hdu1198并查集

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之前忘了并查集使用的方式，导致我把图中的11张图看成了各自的一类，然后就悲剧了=-=

简述这道题的意思，就是有11种suqare然后将这些可以组成各种大小的田地。现在它给你一块由这些suqare组成的田地

，因为每块地都有pipe，要保证这些田都被灌溉你要判断这块地需要多少个wellspring。

首先说下并查集的思路：题目要求是输入m行，n列，而且m和n是不超过50，算出来是2500，说明使用并查集是可以解决的

时间复杂度是O（m*n），并查集最需要注意的地方就是要注意初始化。这道题wa了4次，前两次是没找出自己的并查集是写错的

，后两次改成正确的并查集，然而却忘了初始化_(:зゝ∠)_。

#include<iostream>#include<fstream>#include<cstring>using namespace std;struct Squares{    bool up;    bool down;    bool left;    bool right;    int root;};int p[2510];Squares S[12];int disjoint[12] = { 0 };bool isjudge[2510] = { 0 };int m, n;int map[60][60];void init(){    S[1].up = true;    S[1].down = false;    S[1].left = true;    S[1].right = false;    S[1].root = 1;    S[2].up = true;    S[2].down = false;    S[2].left = false;    S[2].right = true;    S[2].root = 2;    S[3].up = false;    S[3].down = true;    S[3].left = true;    S[3].right = false;    S[3].root = 3;    S[4].up = false;    S[4].down = true;    S[4].left = false;    S[4].right = true;    S[4].root = 4;    S[5].up = true;    S[5].down = true;    S[5].left = false;    S[5].right = false;    S[5].root = 5;    S[6].up = false;    S[6].down = false;    S[6].left = true;    S[6].right = true;    S[6].root = 6;    S[7].up = true;    S[7].down = false;    S[7].left = true;    S[7].right = true;    S[7].root = 7;    S[8].up = true;    S[8].down = true;    S[8].left = true;    S[8].right = false;    S[8].root = 8;    S[9].up = false;    S[9].down = true;    S[9].left = true;    S[9].right = true;    S[9].root = 9;    S[10].up = true;    S[10].down = true;    S[10].left = false;    S[10].right = true;    S[10].root = 10;    S[11].up = true;    S[11].down = true;    S[11].left = true;    S[11].right = true;    S[11].root = 11;}void initroot(){    for (int i = 0; i <= m*n; i++)    {        p[i] = i;    }}int findroot(int a){    int root = a;    while (p[root] != root)        root = p[root];    int i = a, j;    while (i != root)    {        j = p[i];        p[i] = root;        i = j;    }    return root;}void merge(int a, int b){    int fx = findroot(a), fy = findroot(b);    if (fx != fy)    {        p[fx] = fy;    }}int ans;void solve(){    ans = 0;    int loc;    int locup;    int locdown;    int locleft;    int locright;    for (int i = 0; i < m; i++)    {        for (int j = 0; j < n; j++)        {            loc = map[i][j];            if (i - 1 >= 0)//up is right            {                locup = map[i - 1][j];                if (S[locup].down&&S[loc].up)                {                    merge(i*n + j, (i - 1)*n + j);                }            }            if (i + 1 < m)//down is right            {                locdown = map[i + 1][j];                if (S[locdown].up&&S[loc].down)                {                    merge(i *n + j, (i + 1)*n + j);                }            }            if (j + 1 < n)//right is right            {                locright = map[i][j + 1];                if (S[locright].left&&S[loc].right)                {                    merge(i *n + j, i*n + j + 1);                }            }            if (j - 1 >= 0)            {                locleft = map[i][j - 1];                if (S[locleft].right&&S[loc].left)                {                    merge(i*n + j, i*n + j - 1);                }            }        }    }    for (int i = 0; i < m*n; i++)        findroot(i);    for (int i = 0; i < m*n; i++)    {        if (isjudge[p[i]] == false)        {            ans++;            isjudge[p[i]] = true;        }    }    printf("%d\n", ans);}int main(){    init();    while (scanf_s("%d%d", &m, &n))    {        //getchar();        initroot();        memset(isjudge, false, sizeof(isjudge));        memset(map, 0, sizeof(map));        if (m + n == -2)            break;        char c;        for (int i = 0; i < m; i++)        {            for (int j = 0; j < n; j++)            {                cin >> c;                //scanf_s("%c", &c);                map[i][j] = c - 'A' + 1;            }            //getchar();        }        solve();    }}

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