zoj 1008

来源：互联网发布：软件设计师考试编辑：程序博客网时间：2024/05/16 01:09

题目来源：http://acm.zju.edu.cn/onlinejudge/showProblem.do?problemId=8

之前的写法：http://blog.csdn.net/hearthougan/article/details/12357801

#include <iostream>#include <cstdio>#include <cstring>using namespace std;const int MAXN = 25;int Graph[MAXN][4];//最多总共有25种类型int kcount[MAXN];//存放每种类型的个数int iTable[MAXN];//存放放置的方案int n, ikinds;//ikinds表示方块的种类数bool DFS(int iPos){    if(iPos == n*n)        return true;    for(int i = 0; i < ikinds; ++i)    {        if(kcount[i] == 0)            continue ;        if(iPos % n)//如果该位置不是最左边            if(Graph[iTable[iPos-1]][1] != Graph[i][3])                continue ;        if(iPos >= n)//如果该位置不是在最上方一行            if(Graph[iTable[iPos-n]][2] != Graph[i][0])                continue ;        kcount[i]--;        iTable[iPos] = i;        if(DFS(iPos+1))            return true;        kcount[i]++;//回溯时恢复现场    }    return false;}int main(){    int iTop, iRight, iBottom, ileft;    int i, j, kcase = 1;    while(scanf("%d", &n) && n)    {        if(kcase > 1)            printf("\n");        ikinds = 0;        for(i = 0; i < n*n; ++i)        {            scanf("%d %d %d %d", &iTop, &iRight, &iBottom, &ileft);            for(j = 0; j < ikinds; ++j)//查找输入的方块中是否有相同的类型            {                if(Graph[j][0] == iTop && Graph[j][1] == iRight && Graph[j][2] == iBottom && Graph[j][3] == ileft)                {                    kcount[j]++;                    break;                }            }            if(j == ikinds)//没有相同的类型            {                Graph[j][0] = iTop;                Graph[j][1] = iRight;                Graph[j][2] = iBottom;                Graph[j][3] = ileft;                kcount[j] = 1;                ikinds++;            }        }        if(DFS(0))            printf("Game %d: Possible\n", kcase++);        else            printf("Game %d: Impossible\n", kcase++);    }    return 0;}

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