DFS————深度优先搜索2.例题

1.组合的输出

#include
#include
#include
using namespace std;
int m,n;

int a[22];

void input()
{
	cin>>m>>n;
}

void print()
{
	for(int i = 1;i<=n;i++)
	{
		cout<

2.八皇后问题

#include
using namespace std;
int ps[9];
bool l,r;
int num = 1;
void dfs(int b);
void print();


void dfs(int b){
	if(b == 9){
		print();
		return;
	}
	else{
		for(int i = 1;i<=8;i++){
			bool yi = false;
			for(int j = b-1;j>0;j--){
				if((i+b == j+ps[j])||(b-i == j-ps[j])||(i == ps[j])) yi = true;
			}
			if(!yi){
				ps[b] = i;
				dfs(b+1);
			}
		}
	}
}

void print(){
	cout<<"No. "<

3.求先序排列

#include
#include
using namespace std;

struct Tree;

struct Tree{
	char date;
	Tree* leftTree=NULL,*rightTree=NULL;
};

void input(string &zhong,string &hou)
{
	cin>>zhong>>hou;
}

void chli(Tree* &node,string zhong,string hou)
{
		if(zhong == ""||hou == "") return;
		char nodec = hou[hou.size()-1];
		node = new(Tree);
		node->date = nodec;
		string zleft="",zright="";
		bool y = false;
		for(int i = 0;ileftTree,zleft,hleft);
		chli(node->rightTree,zright,hright);
}

void print(Tree* node)
{
	if(node == NULL) return;
	else
	{
		cout<date;
		print(node->leftTree);
		print(node->rightTree);
	}
}


int main(){
	string zhong,hou;
	Tree *head;
	input(zhong,hou);
	chli(head,zhong,hou);
	print(head);
	return 0;
}

4.P1827 [USACO3.4] 美国血统 American Heritage

#include
#include
using namespace std;

struct Tree;

struct Tree{
	char date;
	Tree* leftTree=NULL,*rightTree=NULL;
};

void input(string &zhong,string &hou)
{
	cin>>zhong>>hou;
}

void chli(Tree* &node,string zhong,string hou)
{
		if(zhong == ""||hou == "") return;
		char nodec = hou[0];
		node = new(Tree);
		node->date = nodec;
		string zleft="",zright="";
		bool y = false;
		for(int i = 0;ileftTree,zleft,hleft);
		chli(node->rightTree,zright,hright);
}

void print(Tree* node)
{
	if(node == NULL) return;
	else
	{
		print(node->leftTree);
		print(node->rightTree);
		cout<date;
	}
}


int main(){
	string zhong,hou;
	Tree *head;
	input(zhong,hou);
	chli(head,zhong,hou);
	print(head);
	return 0;
}

5.马走日

#include
#include
#include
#include
#include
#include
#include
#include
using namespace std;

int xj[] = {1,2,2,-1,1,-1,-2,-2},
	yj[] = {2,1,-1,2,-2,-2,1,-1};
bool qipan[9][9];
int T,n,m,cx,cy;
int ans;

void input();
void print();
void dfs(int x,int y);
void run();
bool pd();

int main()
{
	run();
	return 0;
}

void input()
{
	memset(qipan,0,sizeof(qipan));
	ans = 0;
	cin>>n>>m>>cx>>cy;
}

void print()
{
	cout<>T;
	while(T--)
	{
		input();
		dfs(cx,cy);
		print();
	}
}

bool pd(){
	for(int i = 0;i=n||y>=m||qipan[x][y]) return;
		else
		{
			qipan[x][y] = true;
			for(int i = 0;i<8;i++){
				dfs(x+xj[i],y+yj[i]);
			}
			qipan[x][y] = false;
		}
	}
}