BFS

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 代码实现 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 ** Problem Description 大家一定觉的运动以后喝可乐是一件很惬意的事情，但是seeyou却不这么认为。因为每次当seeyou买了可乐以后，阿牛就要求和seeyou一起分享这一瓶可乐，而且一定要喝的和seeyou一样多。但seeyou的手中只有两个杯子，它们的容量分别是N 毫升和M 毫升 可乐的体积为S （S<101）毫升　(正好装满一瓶) ，它们三个之间可以相互倒可乐 (都是没有刻度的，且 S==N+M，101＞S＞0，N＞0，M＞0) 。聪明的ACMER你们说他们能平分吗？如果能请输出倒可乐的最少的次数，如果不能输出"NO"。

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 Problem Description Pass a year learning in Hangzhou, yifenfei arrival hometown Ningbo at finally. Leave Ningbo one year, yifenfei have many people to meet. Especially a good friend Merceki. Yifenfei’s home is at the countryside, but Merceki’s home is in the center of city. So yifenfei made arrangements with Merceki to meet at a KFC. There are many KFC in Ningbo, they want to choose one that let the total time to it be most smallest. Now give you a Ningbo map, Both yifenfei and Merceki can move up, down ,left, right to the adjacent road by cost 11 minutes.

迷宫问题 代码实现 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． 有一点疑惑，就是map的初始值 比如我定义的　map<string,int>d；它的初始的value是什么？随机值？０？还是什么，百度了下，没找到，求指教．

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… Escaped in %d minute(s) 这意思是答案为1输出minute，不为1输出minutes还是说是不是1都输出minute(s)? 试了下，答案是后者。

http://poj.org/problem?id=3278 bfs，用到了stl的queue 代码实现 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 }