刷题记录 kuangbin带你飞专题一：简单搜索

两天肝完
感觉自己的暴力水平上升了
码力大仙！！！

A.POJ1321 棋盘问题

有一个坑点就是可能相邻层没有棋盘，所以直接搜全部

#include<cstdio>
#include<iostream>
#include<cstring>
#include<string>
#include<algorithm>
#include<cmath>
#include<queue>
#include<vector>
using namespace std;
int n,m;
int dx[]={0,1,1,1,-1,-1,-1,0,0};
int dy[]={0,1,-1,0,1,-1,0,1,-1};
char a[15][15];
int zx[1005];
int zy[1005],vis[1005];
int t[20][20];
int ans;
void fuck();
void dfs(int x,int num){if(num==0){ans++;return;}for(int i=x;i<n;i++){for(int j=0;j<n;j++){if(a[i][j]=='#'&&vis[j]==0){vis[j]=1;dfs(i+1,num-1);vis[j]=0;}}}return;
}
int main(){while(scanf("%d%d",&n,&m)){if(n==-1&&m==-1)break;for(int i=0;i<n;i++){scanf("%s",&a[i]);}memset(vis,0,sizeof(vis));ans=0;dfs(0,m);cout<<ans<<endl;}return 0;
}

B.POJ 2251 Dungeon Master

裸三维BFS，没有坑点
就是POJ很神奇，把bfs过程直接暴力就让我MLE？
让我学会了能不写在主函数里就一定写外面

#include<cstdio>
#include<iostream>
#include<cstring>
#include<string>
#include<algorithm>
#include<cmath>
#include<queue>
#include<vector>
using namespace std;
int n,m,k;
char a[35][35][35];
int vis[35][35][35];
int ans;
int sx,sy,sz,zx,zy,zz;
int base[6][3] = { {-1,0,0},{1,0,0},{0,-1,0},{0,1,0},{0,0,-1},{0,0,1} };
struct node{int x,y,z,bu;friend bool operator<(node a,node b){return a.bu>b.bu;}
};
bool zou(int x,int y,int z){if(x<1||x>k)return false;if(y<1||y>m)return false;if(z<1||z>n)return false;return true;
}
int main(){while(scanf("%d%d%d",&n,&m,&k)){if(n==0&&m==0&&k==0)break;int fa=0;memset(vis,0,sizeof(vis));for(int i=1;i<=n;i++){for(int j=1;j<=m;j++){for(int t=1;t<=k;t++){cin>>a[t][j][i];if(a[t][j][i]=='S'){sx=t;sy=j;sz=i;}if(a[t][j][i]=='E'){zx=t;zy=j;zz=i;a[t][j][i]='.';}}}}priority_queue<node>q;q.push((node){sx,sy,sz,0});while(!q.empty()){node x=q.top();q.pop();if(x.x==zx&&x.y==zy&&x.z==zz){printf("Escaped in %d minute(s).\n",x.bu);fa=1;break;}for(int i=0;i<6;i++){int xx=x.x+base[i][0];int yy=x.y+base[i][1];int zz=x.z+base[i][2];if(zou(xx,yy,zz)&&a[xx][yy][zz]!='#'&&vis[xx][yy][zz]==0){vis[xx][yy][zz]=1;q.push((node){xx,yy,zz,x.bu+1});}}}if(fa==0)printf("Trapped!\n");}return 0;
}

C.POJ 3278 Catch That Cow

一维BFS
控制好边界就行，纯手速秒杀题
（某秒因为边界RE了十多发）

#include<cstdio>
#include<iostream>
#include<cstring>
#include<string>
#include<algorithm>
#include<cmath>
#include<queue>
#include<vector>
using namespace std;
int n,m,k;
int a[100005];
struct node
{int x,num;
};
int vis[100005];
int main(){cin>>n>>m;queue<node>q;q.push((node){n,0});vis[n]=1;while(!q.empty()){node x=q.front();q.pop();if(x.x==m){cout<<x.num;return 0;}if(x.x+1<=100000&&x.x+1>=0&&vis[x.x+1]==0){q.push((node){x.x+1,x.num+1});vis[x.x+1]=1;}if(x.x-1<=100000&&x.x-1>=0&&vis[x.x-1]==0){q.push((node){x.x-1,x.num+1});vis[x.x-1]=1;}if(x.x*2<=100000&&x.x*2>=0&&vis[x.x*2]==0){q.push((node){x.x*2,x.num+1});vis[x.x*2]=1;}}return 0;
}

D.POJ 3279 Fliptile

经典开关问题，让我第一次接触到了状态压缩hhh
如果会状压好像也就是个暴力了

#include <cstdio>
#include <cstring>
#include <queue>
#include <iostream>
using namespace std;
int m,n;
int a[20][20],cal[20][20],out[20][20];
const int dx[]={0,0,1,-1,0};
const int dy[]={1,-1,0,0,0};
int zt(int x,int y){int temp=a[x][y];for(int i=0;i<5;i++){int xx=x+dx[i];int yy=y+dy[i];if(xx<1||yy<1||xx>n||yy>m)continue;temp+=cal[xx][yy];}return temp%2;
}
int dfs(){for(int i=2;i<=n;i++){for(int j=1;j<=m;j++){if(zt(i-1,j)){cal[i][j]=1;}}}for(int i=1;i<=m;i++){if(zt(n,i))return -1;}int res=0;for(int i=1;i<=n;i++){for(int j=1;j<=m;j++){res+=cal[i][j];}}return res;
}
int main(){cin>>n>>m;for(int i=1;i<=n;i++){for(int j=1;j<=m;j++){cin>>a[i][j];}}int fa=0;int ans=0x3f3f3f3f;for(int i=0;i<1<<m;i++){memset(cal,0,sizeof(cal));for(int j=1;j<=m;j++){cal[1][m-j+1]=i>>(j-1)&1;}int c=dfs();if(c>=0&&c<ans){fa=1;ans=c;memcpy(out,cal,sizeof(cal));}}if(!fa)cout<<"IMPOSSIBLE"<<endl;else{for(int i=1;i<=n;i++){for(int j=1;j<=m;j++){if(j!=1)cout<<" ";cout<<out[i][j];}cout<<endl;}}return 0;
}

E.POJ 1426 Find The Multiple

思路就是直接暴力构造搜 *10 和 *10+1
判断是否能整除

#include<cstdio>
#include<iostream>
#include<cstring>
#include<string>
#include<algorithm>
#include<cmath>
#include<queue>
#include<vector>
#define int long long
using namespace std;
int n,m,k,c;
void dfs(unsigned long long x,int t){if(t>19)return;if(c==1)return;if(x%n==0){cout<<x<<endl;c=1;return;}dfs(x*10,t+1);dfs(x*10+1,t+1);return;
}
signed main(){while(1){cin>>n;c=0;if(n==0)break;dfs(1,0);}return 0;
}

F.POJ 3126 Prime Path

一个有点意思的搜索，不过难度不大，思路就先跑一遍欧拉筛，然后写一个更换数字的函数，暴力循环判断是否为素数，加入队列中就ok了

#include<cstdio>
#include<iostream>
#include<cstring>
#include<string>
#include<algorithm>
#include<cmath>
#include<queue>
#include<vector>
#define int long long
using namespace std;
int n,m,k,c;
int prime[10005];
bool sf[10005];
const int maxn=10005;
int vis[10005];
int cnt=1;
void sushu(){         int num=0;       memset(sf,true,sizeof(sf));for(int i=2;i<=maxn;i++){          if(sf[i]) prime[++num]=i;     for(int j=1;j<=num;j++){     if(i*prime[j]>maxn) break; sf[i*prime[j]]=false;      if(i%prime[j]==0) break;   }}sf[1]=false;sf[0]=false;
}
int bian(int x,int t,int c){int ge,shi,bai,qian;ge=x%10;x/=10;shi=x%10;x/=10;bai=x%10;qian=x/=10;if(t==1)return c*1000+bai*100+shi*10+ge;else if(t==2)return qian*1000+c*100+shi*10+ge;else if(t==3)return qian*1000+bai*100+c*10+ge;else if(t==4)return qian*1000+bai*100+shi*10+c;
}
struct node{int x,num;
};
void bfs(int x){queue<node>q;q.push((node){x,0});vis[x]=1;while(!q.empty()){node u=q.front();q.pop();if(u.x==m){cout<<u.num<<endl;return;}for(int i=1;i<=4;i++){for(int j=0;j<=9;j++){if(i==4&&(j==2||j==4||j==6||j==8))continue;if(i==1&&j==0)continue;int tt=bian(u.x,i,j);if(sf[tt]==true&&vis[tt]==0){q.push((node){tt,u.num+1});vis[tt]=1;}}}}
}
signed main(){cin>>k;sushu();
//  if(sf[50])cout<<"1";
//  else cout<<"0";while(k--){cin>>n>>m;if(n==m){cout<<"0"<<endl;continue;}memset(vis,0,sizeof(vis));bfs(n);}return 0;
}

G.POJ 3087 Shuffle’m Up

傻逼题，难度不够，英语来凑，直接模拟就行

#include <cstdio>
#include <cstring>
#include <queue>
#include <iostream>
#include <map>
using namespace std;
int n,c;
string a,b,de;
void dfs(int x){map<string,int>m;string t=a+b;int step=0;while(m[t]==0&&t!=de){int k=0;m[t]=1;step++;string str=t;for(int i=0;i<c;i++){str[k]=t[i+c];k++;str[k]=t[i];k++;}t=str;}if(t==de){cout<<x<<" "<<step<<endl;}else{cout<<x<<" "<<"-1"<<endl;}
}
int main(){cin>>n;int kk=0;while(n--){kk++;cin>>c;cin>>a>>b>>de;dfs(kk);}return 0;
}

H.POJ 3414 Pots

这题真的是个超级码力题
写完就感觉自己码力很猛
暴力判断八种情况，我这里开结构体里面放了一个数组表示第一和第二个杯子，并用vector来保存路径

#include <iostream>
#include <cstdio>
#include <fstream>
#include <algorithm>
#include <cmath>
#include <deque>
#include <vector>
#include <queue>
#include <string>
#include <cstring>
#include <map>
#include <stack>
#include <set>
using namespace std;
void fuck();
void shit();
int a,b,c,n,m;
bool vis[102][102];
struct node{int w[2],num;vector<string>v;
};
int fill(int x,int c){return c;
}
int drop(int x){return 0;
}
void bfs(int x,int b,int c){node t;int fa=0;t.w[0]=0;t.w[1]=0;t.num=0;vis[0][0]=false;queue<node>q;q.push(t);while(!q.empty()){node t=q.front();q.pop();if(t.w[0]==c||t.w[1]==c){cout<<t.num<<endl;for(int i=0;i<t.v.size();i++){cout<<t.v[i]<<endl;}fa=1;return;}if(vis[fill(1,a)][t.w[1]]==true){vis[fill(1,a)][t.w[1]]=false;node a=t;a.w[0]=fill(1,x);a.num++;a.v.push_back("FILL(1)");q.push(a);}if(vis[t.w[0]][fill(2,b)]==true){vis[t.w[0]][fill(2,b)]=false;node a=t;a.w[1]=fill(2,b);a.num++;a.v.push_back("FILL(2)");q.push(a);}if(vis[0][t.w[1]]==true){vis[0][t.w[1]]=false; node a=t;a.w[0]=0;a.num++;a.v.push_back("DROP(1)");q.push(a);}if(vis[t.w[0]][0]==true){vis[t.w[0]][0]=false;node a=t;a.w[1]=0;a.num++;a.v.push_back("DROP(2)");q.push(a);}if(t.w[0]+t.w[1]>b){int cha=t.w[0]+t.w[1]-b;if(vis[cha][b]==true){vis[cha][b]=false;node a=t;a.w[0]=cha;a.w[1]=b;a.num++;a.v.push_back("POUR(1,2)");q.push(a);}}else{int cha=t.w[0]+t.w[1];if(vis[0][cha]==true){vis[0][cha]=false;node a=t;a.w[0]=0;a.w[1]=cha;a.num++;a.v.push_back("POUR(1,2)");q.push(a);}}if(t.w[0]+t.w[1]>x){int cha=t.w[0]+t.w[1]-x;if(vis[x][cha]==true){vis[x][cha]=false;node a=t;a.w[0]=x;a.w[1]=cha;a.num++;a.v.push_back("POUR(2,1)");q.push(a);}}else{int cha=t.w[0]+t.w[1];if(vis[cha][0]==true){vis[cha][0]=false;node a=t;a.w[0]=cha;a.w[1]=0;a.num++;a.v.push_back("POUR(2,1)");q.push(a);}}}if(!fa)cout<<"impossible";return;
}
int main(){cin>>a>>b>>c;memset(vis,true,sizeof(vis));bfs(a,b,c);
}

I.FZU 2150 Fire Game

码力屑题，让我自闭了一下午，不想写

J.UVA 11624 Fire!

对火和人分别bfs，并记录时间，需要注意的坑点是可能有多团火，然后注意边界

#include<bits/stdc++.h>
using namespace std;
const int dx[4]={1,-1,0,0};
const int dy[4]={0,0,-1,1};
const int inf=99999999;
int vis[1505][1505],f[1505][1505];
char a[1505][1505];
int n,m,c,tt;
struct node{int x,y,t;
}gg;
int main(){int t;int sx,sy;cin>>t;while(t--){cin>>n>>m;c=0;queue<node>fire;queue<node>q;memset(vis,0,sizeof(vis));memset(a,'.',sizeof(a));for(int i=1;i<=n;i++){for(int j=1;j<=m;j++){cin>>a[i][j];f[i][j]=inf;if(a[i][j]=='J'){sx=i;sy=j;}if(a[i][j]=='F'){fire.push((node){i,j,0});vis[i][j]=1;}}}while(!fire.empty()){node s=fire.front();fire.pop();for(int i=0;i<4;i++){int xx=s.x+dx[i];int yy=s.y+dy[i];if(xx>=1&&yy>=1&&xx<=n&&yy<=m&&vis[xx][yy]==0&&a[xx][yy]=='.'){vis[xx][yy]=1;f[xx][yy]=s.t+1;fire.push((node){xx,yy,s.t+1});}}}memset(vis,0,sizeof(vis));q.push((node){sx,sy,0});vis[sx][sy]=1;while(!q.empty()){node s=q.front();q.pop();if((s.x<=1||s.y<=1||s.x>=n||s.y>=m)&&s.t<f[s.x][s.y]){c++;cout<<s.t+1<<endl;break;}for(int i=0;i<4;i++){int xx=s.x+dx[i];int yy=s.y+dy[i];if(xx>=1&&yy>=1&&xx<=n&&yy<=m&&vis[xx][yy]==0&&a[xx][yy]=='.'){vis[xx][yy]=1;q.push((node){xx,yy,s.t+1});}}}if(!c){cout<<"IMPOSSIBLE"<<endl;}}
}

K.POJ 3984 迷宫问题

弱智题

L.HDU 1241 Oil Deposits

问有多少个联通块
直接循环，dfs走过的点变为普通路，dfs的次数就是答案

#include<cstdio>
#include<iostream>
#include<cstring>
#include<string>
#include<algorithm>
#include<cmath>
#include<queue>
#include<vector>
#define int long long
using namespace std;
int n,m,k,c;
int prime[10005];
bool sf[10005];
const int maxn=10005;
int vis[105][105];
char a[1005][1005];
const int dx[]={1,1,1,-1,-1,-1,0,0};
const int dy[]={1,-1,0,1,-1,0,1,-1};
int cnt=1;
void dfs(int x,int y){for(int i=0;i<8;i++){int xx=x+dx[i];int yy=y+dy[i];if(xx>=1&&yy>=1&&xx<=n&&yy<=m&&a[xx][yy]=='@'){a[xx][yy]='*';dfs(xx,yy);}}return;
}
signed main(){while(1){cin>>n>>m;int ans=0;if(n==0&&m==0)break;for(int i=1;i<=n;i++){for(int j=1;j<=m;j++){cin>>a[i][j];}}for(int i=1;i<=n;i++){for(int j=1;j<=m;j++){if(a[i][j]=='@'){a[i][j]='*';dfs(i,j);//           a[i][j]='*';ans++;}}}cout<<ans<<endl;}
}

M.HDU 1495 非常可乐

又是一个码力屑题，纯暴力bfs
如果变量没写好可能容易自闭

#include<cstdio>
#include<iostream>
#include<cstring>
#include<string>
#include<algorithm>
#include<cmath>
#include<queue>
#include<vector>
using namespace std;
int n,m,k,c;
int v[5];
struct node{int v[3],num;node(int x,int y,int z,int st){v[0]=x;v[1]=y;v[2]=z;num=st;}
};
bool vis[102][102][102];
const int d1[6] = {0, 0, 1, 1, 2, 2};
const int d2[6] = {1, 2, 0, 2, 0, 1};
void bfs(){queue<node>q;memset(vis,true,sizeof(vis));q.push(node{v[0],0,0,0});vis[v[0]][0][0]=false;while(!q.empty()){node x=q.front();q.pop();int cnt=0;int dd=v[0]/2;for(int i=0;i<3;i++){if(x.v[i]==dd){cnt++;}if(cnt>=2){cout<<x.num<<endl;return;}}for(int i=0;i<6;i++){node in=x;in.num++;if(x.v[d1[i]]){int cha=v[d2[i]]-x.v[d2[i]];if(cha>=x.v[d1[i]]){in.v[d2[i]]+=in.v[d1[i]];in.v[d1[i]]=0;}else{in.v[d1[i]]-=cha;in.v[d2[i]]+=cha;}if(vis[in.v[0]][in.v[1]][in.v[2]]==true){q.push(in);vis[in.v[0]][in.v[1]][in.v[2]]=false;}}}}printf("NO\n");
}
signed main(){while(scanf("%d%d%d",&v[0],&v[1],&v[2])){if(v[0]==0&&v[1]==0&&v[2]==0)break;if(v[0]%2==1){cout<<"NO"<<endl;continue;}bfs();}return 0;
}

N.HDU 2612 Find a way

跑两个bfs，记录两人到达每一个KFC的时间，最后循环求出最小时间（今晚刚吃KFC）

#include<cstdio>
#include<iostream>
#include<cstring>
#include<string>
#include<algorithm>
#include<cmath>
#include<queue>
#include<vector>
using namespace std;
int n,m,k,c;
const int maxn=10005;
int vis[205][205];
int cnt,cnt1,cnt2;
char a[1005][1005];
const int dx[]={1,-1,0,0};
const int dy[]={0,0,1,-1};
struct node{int x,y,num;
};
int dis[205][205];
signed main(){while(scanf("%d%d",&n,&m)!=EOF){int ans=99999999;vector<int>v1,v2;c=0;if(n==0&&m==0)break;cnt=0,cnt1=0,cnt2=0;int sx1,sy1,sx2,sy2;memset(dis,0,sizeof(dis));for(int i=1;i<=n;i++){for(int j=1;j<=m;j++){cin>>a[i][j];vis[i][j]=0;if(a[i][j]=='Y'){sx1=i;sy1=j;}if(a[i][j]=='M'){sx2=i;sy2=j;}if(a[i][j]=='@'){v1.push_back(j);v2.push_back(i);}}}queue<node>q;q.push((node){sx1,sy1,0});vis[sx1][sy1]=1;while(!q.empty()){node x=q.front();q.pop();if(a[x.x][x.y]=='@'){dis[x.x][x.y]+=x.num;}for(int j=0;j<4;j++){int xx=dx[j]+x.x;int yy=dy[j]+x.y;if(xx>=1&&yy>=1&&xx<=n&&yy<=m&&vis[xx][yy]==0&&a[xx][yy]!='#'){vis[xx][yy]=1;q.push((node){xx,yy,x.num+1});}}}while(!q.empty()){q.pop();}memset(vis,0,sizeof(vis));q.push((node){sx2,sy2,0});vis[sx2][sy2]=1;while(!q.empty()){node x=q.front();q.pop();if(a[x.x][x.y]=='@'){dis[x.x][x.y]+=x.num;}for(int j=0;j<4;j++){int xx=dx[j]+x.x;int yy=dy[j]+x.y;if(xx>=1&&yy>=1&&xx<=n&&yy<=m&&vis[xx][yy]==0&&a[xx][yy]!='#'){vis[xx][yy]=1;q.push((node){xx,yy,x.num+1});}}}ans=99999999;for(int i=1;i<=n;i++){for(int j=1;j<=m;j++){if(dis[i][j]!=0)ans=min(ans,dis[i][j]);}}cout<<ans*11<<endl;}
}

虽然这个搜索专题难度不大但是码力是真的不小…
搜索（×）锻炼暴力水平（√）
下一个专题：最短路
加油！！！