hdu 1305 Immediate Decodability

字典树。建树的过程中，一边建树一边判断有没有重复前缀的。

#include<stdio.h>
#include<string.h>
#include<math.h>
#include<algorithm>
using namespace std;
char s[1000]; int i, j, summ;
struct nn{ int tot, ling, yi; }dt[50000];
int main()
{
    int aa = 1;
    while (~scanf("%s", s))
    {
        if (strcmp("9", s) == 0) break;
        summ = 0;
        int flag = 0;
        for (i = 0; i < 50000; i++) dt[i].ling = -1, dt[i].yi = -1, dt[i].tot = 0;

        int t = 0;
        int yy = strlen(s);
        for (i = 0; s[i]; i++)
        {
            int jian = 0;
            if (s[i] == ‘0‘)
            {
                if (dt[t].ling == -1)//建立节点
                {
                    summ++;
                    dt[t].ling = summ;
                    dt[summ].tot = 1;
                    jian = 1;
                }
                t = dt[t].ling; //跳到这个节点
                if ( (jian == 0 && dt[t].tot == 1 && i == yy - 1)) flag = 1;
                if (i == yy - 1) dt[t].tot = 5201314;
            }
            else if (s[i] == ‘1‘)
            {
                if (dt[t].yi == -1)//建立节点
                {
                    summ++;
                    dt[t].yi = summ;
                    dt[summ].tot = 1;
                    jian = 1;
                }
                t = dt[t].yi;//跳到这个节点
                if ( (jian == 0 && dt[t].tot == 1 && i == yy - 1)) flag = 1;
                if (i == yy - 1) dt[t].tot = 5201314;
            }
        }

        while (1)
        {
            scanf("%s", s);
            if (strcmp("9", s) == 0) break;
            int t = 0;
            int yy = strlen(s);
            for (i = 0; s[i]; i++)
            {
                int jian = 0;
                if (s[i] == ‘0‘)
                {
                    if (dt[t].ling == -1)//建立节点
                    {
                        summ++;
                        dt[t].ling = summ;
                        dt[summ].tot = 1;
                        jian = 1;
                    }
                    t = dt[t].ling; //跳到这个节点
                    if (dt[t].tot == 5201314 || (jian == 0 && dt[t].tot == 1 && i == yy - 1)) flag = 1;
                    if (i == yy - 1) dt[t].tot = 5201314;
                }
                else if (s[i] == ‘1‘)
                {
                    if (dt[t].yi == -1)//建立节点
                    {
                        summ++;
                        dt[t].yi = summ;
                        dt[summ].tot = 1;
                        jian = 1;
                    }
                    t = dt[t].yi;//跳到这个节点
                    if (dt[t].tot == 5201314 || (jian == 0 && dt[t].tot == 1 && i == yy - 1)) flag = 1;
                    if (i == yy - 1) dt[t].tot = 5201314;
                }
            }
        }
        if (flag == 0) printf("Set %d is immediately decodable\n", aa++);
        else if (flag == 1) printf("Set %d is not immediately decodable\n", aa++);
    }
    return 0;
}