[kuangbin带你飞]专题二 搜索进阶
这一题就是经典八数码。算法难点在于:
1.判重:map,hash,cantor康托展开
2.搜索:bfs dbfs A*
我是直接暴力预处理,从12345678X开始进行所有情况处理,然后结果反序输出。节点用int存储9位数的序列。
#include <iostream> #include <queue> #include <map> #include <algorithm> using namespace std; map<int,string> ans; int n[10]; struct node{ int num,nine; node(int nn,int nni):num(nn),nine(nni) {}; }; int change(int a,int b,int c){ int u = c/n[a]%10, v = c/n[b]%10; //cout<<a<<" "<<b<<" "<<c<<endl; //cout<<u<<" "<<v<<" "<<u*n[a]<<" "<<v*n[b]<<" "<<v*n[a]<<" "<<u*n[b]<<endl; return c - u*n[a] - v*n[b] + v*n[a] + u*n[b]; } void init(){ n[8] = 1;for(int i=7;i>=0;i--) n[i] = n[i+1]*10; //for(int i=0;i<9;i++) cout<<n[i]<<endl; queue<node> q; ans[123456789] = ""; q.push(node(123456789,8)); while(!q.empty()){ node p = q.front();q.pop(); int num = p.num , nine = p.nine; int tnt; //--------------------------------------------- switch(nine){ case 0: tnt = change(0,1,num); if(ans.count(tnt)==0){ ans[tnt] = ans[num] + "l"; q.push(node(tnt,1)); } tnt = change(0,3,num); if(ans.count(tnt)==0){ ans[tnt] = ans[num] + "u"; q.push(node(tnt,3)); } break; case 1: tnt = change(1,0,num); if(ans.count(tnt)==0){ ans[tnt] = ans[num] + "r"; q.push(node(tnt,0)); } tnt = change(1,2,num); if(ans.count(tnt)==0){ ans[tnt] = ans[num] + "l"; q.push(node(tnt,2)); } tnt = change(1,4,num); if(ans.count(tnt)==0){ ans[tnt] = ans[num] + "u"; q.push(node(tnt,4)); } break; case 2: tnt = change(2,1,num); if(ans.count(tnt)==0){ ans[tnt] = ans[num] + "r"; q.push(node(tnt,1)); } tnt = change(2,5,num); if(ans.count(tnt)==0){ ans[tnt] = ans[num] + "u"; q.push(node(tnt,5)); } break; case 3: tnt = change(3,0,num); if(ans.count(tnt)==0){ ans[tnt] = ans[num] + "d"; q.push(node(tnt,0)); } tnt = change(3,4,num); if(ans.count(tnt)==0){ ans[tnt] = ans[num] + "l"; q.push(node(tnt,4)); } tnt = change(3,6,num); if(ans.count(tnt)==0){ ans[tnt] = ans[num] + "u"; q.push(node(tnt,6)); } break; case 4: tnt = change(4,1,num); if(ans.count(tnt)==0){ ans[tnt] = ans[num] + "d"; q.push(node(tnt,1)); } tnt = change(4,3,num); if(ans.count(tnt)==0){ ans[tnt] = ans[num] + "r"; q.push(node(tnt,3)); } tnt = change(4,5,num); if(ans.count(tnt)==0){ ans[tnt] = ans[num] + "l"; q.push(node(tnt,5)); } tnt = change(4,7,num); if(ans.count(tnt)==0){ ans[tnt] = ans[num] + "u"; q.push(node(tnt,7)); } break; case 5: tnt = change(5,2,num); if(ans.count(tnt)==0){ ans[tnt] = ans[num] + "d"; q.push(node(tnt,2)); } tnt = change(5,4,num); if(ans.count(tnt)==0){ ans[tnt] = ans[num] + "r"; q.push(node(tnt,4)); } tnt = change(5,8,num); if(ans.count(tnt)==0){ ans[tnt] = ans[num] + "u"; q.push(node(tnt,8)); } break; case 6: tnt = change(6,3,num); if(ans.count(tnt)==0){ ans[tnt] = ans[num] + "d"; q.push(node(tnt,3)); } tnt = change(6,7,num); if(ans.count(tnt)==0){ ans[tnt] = ans[num] + "l"; q.push(node(tnt,7)); } break; case 7: tnt = change(7,6,num); if(ans.count(tnt)==0){ ans[tnt] = ans[num] + "r"; q.push(node(tnt,6)); } tnt = change(7,4,num); if(ans.count(tnt)==0){ ans[tnt] = ans[num] + "d"; q.push(node(tnt,4)); } tnt = change(7,8,num); if(ans.count(tnt)==0){ ans[tnt] = ans[num] + "l"; q.push(node(tnt,8)); } break; case 8: tnt = change(8,5,num); if(ans.count(tnt)==0){ ans[tnt] = ans[num] + "d"; q.push(node(tnt,5)); } tnt = change(8,7,num); if(ans.count(tnt)==0){ ans[tnt] = ans[num] + "r"; q.push(node(tnt,7)); } break; default:cout<<nine<<endl; } //--------------------------------------------- } } int main(){ init(); char _; while(cin>>_){ int cnt=0; if(_=='x') _='9'; cnt+=(_-'0')*n[0]; for(int i=1;i<9;i++) { cin>>_; if(_=='x') _='9'; cnt+=(_-'0')*n[i]; } //cout<<cnt<<endl; //cout<<ans[cnt]<<endl; if(ans.count(cnt)==0) puts("unsolvable"); else{ string strng = ans[cnt]; //cout<<strng<<endl; for(int i=strng.length()-1;i>=0;i--){ cout<<strng[i]; }cout<<endl; } } return 0; }
这里是要求是给出 起点和终点序列 然后得出最短操作 而且相同长度要求最小字典顺序(重点,我这里wrong了很多次 看题目才知道)
第一遍:就是A* + map 尝试 然后 timelimit。
第二遍:看其他人代码,有两种,一种就是设立limit的dfs;这几天了解一下 这是IDA*算法。迭代加深的A*算法。
f(x) = 层数 + 与目标序列的相同字符曼哈顿距离之和;然后从第一个输入的序列的f(x)为limit,然后limit逐步以最小量增加为dextd直到找到答案位置。
这里我直接给出大牛代码的链接。https://www.cnblogs.com/DOLFAMINGO/p/7538577.html
#include <iostream> #include <cmath> #include <algorithm> #include <cstdio> using namespace std; //int的取值范围为: -2^31——2^31-1,即-2147483648——2147483647 const int INF = 2e9; int Y[10],M[10],op=1,nextd; char C[1234]; int changeId[9][4] = {{3,-1,1,-1},{4,0,2,-1},{5,1,-1,-1}, {6,-1,4,0},{7,3,5,1},{8,4,-1,2}, {-1,-1,7,3},{-1,6,8,4},{-1,7,-1,5}}; char d[5] = "dlru"; int forLimit(int *a){ int ans = 0; for(int i=0;i<9;i++) if(a[i]!=0) ans += abs(i/3 - M[a[i]]/3) + abs(i%3 - M[a[i]]%3); return ans; } bool IDAstar(int k,int step,int pre,int limit){ int h = forLimit(Y); if(h==0){ printf("Case %d: %d ",op++,step); C[step] = '