hdu 1198 并查集应用

   其实可能是因为知道是用并查集做的原因啦，一下就看出题意了，明显是并查集思想，每次输入地图中的一块，检测这一块与它顶头的那块可不可以相通

如果可以合并集合;同理检测其与它左边的那一块;最后遍历一遍看有多少个根结点即要多少个水源

下面是代码，有点乱

#include<iostream>#include<vector>#define M 55#define N 55using namespace std;class elem{public:bool up;bool down;bool right;bool left;};class cor{public:int x;int y;};cor father[M][N];int n,m;void init(vector<elem>& farm){//初始化给出的11块地的样子保存下来，下标为0的不要elem x;x.up = false; x.down = false;x.right = false;x.left = false;farm.push_back(x);//0x.up = true; x.down = false;x.right = false;x.left = true;farm.push_back(x);//1x.up = true;x.down = false;x.right = true;x.left = false;farm.push_back(x);//2x.up = false;x.down = true;x.right = false;x.left = true;farm.push_back(x);//3x.up = false;x.down = true;x.right = true;x.left = false;farm.push_back(x);//4x.up = true;x.down = true;x.right = false;x.left = false;farm.push_back(x);//5x.up = false;x.down = false;x.right = true;x.left = true;farm.push_back(x);//6x.up = true;x.down = false;x.right = true;x.left = true;farm.push_back(x);//7x.up = true;x.down = true;x.right = false;x.left = true;farm.push_back(x);//8x.up = false;x.down = true;x.right = true;x.left = true;farm.push_back(x);//9x.up = true;x.down = true;x.right = true;x.left = false;farm.push_back(x);//10x.up = true;x.down = true;x.right = true;x.left = true;farm.push_back(x);//11}int change(char x){//索引return x-'A'+1;}cor find_father(int x,int y){//找根结点if(x == father[x][y].x && y == father[x][y].y){cor root; root.x = x; root.y = y;return root;}else{father[x][y] = find_father(father[x][y].x,father[x][y].y);}return father[x][y];}void merge(cor a,cor b){合并father[a.x][a.y] = b;}void process(int x,int y,elem map[M][N]){//两个方向的处理int _x = x-1;int _y = y;if(_x >= 1 && _x <= n && _y >= 1 && _y <= m && map[x][y].up && map[_x][_y].down){//上下两个图都有pipecor a = find_father(x,y);cor b = find_father(_x,_y);if(!(a.x == b.x && a.y == b.y))    merge(a,b);}_x = x; _y = y-1;if(_x >= 1 && _x <= n && _y >= 1 && _y <= m && map[x][y].left && map[_x][_y].right){//左右cor a = find_father(x,y);cor b = find_father(_x,_y);if(!(a.x == b.x && a.y == b.y))    merge(a,b);}}int main(){vector<elem> farm;//保存11个初始地图elem map[M][N];init(farm);while(cin >>n >>m && (n > 0 && m > 0)){for(int i = 1;i <= n;i++){for(int j = 1;j <= m;j++){    father[i][j].x = i;father[i][j].y = j;}}for(int i = 1;i <= n;i++){for(int j = 1; j <= m;j++){char x;cin >>x;map[i][j] = farm[change(x)];process(i,j,map);}}int count = 0;for(int i = 1;i <= n;i++){for(int j = 1;j <= m;j++){if(i == father[i][j].x && father[i][j].y == j)count++;}}cout <<count <<endl;}return 0;}