参考书籍《算法竞赛入门到进阶》
poj2386 题目链接 http://poj.org/problem?id=2386
注意理解题目中的所谓的池塘,记录搜索次数即可。
AC代码:
![](https://images.cnblogs.com/OutliningIndicators/ContractedBlock.gif)
1 #include <iostream> 2 #include <string.h> 3 using namespace std; 4 const int maxn=150; 5 char a[maxn][maxn]; 6 int dx[8]={1,1,1,-1,-1,-1,0,0}; 7 int dy[8]={1,-1,0,1,-1,0,1,-1}; 8 int n,m; 9 void dfs(int x,int y) 10 { 11 a[x][y]='.'; 12 for (int i = 0; i < 8; ++i) 13 { 14 int nx = x+dx[i]; 15 int ny = y+dy[i]; 16 if (nx>=0&&nx<n&&ny>=0&&ny<m&&a[nx][ny]=='W') dfs(nx,ny); 17 } 18 return ; 19 } 20 int main(int argc, char const *argv[]) 21 { 22 cin>>n>>m; 23 int res=0; 24 for (int i = 0; i < n; ++i) 25 { 26 for (int j = 0; j < m; ++j) 27 { 28 cin>>a[i][j]; 29 } 30 } 31 for (int i = 0; i < n; ++i) 32 { 33 for (int j = 0; j < m; ++j) 34 { 35 if (a[i][j]=='W') 36 { 37 dfs(i,j); 38 res++; 39 } 40 } 41 } 42 cout<<res<<endl; 43 return 0; 44 }
hdu2553 题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=2553
经典深搜问题
AC代码:
![](https://images.cnblogs.com/OutliningIndicators/ContractedBlock.gif)
1 #include <bits/stdc++.h> 2 using namespace std; 3 int n,tot=0; 4 int col[12]; 5 bool check(int c,int r) 6 { 7 for (int i = 0; i < r; ++i) 8 { 9 if (col[i]==c||(abs(col[i]-c)==abs(i-r))) return false; 10 } 11 return true; 12 } 13 void dfs(int r) 14 { 15 if (r==n) 16 { 17 tot++; 18 return ; 19 } 20 for (int c = 0; c < n; ++c) 21 { 22 if (check(c,r)) 23 { 24 col[r]=c; 25 dfs(r+1); 26 } 27 } 28 } 29 int main(int argc, char const *argv[]) 30 { 31 int ans[12]={0}; 32 for (n = 0; n <= 10; ++n) 33 { 34 memset(col,0,sizeof(col)); 35 tot=0; 36 dfs(0); 37 ans[n]=tot; 38 } 39 while(cin>>n) 40 { 41 if (n==0) return 0; 42 cout<<ans[n]<<endl; 43 } 44 return 0; 45 }
poj2531 题目链接:http://poj.org/problem?id=2531
题目要求两个集合中点的距离和最大,我们先假设都在一个集合,这样距离和为0,然后一个一个地往另外一个集合放,距离和变小则返回,变大继续,过程中不断更新最大值。
AC代码:
![](https://images.cnblogs.com/OutliningIndicators/ContractedBlock.gif)
1 #include <iostream> 2 #include <string.h> 3 using namespace std; 4 const int inf=0x3f3f3f3f; 5 int a[25][25]; 6 int g[25]; 7 int res; 8 int n; 9 void dfs(int step,int sum) 10 { 11 g[step]=1; 12 int num=sum; 13 for (int i = 0; i < n; ++i) 14 { 15 if (g[i]==1) num-=a[step][i]; 16 else num+=a[step][i]; 17 } 18 res=max(res,num); 19 for (int i = step+1; i < n; ++i) 20 { 21 if (num>sum) 22 { 23 dfs(i,num); 24 g[i]=0; 25 } 26 } 27 } 28 int main(int argc, char const *argv[]) 29 { 30 while(cin>>n) 31 { 32 memset(a,0,sizeof(a)); 33 memset(g,0,sizeof(g)); 34 for (int i = 0; i < n; ++i) 35 { 36 for (int j = 0; j < n; ++j) 37 { 38 cin>>a[i][j]; 39 } 40 } 41 res=-0x3f3f3f3f; 42 dfs(0,0); 43 cout<<res; 44 } 45 return 0; 46 }
poj1416 题目链接:http://poj.org/problem?id=1416
预处理能被切成的情况,然后深搜,借鉴了网上的AC代码
AC代码:
![](https://images.cnblogs.com/OutliningIndicators/ContractedBlock.gif)
1 #include <iostream> 2 #include <algorithm> 3 using namespace std; 4 #define MAXN 10000 5 int a,len,flag,rejected,k; 6 string b; 7 struct ss 8 { 9 int goal; 10 string str; 11 }score[MAXN],edge[10][10]; 12 bool dfs(int now,int ret,string tmp_str) 13 { 14 if (ret>a||now>len) return false; 15 if (ret<=a&&now==len) 16 { 17 score[k].goal=ret; 18 score[k].str=tmp_str; 19 k++; 20 flag=1; 21 return true; 22 } 23 int next=now+1; 24 for (int i = next; i <=len; ++i) dfs(i,ret+edge[next][i].goal,tmp_str+" "+edge[next][i].str); 25 } 26 void build_map() 27 { 28 for (int i = 1; i <= len; ++i) 29 { 30 for (int j = i; j <= len; ++j) 31 { 32 if (i==j) 33 { 34 edge[i][j].goal=b[i]-'0'; 35 edge[i][j].str=b[i]; 36 } 37 else 38 { 39 edge[i][j].goal=edge[i][j-1].goal*10+b[j]-'0'; 40 edge[i][j].str=edge[i][j-1].str+b[j]; 41 } 42 } 43 } 44 } 45 bool cmp(ss aa,ss bb) 46 { 47 return aa.goal>bb.goal; 48 } 49 int main(int argc, char const *argv[]) 50 { 51 while(cin>>a>>b) 52 { 53 if (!a&&b[0]=='0') break; 54 len = b.length(); 55 b=" "+b; 56 build_map(); 57 flag=rejected=0,k=1; 58 dfs(0,0,""); 59 if (k>1) sort(score+1,score+k,cmp); 60 if (score[1].goal==score[2].goal&&score[1].goal!=0) rejected=1; 61 if (flag&&!rejected) cout<<score[1].goal<<score[1].str<<endl; 62 else if (rejected) cout<<"rejected"<<endl; 63 else cout<<"error"<<endl; 64 } 65 return 0; 66 }
poj2676 题目链接:http://poj.org/problem?id=2676
九宫格,开三个数组,一个用来存同一行该数是否被用过,一个用来存同一列该数是否被用过,剩下一个用来存该宫格里是否被用过
AC代码:
![](https://images.cnblogs.com/OutliningIndicators/ContractedBlock.gif)
1 #include <iostream> 2 #include <string.h> 3 using namespace std; 4 int mapp[10][10];//初始图 5 bool h[10][10];//行 6 bool l[10][10];//列 7 bool g[10][10];//格 8 bool flage=false; 9 bool dfs(int x,int y) 10 { 11 if (x==9) 12 { 13 flage=true; 14 return true; 15 } 16 if (mapp[x][y]) 17 { 18 if(y==8) dfs(x+1,0); 19 else dfs(x,y+1); 20 if (flage) return true; 21 else return false; 22 } 23 else 24 { 25 int k=x/3*3+y/3; 26 for (int i = 1; i <= 9; ++i)//遍历1-9的9个数字 27 { 28 if (!h[x][i]&&!l[y][i]&&!g[k][i]) 29 { 30 mapp[x][y]=i; 31 h[x][i]=true; 32 l[y][i]=true; 33 g[k][i]=true; 34 if(y==8) dfs(x+1,0); 35 else dfs(x,y+1); 36 if (flage) 37 { 38 return true; 39 } 40 else 41 { 42 mapp[x][y]=0; 43 h[x][i]=false; 44 l[y][i]=false; 45 g[k][i]=false; 46 } 47 48 } 49 } 50 } 51 return false; 52 } 53 int main(int argc, char const *argv[]) 54 { 55 int t; 56 cin>>t; 57 while(t--) 58 { 59 memset(h,false,sizeof(h)); 60 memset(l,false,sizeof(l)); 61 memset(g,false,sizeof(g)); 62 flage=false; 63 char c; 64 for (int i = 0; i < 9; ++i) 65 { 66 for (int j = 0; j < 9; ++j) 67 { 68 cin>>c; 69 mapp[i][j]=c-'0'; 70 if (mapp[i][j]) 71 { 72 int k=i/3*3+j/3; 73 h[i][mapp[i][j]]=true; 74 l[j][mapp[i][j]]=true; 75 g[k][mapp[i][j]]=true; 76 } 77 } 78 } 79 if(dfs(0,0)) 80 { 81 for (int i = 0; i < 9; ++i) 82 { 83 for (int j = 0; j < 9; ++j) 84 { 85 cout<<mapp[i][j]; 86 } 87 cout<<endl; 88 } 89 } 90 } 91 return 0; 92 }
poj1129 题目链接:http://poj.org/problem?id=1129
本题我没有使用深搜,用四色定理直接染色输出染色数即可,注意染同一颜色的是与其前被染的都不相邻才行,这是写代码时容易错的一个地方。
附样例:
5
A:BE
B:AC
C:BD
D:CE
E:AD
该样例应该输出3
AC代码:
![](https://images.cnblogs.com/OutliningIndicators/ContractedBlock.gif)
1 #include <iostream> 2 #include <string.h> 3 #include <algorithm> 4 using namespace std; 5 struct node 6 { 7 char a; 8 int number; 9 bool f; 10 }aa[100]; 11 bool mapp[30][30]; 12 node bb[100]; 13 int cmp(const node &a,const node &b) 14 { 15 return a.number>b.number; 16 } 17 int main(int argc, char const *argv[]) 18 { 19 int t; 20 while(cin>>t) 21 { 22 memset(mapp,false,sizeof(mapp)); 23 memset(aa,0,sizeof(aa)); 24 string ss; 25 if (t==0) break; 26 else 27 { 28 for (int i = 0; i < t; ++i) 29 { 30 cin>>ss; 31 int len=ss.length(); 32 aa[i].a=ss[0]; 33 aa[i].number=len-2; 34 aa[i].f=false; 35 for (int j = 2; j < len; ++j) 36 { 37 mapp[ss[0]-'A'][ss[j]-'A']=true; 38 } 39 } 40 sort(aa,aa+t,cmp); 41 int num=0; 42 for (int i = 0; i < t; ++i) 43 { 44 if (aa[i].f==false) 45 { 46 int kk=0; 47 memset(bb,0,sizeof(bb)); 48 aa[i].f=true; 49 bb[kk].a=aa[i].a; 50 kk++; 51 num++; 52 //cout<<num<<" "<<aa[i].a<<endl; 53 for (int j = 0; j < t; ++j) 54 { 55 if (aa[i].a-'A'==j) continue; 56 bool flage=true; 57 for (int ii = 0; ii < kk; ++ii) 58 { 59 if (mapp[bb[ii].a-'A'][j]!=false) 60 { 61 flage=false; 62 } 63 } 64 if (flage) 65 { 66 for (int k = 0; k < t; ++k) 67 { 68 if(aa[k].a==j+'A'&&aa[k].f==false) 69 { 70 aa[k].f=true; 71 bb[kk].a=aa[k].a; 72 kk++; 73 //cout<<aa[k].a<<endl; 74 break; 75 } 76 } 77 } 78 } 79 } 80 } 81 if (num==1) 82 { 83 cout<<"1 channel needed."<<endl; 84 } 85 else 86 { 87 cout<<num<<" channels needed."<<endl; 88 } 89 } 90 } 91 return 0; 92 }
hdu1175 题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=1175
注意记录转弯次数,判断是否在同一直线上,直接深搜即可
AC代码:
![](https://images.cnblogs.com/OutliningIndicators/ContractedBlock.gif)
1 #include <bits/stdc++.h> 2 using namespace std; 3 int maze[1010][1010]; 4 bool vis[1010][1010]; 5 bool flage; 6 int dx[4]={0,0,1,-1}; 7 int dy[4]={1,-1,0,0}; 8 int n,m,q,x1,y1,x2,y2; 9 void dfs(int x,int y,int d,int t) 10 { 11 if (t>2||flage) return ; 12 if (t==2&&(x-x2)!=0&&(y-y2)!=0) return ; 13 if (x==x2&&y==y2&&t<=2) 14 { 15 flage=1; 16 return ; 17 } 18 for (int i = 0; i < 4; ++i) 19 { 20 int xx=x+dx[i]; 21 int yy=y+dy[i]; 22 if (xx<1||xx>n||yy<1||yy>m||vis[xx][yy]) continue; 23 if (maze[xx][yy]==0||(xx==x2&&yy==y2)) 24 { 25 vis[xx][yy]=1; 26 if (d==-1||d==i) dfs(xx,yy,i,t); 27 else dfs(xx,yy,i,t+1); 28 vis[xx][yy]=0; 29 } 30 } 31 return ; 32 } 33 int main(int argc, char const *argv[]) 34 { 35 while(cin>>n>>m) 36 { 37 if (n+m==0) break; 38 memset(maze,0,sizeof(maze)); 39 for (int i = 1; i <= n; ++i) 40 { 41 for (int j = 1; j <= m; ++j) 42 { 43 cin>>maze[i][j]; 44 } 45 } 46 cin>>q; 47 for (int i = 0; i < q; ++i) 48 { 49 cin>>x1>>y1>>x2>>y2; 50 memset(vis,0,sizeof(vis)); 51 flage=0; 52 if (maze[x1][y1]==maze[x2][y2]&&maze[x1][y1]) 53 { 54 dfs(x1,y1,-1,0); 55 } 56 if (flage) cout<<"YES"<<endl; 57 else cout<<"NO"<<endl; 58 } 59 } 60 return 0; 61 }
hdu5113 题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=5113
注意理解题目,已经输出行末不能有空格
AC代码:
![](https://images.cnblogs.com/OutliningIndicators/ContractedBlock.gif)
1 #include <bits/stdc++.h> 2 using namespace std; 3 int visited[10][10]; 4 int a[100]; 5 int n,m,k; 6 bool flag=false; 7 void dfs(int x,int y,int nn) 8 { 9 for (int i = 1; i <= k; ++i) 10 { 11 if ((nn+1)<2*a[i]) return ; 12 } 13 if (nn==0) 14 { 15 flag=true; 16 return ; 17 } 18 else 19 { 20 for (int i = 1; i <= k; ++i) 21 { 22 if (visited[x-1][y]!=i&&visited[x][y-1]!=i&&a[i]!=0) 23 { 24 visited[x][y]=i; 25 a[i]--; 26 if (y==m) dfs(x+1,1,nn-1); 27 else dfs(x,y+1,nn-1); 28 if (flag) return ; 29 else 30 { 31 visited[x][y]=0; 32 a[i]++; 33 } 34 } 35 36 } 37 } 38 return ; 39 } 40 int main(int argc, char const *argv[]) 41 { 42 int t; 43 cin>>t; 44 for (int kk = 1; kk <= t; ++kk) 45 { 46 flag=false; 47 memset(visited,0,sizeof(visited)); 48 memset(a,0,sizeof(a)); 49 cin>>n>>m>>k; 50 for (int i = 1; i <= k; ++i) 51 { 52 cin>>a[i]; 53 } 54 dfs(1,1,n*m); 55 if(flag) 56 { 57 cout<<"Case #"<<kk<<":"<<endl; 58 cout<<"YES"<<endl; 59 for (int i = 1; i <= n; ++i) 60 { 61 for (int j = 1; j <= m; ++j) 62 { 63 if (j!=1) cout<<" "; 64 cout<<visited[i][j]; 65 } 66 cout<<endl; 67 } 68 } 69 else 70 { 71 cout<<"Case #"<<kk<<":"<<endl; 72 cout<<"NO"<<endl; 73 } 74 } 75 return 0; 76 }
poj3134 题目链接:http://poj.org/problem?id=3134
(按书上写的)
![](https://images.cnblogs.com/OutliningIndicators/ContractedBlock.gif)
1 #include <iostream> 2 //#include <stdlib.h> 3 using namespace std; 4 int val[1010]; 5 int pos,n; 6 bool ida(int now,int depth) 7 { 8 if (now>depth) return false; 9 if (val[pos]<<(depth-now)<n) return false; 10 if (val[pos] == n) return true; 11 pos++; 12 for (int i = 0; i < pos; ++i) 13 { 14 val[pos] = val[pos-1] + val[i]; 15 if (ida(now + 1,depth)) return true; 16 val[pos] = val[pos-1] - val[i]; 17 if (ida(now + 1,depth)) return true; 18 } 19 pos --; 20 return false; 21 } 22 int main(int argc, char const *argv[]) 23 { 24 int t; 25 while(cin>>n&&n) 26 { 27 int depth; 28 for (depth = 0; ; ++depth) 29 { 30 val[pos = 0] = 1; 31 if (ida(0,depth)) break; 32 } 33 cout<<depth<<endl; 34 } 35 return 0; 36 }