UVa 11210

题目链接

分析：
身为一个麻将老手，我都不一定能”听“全牌
一共只有34种牌，我们可以考虑枚举每一张牌，
这样问题就可以转化成判断14张牌是否可以“和”

为此，我们可以枚举将牌，之后每次选三张作为刻子或者顺子
如图：选将有5种方式

为了快速的选出将，顺，刻，我们可以用一个34维向量表示每一张牌有多少，
除了第一次枚举将牌，我们每次只用考虑最小的牌在哪一个顺子或刻子里就好了

tip

每种牌只有4张

//这里写代码片#include<cstdio>#include<cstring>#include<iostream>using namespace std;const  char *mahjong[]={"1T","2T","3T","4T","5T","6T","7T","8T","9T","1S",                        "2S","3S","4S","5S","6S","7S","8S","9S","1W","2W",                        "3W","4W","5W","6W","7W","8W","9W","DONG","NAN","XI",                        "BEI","ZHONG","FA","BAI"};int mj[15],c[40];int cl(char *s){    for (int i=0;i<34;i++)        if (strcmp(mahjong[i],s)==0) return i;    return -1;}int solve(int num)                                              //当前已经有的3元组 {    int i;    for (i=0;i<34;i++)                                          //刻子         if (c[i]>=3)        {            if (num==4) return 1;            c[i]-=3;            if (solve(num+1)) return 1;            c[i]+=3;        }    for (i=0;i<27;i++)                                          //顺子         if (i%9<=6&&c[i+1]>=1&&c[i+2]>=1&&c[i]>=1)              //没有以8,9开头的顺子        {            if (num==4) return 1;            c[i]--; c[i+1]--; c[i+2]--;            if (solve(num+1)) return 1;            c[i]++; c[i+1]++; c[i+2]++;        }     return 0;}int check(){    for (int i=0;i<34;i++)        //每种牌都可以作为将    {        if (c[i]<2) continue;        c[i]-=2;        if (solve(1)) return 1;        c[i]+=2;    }     return 0;}int main(){    int cas=0;    char s[10];    while (scanf("%s",&s)==1)    {        if (s[0]=='0') break;        printf("Case %d:",++cas);        mj[0]=cl(s);        for (int i=1;i<13;i++)        {            scanf("%s",&s);            mj[i]=cl(s);        }        bool ff=0;        for (int i=0;i<34;i++)        {            memset(c,0,sizeof(c));            for (int j=0;j<13;j++) c[mj[j]]++;            if (c[i]==4) continue;            c[i]++;            if (check())            {                ff=1;                printf(" %s",mahjong[i]);            }            c[i]--;        }        if (!ff) printf(" Not ready");        printf("\n");    }    return 0;}