
public:
int l;        //左边节
int r;        //右边界
int u;        //上边界
int d;        //下边界
}tmp[12], num
;

int parent
;
int rank
;
int map[55][55];

void init()    //记录11种 pipes type。
{
for(int i = 1; i <= 11; i++)
{
tmp[i].l = 0;
tmp[i].r = 0;
tmp[i].u = 0;
tmp[i].d = 0;
}

tmp[1].l = 1;
tmp[1].u = 1;

tmp[2].r = 1;
tmp[2].u = 1;

tmp[3].l = 1;
tmp[3].d = 1;

tmp[4].r = 1;
tmp[4].d = 1;

tmp[5].u = 1;
tmp[5].d = 1;

tmp[6].l = 1;
tmp[6].r = 1;

tmp[7].l = 1;
tmp[7].r = 1;
tmp[7].u = 1;

tmp[8].l = 1;
tmp[8].u = 1;
tmp[8].d = 1;

tmp[9].l = 1;
tmp[9].r = 1;
tmp[9].d = 1;

tmp[10].u = 1;
tmp[10].r = 1;
tmp[10].d = 1;

tmp[11].l = 1;
tmp[11].r = 1;
tmp[11].u = 1;
tmp[11].d = 1;
}

int find(int x)        //并查集非递归路径压缩查找，个人习惯这样写，因为递归的路径压缩可能会Ovre Stack(只是可能，大牛勿喷。。。)
{
int k, r, j;
r = x;
while(r != parent[r])
r = parent[r];
k = x;
while(k != r)
{
j = parent[k];
parent[k] = r;
k = j;
}
return r;
}

void Union(int a, int b)    //合并
{
int x = find(a);
int y = find(b);

if(x == y)    return;

if(rank[x] > rank[y])
parent[y] = x;
else
{
parent[x] = y;
if(rank[x] == rank[y])
++rank[y];
}
}

int main()
{
//freopen("data.in", "r", stdin);

int n, m;
init();
while(scanf("%d%d", &n, &m) != EOF)
{
if(n < 0 || m < 0)
break;
getchar();
char c;
int x = 1, t, i, j;
for(i = 1; i <= n; i++)
{
for(j = 1; j <= m; j++)
{
scanf("%c", &c);
t = c-'A'+1;
map[i][j] = x;        //重新编号，起始号为1
num[x].l = tmp[t].l;    //继承对应的 pipes type
num[x].r = tmp[t].r;
num[x].u = tmp[t].u;
num[x].d = tmp[t].d;
x++;
}
getchar();
}

/*for(i = 1; i <= n; i++)
{
for(j = 1; j <= m; j++)
printf("%d ", map[i][j]);
printf("\n");
}*/

for(i = 1; i <= n; i++)
for(j = 1; j <= m; j++)
{
parent[map[i][j]] = map[i][j];         //初始化parent[],rank[];
rank[map[i][j]] = 0;
}
for(i = 1; i <= n; i++)
{
for(j = 1; j <= m; j++)
{
if(num[map[i][j]].l == num[map[i][j-1]].r && num[map[i][j]].l != 0 && j > 1)        //左边
Union(map[i][j], map[i][j-1]);

if(num[map[i][j]].r == num[map[i][j+1]].l && num[map[i][j]].r != 0 && j+1 <= m)        //右边
Union(map[i][j], map[i][j+1]);

if(num[map[i][j]].u == num[map[i-1][j]].d && num[map[i][j]].u != 0 && i > 1)        //上边
Union(map[i][j], map[i-1][j]);

if(num[map[i][j]].d == num[map[i+1][j]].u && num[map[i][j]].d != 0 && i+1 <= n)        //下边
Union(map[i][j], map[i+1][j]);
}
}
int cnt = 0;
for(i = 1; i <= n; i++)
for(j = 1; j <= m; j++)
if(find(map[i][j]) == map[i][j])    //统计个数
cnt++;
printf("%d\n", cnt);
}
return 0;
}