hdoj 2514 Another Eight Puzzle(DFS)

略恶心，，乱搞居然一遍A掉了。

Another Eight Puzzle

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)

Total Submission(s): 792    Accepted Submission(s): 494

Problem Description

Fill the following 8 circles with digits 1~8,with each number exactly once . Conntcted circles cannot be filled with two consecutive numbers.
There are 17 pairs of connected cicles:
A-B , A-C, A-D
B-C, B-E, B-F
C-D, C-E, C-F, C-G
D-F, D-G
E-F, E-H
F-G, F-H
G-H

Filling G with 1 and D with 2 (or G with 2 and D with 1) is illegal since G and D are connected and 1 and 2 are consecutive .However ,filling A with 8 and B with 1 is legal since 8 and 1 are not consecutive .

In this problems,some circles are already filled,your tast is to fill the remaining circles to obtain a solution (if possivle).

 

Input

The first line contains a single integer T(1≤T≤10),the number of test cases. Each test case is a single line containing 8 integers 0~8,the numbers in circle A~H.0 indicates an empty circle.

 

Output

For each test case ,print the case number and the solution in the same format as the input . if there is no solution ,print “No answer”.If there more than one solution,print “Not unique”.

 

Sample Input

37 3 1 4 5 8 0 07 0 0 0 0 0 0 01 0 0 0 0 0 0 0

 

Sample Output

Case 1: 7 3 1 4 5 8 6 2Case 2: Not uniqueCase 3: No answer

 

Source

ECJTU 2008 Autumn Contest

 

想清楚关系，，深搜搞定。

#include<iostream>#include<cstdio>#include<cstring>#include<cmath>using namespace std;const int N = 10;int flag[20][2] = {1, 2, 1, 3, 1, 4, 2, 3, 2, 5, 2, 6, 3, 4, 3, 5, 3, 6, 3, 7,                    4, 6, 4, 7, 5, 6, 5, 8, 6, 7, 6, 8, 7, 8};int sign[N][N];int num[N], res[N];int vis[N];int cnt;void init(){    memset(sign, 0, sizeof(sign));    for(int i = 0; i < 17; i++){        int tx = flag[i][0], ty = flag[i][1];        sign[tx][ty] = sign[ty][tx] = 1;    }}bool judge(int k, int t){    for(int i = 1; i < k; i++){        if(sign[i][k] && abs(num[i] - t) == 1) return false;    }    return true;}void dfs(int cur){    if(cnt > 1) return;    if(cur > 8){        cnt++;        for(int i = 1; i <= 8; i++) res[i] = num[i];        return;    }    if(num[cur]) dfs(cur + 1);    else{        for(int i = 1; i <= 8; i++){            if(!vis[i] && judge(cur, i)){                vis[i] = 1;                num[cur] = i;                dfs(cur + 1);                num[cur] = 0;                vis[i] = 0;            }        }    }}int main(){    int T;    init();    scanf("%d", &T);    int cas = 1;    while(T--){        int t;        memset(vis, 0, sizeof(vis));        memset(num, 0, sizeof(num));        for(int i = 1; i <= 8; i++){            scanf("%d", &num[i]);            vis[num[i]] = 1;        }        cnt = 0;        dfs(1);        printf("Case %d: ", cas++);        if(cnt > 1) puts("Not unique");        else if(cnt == 0) puts("No answer");        else{            for(int i = 1; i < 8; i++) printf("%d ", res[i]);            printf("%d\n", res[8]);        }    }    return 0;}