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;

}