蓝桥杯:标题：剪格子

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p1.jpg p2.jpg

如图p1.jpg所示，3 x 3 的格子中填写了一些整数。

提交时，注意选择所期望的编译器类型。

Thinking:

First of all,you have to realize that the grid which located in the upper left must be a part of the two part divided,thus,we can start DFS from coordinate(0,0).

Detail in the code notes.

#include<iostream>
#include<string>
#include<algorithm>
using namespace std;
int visited[10][10]={0},sum=0,direction[4][2]={{-1,0},{0,-1},{1,0},{0,1}},total=0;
int row,col,num[10][10]={0},nSeq=0;
string reserve[50];
bool judge(int count)//To judge whether the sequence is repeat.
{
	int i,j,k=0,l;
	for(i=0;i<row;i++)
		for(j=0;j<col;j++)
			if(visited[i][j]==1)
				reserve[nSeq].push_back(i*col+j);//push the coordinate into reserve[]
	sort(reserve[nSeq].begin(),reserve[nSeq].end());//and sort it smaller to bigger. 
	nSeq++;//nSeq represent n Sequences.
	if(nSeq!=1)//if nSeq is equal to 1,you dont' have to judge it. 
	{
		for(l=0;l<nSeq-1;l++)
			if(reserve[l]==reserve[nSeq-1])
			{	
				reserve[nSeq-1].clear();
				nSeq--;
				return false;
			}
	}
	return true;
}
int minLength()//If the reserve[] have more than 1 sequence,you have to find the shortest sequence.
{
	int i,j,minLen=999;
	for(i=0;i<nSeq;i++)//easy to understand,no more explains.
		if(reserve[i].length()<minLen)
			minLen=reserve[i].length();
	return minLen;
}
void DFS(int x,int y,int count)
{
	int next_x,next_y,i;
	if(count==sum)
	{
		if(judge(count))
			total++;
		return;
	}
	else if(count>sum)
		return;
	for(i=0;i<4;i++)
	{
		next_x=x+direction[i][0];
		next_y=y+direction[i][1];
		if(next_x<0||next_y<0||next_x>=row||next_y>=col)//boundary condition
			continue;
		else if(visited[next_x][next_y]==0)//if the point hasn't been visited.
		{
			visited[next_x][next_y]=1;//set the point has been visited.
			DFS(next_x,next_y,count+num[next_x][next_y]);
			visited[next_x][next_y]=0;//set the point hasn't been visited.
		}
	}
}
int main()
{
	int i,j;
	cin>>row>>col;
	for(i=0;i<row;i++)
		for(j=0;j<col;j++)
		{
			cin>>num[i][j];
			sum+=num[i][j];
		}
	if(sum%2!=0)//if sum is odd ,quit.
	{
		cout<<"0"<<endl;
		exit(0);
	}
	sum/=2;
	visited[0][0]=1;
	DFS(0,0,num[0][0]);
	if(total!=0)//if the path exist
		cout<<minLength()<<endl;
	else
		cout<<"0"<<endl;
	return 0;
}

蓝桥杯:标题：剪格子

标签：

原文地址：http://blog.csdn.net/lc0817/article/details/42833823