Bfs on 111qqz's blog

BZOJ 1656: [Usaco2006 Jan] The Grove 树木(神奇的bfs之射线法)

Fri, 15 Apr 2016 07:23:53 +0000

1656: [Usaco2006 Jan] The Grove 树木
#

Time Limit: 5 Sec Memory Limit: 64 MB Submit: 143 Solved: 88 [Submit][Status][Discuss]

BZOJ 1646: [Usaco2007 Open]Catch That Cow 抓住那只牛 (BFS)

Sun, 10 Apr 2016 13:07:12 +0000

1646: [Usaco2007 Open]Catch That Cow 抓住那只牛
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Time Limit: 5 Sec Memory Limit: 64 MB Submit: 915 Solved: 441 [Submit][Status][Discuss]

BZOJ 1644: [Usaco2007 Oct]Obstacle Course 障碍训练课 (BFS,DP)

Sun, 10 Apr 2016 12:41:46 +0000

1644: [Usaco2007 Oct]Obstacle Course 障碍训练课
#

Time Limit: 5 Sec Memory Limit: 64 MB Submit: 451 Solved: 226 [Submit][Status][Discuss]

BZOJ 1632: [Usaco2007 Feb]Lilypad Pond (BFS,dp)

Wed, 06 Apr 2016 15:37:13 +0000

1632: [Usaco2007 Feb]Lilypad Pond
#

Time Limit: 5 Sec Memory Limit: 64 MB Submit: 496 Solved: 153 [Submit][Status][Discuss]

BZOJ 1627: [Usaco2007 Dec]穿越泥地 (BFS)

Mon, 04 Apr 2016 07:00:06 +0000

1627: [Usaco2007 Dec]穿越泥地
#

Time Limit: 5 Sec Memory Limit: 64 MB Submit: 624 Solved: 411 [Submit][Status][Discuss]

BZOJ1619: [Usaco2008 Nov]Guarding the Farm 保卫牧场（BFS）

Sun, 03 Apr 2016 17:23:04 +0000

1619: [Usaco2008 Nov]Guarding the Farm 保卫牧场
#

Time Limit: 5 Sec Memory Limit: 64 MB Submit: 661 Solved: 292 [Submit][Status][Discuss]

BZOJ 1611: [Usaco2008 Feb]Meteor Shower流星雨 (BFS)

Sat, 02 Apr 2016 07:06:38 +0000

1611: [Usaco2008 Feb]Meteor Shower流星雨
#

Time Limit: 5 Sec Memory Limit: 64 MB Submit: 1239 Solved: 537 [Submit][Status][Discuss]

bzoj 1602: [Usaco2008 Oct]牧场行走 (bfs,优先队列)

Thu, 31 Mar 2016 12:57:27 +0000

Description

N头牛（2<=n<=1000）别人被标记为1到n，在同样被标记1到n的n块土地上吃草，第i头牛在第i块牧场吃草。这n块土地被n-1条边连接。奶牛可以在边上行走，第i条边连接第Ai，Bi块牧场，第i条边的长度是Li（1<=Li<=10000）。这些边被安排成任意两头奶牛都可以通过这些边到达的情况，所以说这是一棵树。这些奶牛是非常喜欢交际的，经常会去互相访问,他们想让你去帮助他们计算Q(1<=q<=1000)对奶牛之间的距离。

bc #77 ||hdu 5652 India and China Origins (图的动态连通性问题，并查集or 二分+bfs验证连通性)

Mon, 28 Mar 2016 12:37:40 +0000

题目链接题意：没图不好描述，有中文题面中文题面，直接看吧。思路：据说这道题有三种做法。当时比赛一种都不会。

codeforces 520 B. Two Buttons (bfs)

Sun, 16 Aug 2015 20:34:00 +0000

做过一道类似的题

因为是问最短，很容易想到是bfs

对于点x，可以到达点x-1,和点2*x

poj 2688 Cleaning Robot (tsp问题)

Sat, 15 Aug 2015 20:16:00 +0000

＿＿＿＿＿＿

好蠢，竟然没看出来这道题的不同之处，以为就是个搜

然后样例什么的都过了．．．

I - Fire Game (两个点开始的bfs)

Mon, 27 Jul 2015 11:01:00 +0000

http://acm.hust.edu.cn/vjudge/contest/view.action?cid=83084#problem/I

I - Fire Game

**Time Limit:**1000MS **Memory Limit:**32768KB 64bit IO Format:%I64d & %I64u

Submit Status

Description

Fat brother and Maze are playing a kind of special (hentai) game on an N*M board (N rows, M columns). At the beginning, each grid of this board is consisting of grass or just empty and then they start to fire all the grass. Firstly they choose two grids which are consisting of grass and set fire. As we all know, the fire can spread among the grass. If the grid (x, y) is firing at time t, the grid which is adjacent to this grid will fire at time t+1 which refers to the grid (x+1, y), (x-1, y), (x, y+1), (x, y-1). This process ends when no new grid get fire. If then all the grid which are consisting of grass is get fired, Fat brother and Maze will stand in the middle of the grid and playing a MORE special (hentai) game. (Maybe it’s the OOXX game which decrypted in the last problem, who knows.)

poj 3414 pots (bfs+路径记录)

Mon, 27 Jul 2015 08:44:00 +0000

好爽,一遍ac

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/*************************************************************************
> File Name: code/2015summer/searching/H.cpp
> Author: 111qqz
> Email: rkz2013@126.com
> Created Time: 2015年07月27日 星期一 09时11分28秒
************************************************************************/
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#include<iostream>
#include<iomanip>
#include<cstdio>
#include<algorithm>
#include<cmath>
#include<cstring>
#include<string>
#include<map>
#include<set>
#include<queue>
#include<vector>
#include<stack>
#define y0 abc111qqz
#define y1 hust111qqz
#define yn hez111qqz
#define j1 cute111qqz
#define tm crazy111qqz
#define lr dying111qqz
using namespace std;
#define REP(i, n) for (int i=0;i<int(n);++i)
typedef long long LL;
typedef unsigned long long ULL;
const int N=1E2+5;
int A,B,C;
int d[N][N];
bool flag;
struct node
{
int d,opt,par,prea,preb;
}q[N][N];
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void print(int x,int y)
{
// cout<<"x:"<<x<<"y:"<<y<<endl;
if (q[x][y].prea!=-1&&q[x][y].preb!=-1)
{
// 	 cout<<"who is 111qqz"<<endl;
 print(q[x][y].prea,q[x][y].preb);
 if (q[x][y].opt==1){
	printf("FILL(%d)\n",q[x][y].par);
 }
 if (q[x][y].opt==2)
 {
	printf("DROP(%d)\n",q[x][y].par);
 }
 if (q[x][y].opt==3)
 {
	printf("POUR(%d,%d)\n",q[x][y].par,3-q[x][y].par);
 }
}
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}
void bfs()
{
memset(q,-1,sizeof(q));
queue<int>a;
queue<int>b;
a.push(0);
b.push(0);
q[0][0].d=0;
while (!a.empty()&&!b.empty())
{
 int av = a.front();a.pop();
 int bv = b.front();b.pop();
// 	 cout<<"av:"<<av<<"bv:"<<bv<<endl;
 if (av==C||bv==C)
 {
	flag = true;
	//cout<<"yeah~~~~~~~~~~~~~~~"<<endl;
	cout<<q[av][bv].d<<endl;
// 		cout<<"prea:"<<q[av][bv].prea<<"preb:"<<q[av][bv].preb<<endl;
	print(av,bv);
 	return;
 }
 if (av<A&&q[A][bv].d==-1)
 {
	q[A][bv].d=q[av][bv].d+1;
	q[A][bv].opt=1;
	q[A][bv].par=1;
	q[A][bv].prea=av;
	q[A][bv].preb=bv;
	a.push(A);
	b.push(bv);
 }
 if (av>0&&q[0][bv].d==-1)
 {
	q[0][bv].d=q[av][bv].d+1;
	q[0][bv].opt=2;
	q[0][bv].par=1;
	q[0][bv].prea=av;
	q[0][bv].preb=bv;
	a.push(0);
	b.push(bv);
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 }
 if (bv<B&&q[av][B].d==-1)
 {
	q[av][B].d=q[av][bv].d+1;
	q[av][B].opt=1;
	q[av][B].par=2;
	q[av][B].prea = av;
	q[av][B].preb = bv;
	a.push(av);
	b.push(B);
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 }
 if (bv>0&&q[av][0].d==-1)
 {
	q[av][0].d=q[av][bv].d+1;
	q[av][0].opt=2;
	q[av][0].par=2;
	q[av][0].prea=av;
	q[av][0].preb=bv;
	a.push(av);
	b.push(0);
 }
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 if (av+bv<=B&&q[0][av+bv].d==-1)
 {
	q[0][av+bv].d=q[av][bv].d+1;
	q[0][av+bv].opt=3;
	q[0][av+bv].par=1;
	q[0][av+bv].prea=av;
	q[0][av+bv].preb=bv;
	a.push(0);
	b.push(av+bv);
 }
 if (av+bv>B&&q[av-(B-bv)][B].d==-1) //把1往2里倒入的两种情况
 {
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	int tmp = av-(B-bv);
	q[tmp][B].d=q[av][bv].d+1;
	q[tmp][B].opt=3;
	q[tmp][B].par=1;
	q[tmp][B].prea=av;
	q[tmp][B].preb=bv;
	a.push(tmp);
	b.push(B);
 }
149
 if (bv+av<=A&&q[av+bv][0].d==-1)
 {
	q[av+bv][0].d=q[av][bv].d+1;
	q[av+bv][0].opt=3;
	q[av+bv][0].par=2;
	q[av+bv][0].prea=av;
	q[av+bv][0].preb=bv;
	a.push(av+bv);
	b.push(0);
 }
 if (bv+av>A&&q[A][bv-(A-av)].d==-1)
 {
	int tmp = bv-(A-av);
	q[A][tmp].d=q[av][bv].d+1;
	q[A][tmp].opt=3;
	q[A][tmp].par=2;
	q[A][tmp].prea=av;
	q[A][tmp].preb=bv;
	a.push(A);
	b.push(tmp);
 }
}
}
int main()
{
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flag = false;
cin>>A>>B>>C;
bfs();
if (!flag)
{
 cout<<"impossible"<<endl;
}
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return 0;
}

hdoj 1495 非常可乐(bfs)

Sun, 26 Jul 2015 14:59:00 +0000

非常可乐
#

**Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 7194 Accepted Submission(s): 2865 **

hdoj 2612 find a way (两次bfs)

Sat, 25 Jul 2015 07:01:00 +0000

Find a way
#

****Time Limit: 3000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 6221 Accepted Submission(s): 2070 **

poj 3984 迷宫问题

Sat, 25 Jul 2015 05:53:00 +0000

迷宫问题

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/*************************************************************************
> File Name: code/2015summer/searching/KK.cpp
> Author: 111qqz
> Email: rkz2013@126.com
> Created Time: 2015年07月25日 星期六 13时33分00秒
************************************************************************/
10
#include<iostream>
#include<iomanip>
#include<cstdio>
#include<algorithm>
#include<cmath>
#include<cstring>
#include<string>
#include<map>
#include<set>
#include<queue>
#include<vector>
#include<stack>
#define y0 abc111qqz
#define y1 hust111qqz
#define yn hez111qqz
#define j1 cute111qqz
#define tm crazy111qqz
#define lr dying111qqz
using namespace std;
#define REP(i, n) for (int i=0;i<int(n);++i)
typedef long long LL;
typedef unsigned long long ULL;
int a[10][10];
int head = 0;
int tail = 1;
int dirx[2]={1,0};
int diry[2]={0,1};
struct node
{
int x,y,pre;
}q[10];
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void print(int x)
{
if (q[x].pre!=-1)
{
 print(q[x].pre);
 printf("(%d, %d)\n",q[x].x,q[x].y);
}
}
void bfs()
{
q[head].x=0;
q[head].y=0;
q[head].pre=-1;
while (head<tail)
{
 if (q[head].x==4&&q[head].y==4)
 {
	print(head);
	return;
 }
 for (int i = 0 ; i < 2 ; i++ )
 {
	int newx=dirx[i]+q[head].x;
	int newy=diry[i]+q[head].y;
	if (newx>=0&&newx<5&&newy>=0&&newy<5&&a[newx][newy]==0)
	{
	 q[tail].x=newx;
	 q[tail].y=newy;
	 q[tail].pre=head;
	 tail++;
	}
 }
 head++;
}
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}
int main()
{
for ( int i = 0 ; i < 5 ; i++ )
{
 for ( int j = 0 ; j < 5; j++)
 {
	cin>>a[i][j];
 }
}
printf("(0, 0)\n");
bfs();
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return 0;
}

poj 3087 Shuffle'm Up (bfs)

Fri, 24 Jul 2015 16:12:00 +0000

http://poj.org/problem?id=3087

用bfs写的，但是其实就是个模拟啊喂！

只有一种操作，何谈最短？　一直往下写就行了．

poj 3126 Prime Path (bfs)

Fri, 24 Jul 2015 12:53:00 +0000

http://poj.org/problem?id=3126

题意是说,给定两个四位素数a b 问从a变换到b,最少需要变换几次. 变换的要求是,每次只能改变一个数字,而且中间过程得到的四位数也必须为素数. 因为提到最少变换几次,容易想到bfs,bfs第一次搜到的一定是最短步数.

poj 2251 Dungeon Master (三维bfs)

Tue, 21 Jul 2015 05:02:00 +0000

http://poj.org/problem?id=2251

简单bfs，只不过是三维的。。。

唯一的坑点在输出上…

poj 3278 catch that cow

Wed, 08 Jul 2015 02:25:00 +0000

http://poj.org/problem?id=3278

bfs,用到了stl的queue

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/* ***********************************************
Author :111qqz
Created Time :2016年02月19日 星期五 15时45分05秒
File Name :3278.cpp
************************************************ */
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#include <algorithm>
#include <cstdio>
#include <iostream>
#include <cstring>
#include <string>
#include <cmath>
#include <map>
#include <stack>
#include <queue>
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using namespace std;
typedef long long LL;
const int inf = 8E8;
const int N=2E5+7;
int d[N];
int n,k;
void bfs()
{
queue<int> q;
memset(d,-1,sizeof(d));
q.push(n);
d[n]=0;
while (!q.empty())
{
int x = q.front();
q.pop();
if ( x==k )
{
break;
}
int next[10];
next[1]=x-1;
next[2]=x+1;
next[3]=2*x;
for ( int i = 1; i <= 3 ; i++ )
{
if (next[i]>=0&&next[i]<=100000&&d[next[i]]==-1)
{
d[next[i]]=d[x]+1;
q.push(next[i]);
}
}
}
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}
int main()
{
while (scanf("%d %d",&n,&k)!=EOF)
{
bfs();
cout<<d[k]<<endl;
}
return 0;
}