POJ 2286 - The Rotation Game ( IDA* )
2018-03-27 11:42
453 查看
题意
一个 ” # ” 字形图形, 每个横/竖条有七个小方块, 只用1/2/3标记. 其中有八个(ABCDEFG)移动方向, 要通过一些移动操作使得中间的八个小方块的标记相同. 求解这个操作方式以及中间八个小方块相同的标记思路
迭代深搜剪枝 : 当前需要被改变的小块数量 + 当前操作深度 > 迭代的最大操作深度
则一定无法达成, 剪枝
if( remain() + d > maxd ) return false; //剪枝
AC代码
#include <iostream> #include <algorithm> #include <cstring> #include <string> #include <vector> #include <cstdio> #include <cmath> #define mst(a) memset(a, 0, sizeof(a)) using namespace std; int n, m; const int maxn = 10; int mrk[10][10] = { //标号 {0,2,6,11,15,20,22}, //A {1,3,8,12,17,21,23}, //B {10,9,8,7,6,5,4}, //C {19,18,17,16,15,14,13}, //D {23,21,17,12,8,3,1}, //E {22,20,15,11,6,2,0}, //F {13,14,15,16,17,18,19}, //G {4,5,6,7,8,9,10} //H }; int mid[] = {6,7,8,11,12,15,16,17}; //中间的八块 int r[] = {5,4,7,6,1,0,3,2}; //反向 int g[30]; char op[1000+10]; //记录操作 int maxd; bool judge(){ //判断中间八个是否相同 for( int i = 1; i < 8; i++ ) if( g[mid[i]] != g[mid[i-1]] ) return false; return true; } int remain(){ int re = 10; for( int i = 1; i <= 3; i++ ){ int cnt = 0; for( int j = 0; j < 8; j++ ) if( g[mid[j]] != i ) cnt++; re = min(re, cnt); } return re; } void mmove( int x ){ //移动 int temp = g[mrk[x][0]]; for( int i = 0; i < 6; i++ ) g[mrk[x][i]] = g[mrk[x][i+1]]; g[mrk[x][6]] = temp; return; } bool dfs( int d ){ if( d == maxd ){ if( judge() ){ op[d] = '\0'; return true; } else return false; } if( remain() + d > maxd ) return false; //剪枝 for( int i = 0; i < 8; i++ ){ char c = 'A' + i; op[d] = c; mmove(i); if( dfs(d+1) ) return true; mmove(r[i]); //还原移动 /* 看到dalao们是这么写还原移动的~ 记录一下 if(i%2==0) mmove((i+5)%8); else mmove((i+3)%8); */ } return false; } void solve(){ for( maxd = 1; ; maxd++ ) if( dfs(0) ) break; printf("%s\n",op); } int main() { while( scanf("%d",&g[0]) == 1 && g[0] ){ for( int i = 1; i <= 23; i++ ) scanf("%d",&g[i]); if( judge() ) puts("No moves needed"); else solve(); printf("%d\n",g[mid[0]]); } return 0; } /* A B 0 1 2 3 H 4 5 6 7 8 9 10 C 11 12 G 13 14 15 16 17 18 19 D 20 21 22 23 F E */
相关文章推荐
- POJ 2286 UVA 1343 The Rotation Game IDA*
- (IDA*)POJ 2286 The Rotation Game
- [poj 2286] The Rotation Game IDA*(dfs)
- POJ-2286 The Rotation Game IDA*
- POJ 2286 The Rotation Game 迭代搜索深度 + A* == IDA*
- POJ 2286 The Rotation Game(IDA*)
- POJ2286 The Rotation Game(IDA*)
- POJ 2286 The Rotation Game (IDA*)
- POJ - 2286 - The Rotation Game (IDA*)
- POJ - 2286 - The Rotation Game (IDA*)
- POJ-2286-The Rotation Game(IDA*入门)
- POJ 2286 The Rotation Game 迭代搜索深度 + A* == IDA*
- POJ 2286 The Rotation Game (IDA*)
- POJ 2286 The Rotation Game IDA*
- The Rotation Game (poj 2286 搜索IDA*)
- POJ 2286 The Rotation Game 搜索-IDA*+迭代加深
- 【poj 2286】The Rotation Game 题意&题解&代码(C++)
- POJ 2286 The Rotation Game
- POJ 2286 - The Rotation Game
- POJ 2286 The Rotation Game