Bfs

1656: [Usaco2006 Jan] The Grove 树木 # Time Limit: 5 Sec Memory Limit: 64 MB Submit: 143 Solved: 88 [Submit][Status][Discuss]

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

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]

1632: [Usaco2007 Feb]Lilypad Pond # Time Limit: 5 Sec Memory Limit: 64 MB Submit: 496 Solved: 153 [Submit][Status][Discuss]

1627: [Usaco2007 Dec]穿越泥地 # Time Limit: 5 Sec Memory Limit: 64 MB Submit: 624 Solved: 411 [Submit][Status][Discuss]

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]

1611: [Usaco2008 Feb]Meteor Shower流星雨 # Time Limit: 5 Sec Memory Limit: 64 MB Submit: 1239 Solved: 537 [Submit][Status][Discuss]

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。

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。

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

＿＿＿＿＿＿ 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．．．

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.)

好爽,一遍ac 1 2 3 /************************************************************************* 4 > File Name: code/2015summer/searching/H.cpp 5 > Author: 111qqz 6 > Email: rkz2013@126.com 7 > Created Time: 2015年07月27日 星期一 09时11分28秒 8 ************************************************************************/ 9 10 #include<iostream> 11 #include<iomanip> 12 #include<cstdio> 13 #include<algorithm> 14 #include<cmath> 15 #include<cstring> 16 #include<string> 17 #include<map> 18 #include<set> 19 #include<queue> 20 #include<vector> 21 #include<stack> 22 #define y0 abc111qqz 23 #define y1 hust111qqz 24 #define yn hez111qqz 25 #define j1 cute111qqz 26 #define tm crazy111qqz 27 #define lr dying111qqz 28 using namespace std; 29 #define REP(i, n) for (int i=0;i<int(n);++i) 30 typedef long long LL; 31 typedef unsigned long long ULL; 32 const int N=1E2+5; 33 int A,B,C; 34 int d[N][N]; 35 bool flag; 36 struct node 37 { 38 int d,opt,par,prea,preb; 39 }q[N][N]; 40 41 void print(int x,int y) 42 { 43 // cout<<"x:"<<x<<"y:"<<y<<endl; 44 if (q[x][y].prea!=-1&&q[x][y].preb!=-1) 45 { 46 // cout<<"who is 111qqz"<<endl; 47 print(q[x][y].prea,q[x][y].preb); 48 if (q[x][y].opt==1){ 49 printf("FILL(%d)\n",q[x][y].par); 50 } 51 if (q[x][y].opt==2) 52 { 53 printf("DROP(%d)\n",q[x][y].par); 54 } 55 if (q[x][y].opt==3) 56 { 57 printf("POUR(%d,%d)\n",q[x][y].par,3-q[x][y].par); 58 } 59 } 60 61 } 62 void bfs() 63 { 64 memset(q,-1,sizeof(q)); 65 queue<int>a; 66 queue<int>b; 67 a.push(0); 68 b.push(0); 69 q[0][0].d=0; 70 while (!a.empty()&&!b.empty()) 71 { 72 int av = a.front();a.pop(); 73 int bv = b.front();b.pop(); 74 // cout<<"av:"<<av<<"bv:"<<bv<<endl; 75 if (av==C||bv==C) 76 { 77 flag = true; 78 //cout<<"yeah~~~~~~~~~~~~~~~"<<endl; 79 cout<<q[av][bv].d<<endl; 80 // cout<<"prea:"<<q[av][bv].prea<<"preb:"<<q[av][bv].preb<<endl; 81 print(av,bv); 82 return; 83 } 84 if (av<A&&q[A][bv].d==-1) 85 { 86 q[A][bv].d=q[av][bv].d+1; 87 q[A][bv].opt=1; 88 q[A][bv].par=1; 89 q[A][bv].prea=av; 90 q[A][bv].preb=bv; 91 a.push(A); 92 b.push(bv); 93 } 94 if (av>0&&q[0][bv].d==-1) 95 { 96 q[0][bv].d=q[av][bv].d+1; 97 q[0][bv].opt=2; 98 q[0][bv].par=1; 99 q[0][bv].prea=av; 100 q[0][bv].preb=bv; 101 a.push(0); 102 b.push(bv); 103 104 } 105 if (bv<B&&q[av][B].d==-1) 106 { 107 q[av][B].d=q[av][bv].d+1; 108 q[av][B].opt=1; 109 q[av][B].par=2; 110 q[av][B].prea = av; 111 q[av][B].preb = bv; 112 a.push(av); 113 b.push(B); 114 115 } 116 if (bv>0&&q[av][0].d==-1) 117 { 118 q[av][0].d=q[av][bv].d+1; 119 q[av][0].opt=2; 120 q[av][0].par=2; 121 q[av][0].prea=av; 122 q[av][0].preb=bv; 123 a.push(av); 124 b.push(0); 125 } 126 127 if (av+bv<=B&&q[0][av+bv].d==-1) 128 { 129 q[0][av+bv].d=q[av][bv].d+1; 130 q[0][av+bv].opt=3; 131 q[0][av+bv].par=1; 132 q[0][av+bv].prea=av; 133 q[0][av+bv].preb=bv; 134 a.push(0); 135 b.push(av+bv); 136 } 137 if (av+bv>B&&q[av-(B-bv)][B].d==-1) //把1往2里倒入的两种情况 138 { 139 140 int tmp = av-(B-bv); 141 q[tmp][B].d=q[av][bv].d+1; 142 q[tmp][B].opt=3; 143 q[tmp][B].par=1; 144 q[tmp][B].prea=av; 145 q[tmp][B].preb=bv; 146 a.push(tmp); 147 b.push(B); 148 } 149 150 if (bv+av<=A&&q[av+bv][0].d==-1) 151 { 152 q[av+bv][0].d=q[av][bv].d+1; 153 q[av+bv][0].opt=3; 154 q[av+bv][0].par=2; 155 q[av+bv][0].prea=av; 156 q[av+bv][0].preb=bv; 157 a.push(av+bv); 158 b.push(0); 159 } 160 if (bv+av>A&&q[A][bv-(A-av)].d==-1) 161 { 162 int tmp = bv-(A-av); 163 q[A][tmp].d=q[av][bv].d+1; 164 q[A][tmp].opt=3; 165 q[A][tmp].par=2; 166 q[A][tmp].prea=av; 167 q[A][tmp].preb=bv; 168 a.push(A); 169 b.push(tmp); 170 } 171 } 172 } 173 int main() 174 { 175 176 flag = false; 177 cin>>A>>B>>C; 178 bfs(); 179 if (!flag) 180 { 181 cout<<"impossible"<<endl; 182 } 183 184 185 return 0; 186 }

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 **

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 **

迷宫问题 1 2 3 4 /************************************************************************* 5 > File Name: code/2015summer/searching/KK.cpp 6 > Author: 111qqz 7 > Email: rkz2013@126.com 8 > Created Time: 2015年07月25日 星期六 13时33分00秒 9 ************************************************************************/ 10 11 #include<iostream> 12 #include<iomanip> 13 #include<cstdio> 14 #include<algorithm> 15 #include<cmath> 16 #include<cstring> 17 #include<string> 18 #include<map> 19 #include<set> 20 #include<queue> 21 #include<vector> 22 #include<stack> 23 #define y0 abc111qqz 24 #define y1 hust111qqz 25 #define yn hez111qqz 26 #define j1 cute111qqz 27 #define tm crazy111qqz 28 #define lr dying111qqz 29 using namespace std; 30 #define REP(i, n) for (int i=0;i<int(n);++i) 31 typedef long long LL; 32 typedef unsigned long long ULL; 33 int a[10][10]; 34 int head = 0; 35 int tail = 1; 36 int dirx[2]={1,0}; 37 int diry[2]={0,1}; 38 struct node 39 { 40 int x,y,pre; 41 }q[10]; 42 43 void print(int x) 44 { 45 if (q[x].pre!=-1) 46 { 47 print(q[x].pre); 48 printf("(%d, %d)\n",q[x].x,q[x].y); 49 } 50 } 51 void bfs() 52 { 53 q[head].x=0; 54 q[head].y=0; 55 q[head].pre=-1; 56 while (head<tail) 57 { 58 if (q[head].x==4&&q[head].y==4) 59 { 60 print(head); 61 return; 62 } 63 for (int i = 0 ; i < 2 ; i++ ) 64 { 65 int newx=dirx[i]+q[head].x; 66 int newy=diry[i]+q[head].y; 67 if (newx>=0&&newx<5&&newy>=0&&newy<5&&a[newx][newy]==0) 68 { 69 q[tail].x=newx; 70 q[tail].y=newy; 71 q[tail].pre=head; 72 tail++; 73 } 74 } 75 head++; 76 } 77 78 } 79 int main() 80 { 81 for ( int i = 0 ; i < 5 ; i++ ) 82 { 83 for ( int j = 0 ; j < 5; j++) 84 { 85 cin>>a[i][j]; 86 } 87 } 88 printf("(0, 0)\n"); 89 bfs(); 90 91 return 0; 92 }

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．

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.

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…

http://poj.org/problem?id=3278 bfs,用到了stl的queue 1 2 3 /* *********************************************** 4 Author :111qqz 5 Created Time :2016年02月19日 星期五 15时45分05秒 6 File Name :3278.cpp 7 ************************************************ */ 8 9 #include <algorithm> 10 #include <cstdio> 11 #include <iostream> 12 #include <cstring> 13 #include <string> 14 #include <cmath> 15 #include <map> 16 #include <stack> 17 #include <queue> 18 19 using namespace std; 20 typedef long long LL; 21 const int inf = 8E8; 22 const int N=2E5+7; 23 int d[N]; 24 int n,k; 25 void bfs() 26 { 27 queue<int> q; 28 memset(d,-1,sizeof(d)); 29 q.push(n); 30 d[n]=0; 31 while (!q.empty()) 32 { 33 int x = q.front(); 34 q.pop(); 35 if ( x==k ) 36 { 37 break; 38 } 39 int next[10]; 40 next[1]=x-1; 41 next[2]=x+1; 42 next[3]=2*x; 43 for ( int i = 1; i <= 3 ; i++ ) 44 { 45 if (next[i]>=0&&next[i]<=100000&&d[next[i]]==-1) 46 { 47 d[next[i]]=d[x]+1; 48 q.push(next[i]); 49 } 50 } 51 } 52 53 54 } 55 int main() 56 { 57 while (scanf("%d %d",&n,&k)!=EOF) 58 { 59 bfs(); 60 cout<<d[k]<<endl; 61 } 62 return 0; 63 }