HDU 5929 Coconuts（离散化+dfs）

2019-11-08 02:43:43

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Coconuts

PRoblem DescriptionTanBig, a friend of Mr. Frog, likes eating very much, so he always has dreams about eating. One day, TanBig dreams of a field of coconuts, and the field looks like a large chessboard which has R rows and C columns. In every cell of the field, there is one coconut. Unfortunately, some of the coconuts have gone bad. For sake of his health, TanBig will eat the coconuts following the rule that he can only eat good coconuts and can only eat a connected component of good coconuts one time(you can consider the bad coconuts as barriers, and the good coconuts are 4-connected, which means one coconut in cell (x, y) is connected to (x - 1, y), (x + 1, y), (x, y + 1), (x, y - 1).Now TanBig wants to know how many times he needs to eat all the good coconuts in the field, and how many coconuts he would eat each time(the area of each 4-connected component). InputThe first line contains apositiveinteger T(T≤10) which denotes the test cases. T test cases begin from the second line. In every test case, the first line contains two integers R and C, 0<R,C≤109 the second line contains an integer n, the number of bad coconuts, 0≤n≤200 from the third line, there comes n lines, each line contains two integers, xi and yi, which means in cell(xi,yi), there is a bad coconut.It is guaranteed that in the input data, the first row and the last row will not have bad coconuts at the same time, the first column and the last column will not have bad coconuts at the same time. OutputFor each test case, output "Case #x:" in the first line, where x denotes the number of test case, one integer k in the second line, denoting the number of times TanBig needs, in the third line, k integers denoting the number of coconuts he would eat each time, you should output them in increasing order. Sample Input

23 321 22 13 312 2 Sample OutputCase #1:21 6Case #2:18
一个求有多少个联通块的题目，数据量比较大，需要对点的坐标分别离散化。
#include <iostream>#include <algorithm>#include <vector>#include <cstring>#include <string>#include <cstdio>using namespace std;const int Size = 1024;int T;//测试数据int Case = 1;int R,C;int Q;vector<long long>Area;//存储每一块联通块的面积int x[Size];int y[Size];int temp[Size];int lenx[Size];//x方向每一段的长度；int leny[Size];//y方向每一段的长度；int Pos[Size];int cntx,cnty;bool book[Size][Size];int dx[] = {1, -1, 0, 0};int dy[] = {0, 0, -1, 1};long long dfs(int x, int y){	long long sum = (long long)lenx[x]*leny[y];	book[x][y] = true;	for(int i=0; i<4; i++)	{		int _x = x+dx[i];		int _y = y+dy[i];		if(!book[_x][_y] && _x>=1 && _x<=cntx && _y>=1 && _y<=cnty)			sum += dfs(_x,_y);	}	return sum;}int main(){	cin>>T;	while(T--)	{		//Initial		Area.clear();		//Input		cin>>R>>C;		cin>>Q;		for(int i=0; i<Q; i++)			cin>>x[i]>>y[i];		//离散化x轴		int index = 0;		temp[index++] = 0;		temp[index++] = R;		for(int i=0; i<Q; i++)			temp[index++] = x[i];//将bad coconuts的x坐标以及边界存起来		sort(temp, temp+index);//对x坐标排序		index = unique(temp, temp+index)-temp;//对x坐标去重		cntx = 0;//x轴离散的段数		for(int i=1; i<index; i++)//求每一段的长度		{			if(temp[i] > temp[i-1]+1)				lenx[++cntx] = temp[i]-(temp[i-1]+1);			lenx[++cntx] = 1;			Pos[i] = cntx;//bad coconuts所在的段数		}		for(int i=0; i<Q; i++)		{			int t = lower_bound(temp,temp+index,x[i])-temp;			x[i] = Pos[t];//离散后更新原坐标		}		//离散化y轴		index = 0;		temp[index++] = 0;		temp[index++] = C;		for(int i=0; i<Q; i++)			temp[index++] = y[i];		sort(temp, temp+index);		index = unique(temp, temp+index)-temp;		cnty = 0;		for(int i=1; i<index; i++)		{			if(temp[i] > temp[i-1]+1)				leny[++cnty] = temp[i]-(temp[i-1]+1);			leny[++cnty] = 1;			Pos[i] = cnty;		}		for(int i=0; i<Q; i++)		{			int t = lower_bound(temp, temp+index, y[i])-temp;			y[i] = Pos[t];		}		//DFS		memset(book, false, sizeof(book));		for(int i=0; i<Q; i++)			book[x[i]][y[i]] = true;		for(int i=1; i<=cntx; i++)			for(int j=1; j<=cnty; j++)				if(!book[i][j])					Area.push_back(dfs(i, j));		//output		sort(Area.begin(), Area.end());		int len = (int)Area.size();		cout<<"Case #"<<(Case++)<<":/n"<<len<<"/n";		for(int i=0; i<len-1; i++)			cout<<Area[i]<<" ";		cout<<Area[len-1]<<endl;	}}

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