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  • UVA 1343 The Rotation Game

    https://vjudge.net/problem/UVA-1343

    题目

    给出一种井字棋盘,只含数字1、2、3,只允许六种移动,问如何移动才能使棋盘中间八格数字相等,输出步数最少的操作,如果有多个步数相同的操作,输出字典序最小的操作

    题解

    每一步操作都有点麻烦,先写个工具模拟所有操作

    先编号,方便输入和操作

    只看A操作

    操作后下面的数占据了上面的位置,然后就可以写程序得到每个数的新位置

    工具代码

    #include<bits/stdc++.h>
    using namespace std;
    #define REP(i,x,y) for(register int i=(x); i<(y); i++)
    int arr[7][7] = {
    {0,0,1,0,2,0,0},
    {0,0,3,0,4,0,0},
    {5,6,7,8,9,10,11},
    {0,0,12,0,13,0,0},
    {14,15,16,17,18,19,20},
    {0,0,21,0,22,0,0},
    {0,0,23,0,24,0,0}
    };
    int arr2[7][7];
    inline void op(int fx, int fy, int dx, int dy) {
    	int ix=fx+dx,iy=fy+dy;
    	while(0<=ix && ix<7 && 0<=iy && iy<7) {
    		arr2[ix][iy]=arr[ix-dx][iy-dy];
    		ix+=dx, iy+=dy;
    	}
    	arr2[fx][fy]=arr[ix-dx][iy-dy];
    }
    int main() {
    	const int fx[] = {0,0,2,4,6,6,4,2};
    	const int fy[] = {2,4,6,6,4,2,0,0};
    	const int dx[] = {1,1,0,0,-1,-1,0,0};
    	const int dy[] = {0,0,-1,-1,0,0,1,1};
    	REP(i,0,8) {
    		printf("{");
    		memcpy(arr2,arr,sizeof arr);
    		op(fx[i],fy[i],dx[i],dy[i]);
    		REP(i,0,7) {
    			REP(j,0,7) {
    				if(arr2[i][j])
    				printf("%d,", arr2[i][j]-1);
    			}
    		}
    		printf("}, // %c
    ", 'A'+i);
    	}
    	return 0;
    }
    

    然后就可以用IDA*写程序代码了……(因为不想写hash= =)

    AC代码

    #include<bits/stdc++.h>
    using namespace std;
    #define REP(i,x,y) for(register int i=(x); i<(y); i++)
    #define _REP(i,x,y) for(i=(x); i<(y); i++)
    #define REPE(i,x,y) for(register int i=(x); i<=(y); i++)
    #define _REPE(i,x,y) for(i=(x); i<=(y); i++)
    #ifdef sahdsg
    #define DBG(a,...) printf(a, ##__VA_ARGS__)
    #else
    #define DBG(a,...) (void)0
    #endif
    
    const int mv[8][24] = {
    {22,1,0,3,4,5,2,7,8,9,10,6,12,13,14,11,16,17,18,19,15,21,20,23}, // A
    {0,23,2,1,4,5,6,7,3,9,10,11,8,13,14,15,16,12,18,19,20,17,22,21}, // B
    {0,1,2,3,5,6,7,8,9,10,4,11,12,13,14,15,16,17,18,19,20,21,22,23}, // C
    {0,1,2,3,4,5,6,7,8,9,10,11,12,14,15,16,17,18,19,13,20,21,22,23}, // D
    {0,3,2,8,4,5,6,7,12,9,10,11,17,13,14,15,16,21,18,19,20,23,22,1}, // E
    {2,1,6,3,4,5,11,7,8,9,10,15,12,13,14,20,16,17,18,19,22,21,0,23}, // F
    {0,1,2,3,4,5,6,7,8,9,10,11,12,19,13,14,15,16,17,18,20,21,22,23}, // G
    {0,1,2,3,10,4,5,6,7,8,9,11,12,13,14,15,16,17,18,19,20,21,22,23} // H
    };
    int arr[24];
    
    int target;
    
    const int vi[]={6,7,8,11,12,15,16,17};
    int md;
    char ans[4][450];
    bool dfs(int *oarr, int d) {
    	
    	int e=0;
    	REP(i,0,8) {
    		if(oarr[vi[i]]!=target) e++;
    	}
    //	DBG("#%d %d
    ", d,e);
    	if(e+d>md) return false;
    	if(e==0) return true;
    	int narr[24];
    	REP(i,0,8) {
    		REP(j,0,24) {
    			narr[mv[i][j]]=oarr[j];
    		}
    		ans[target][d]='A'+i;
    //		putchar(ans[d]);
    		if(dfs(narr, d+1)) return true;
    	}
    	return false;
    }
    int main() {
    	#ifdef sahdsg
    	freopen("in.txt", "r", stdin);
    	#endif
    	while(~scanf("%d", arr) && arr[0]) {
    		REP(i,1,24) {
    			scanf("%d", &arr[i]);
    		}
    		for(md=0;;md++) {
    			bool f = false;
    			for(target=1;target<=3;target++) {
    				ans[target][0]=0;
    				if(dfs(arr,0)) {
    					if(md==0) {
    						puts("No moves needed");
    						printf("%d
    ", target);
    						break;
    					}
    					f=true;
    				}
    			}
    			if(f) {
    				int c=-1;
    				REPE(i,1,3) {
    					if(ans[i][0]!=0 && ((c==-1) || (strcmp(ans[i],ans[c])<0))) {
    						c=i;
    					}
    				}
    				if(c==-1) break;
    				REP(i,0,md) {
    					putchar(ans[c][i]);
    				}
    				putchar('
    ');
    				printf("%d
    ", c);
    				break;
    			}
    		}
    	}
    	return 0;
    }
    
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  • 原文地址:https://www.cnblogs.com/sahdsg/p/10450948.html
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