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hdu1251_�l�计��N��

Tue, 07 Dec 2010 08:34:00 GMT

HDU 1251 �l�计��N��

要看论文准备毕业设计了，好几周都没有�?font face="Times New Roman">ACM了，今天实在手痒了，��去hdu上溜达了一圈，挑几个题做，于是乎就看到了这个题�Q�典型的字典�?w��i)�?/font>

题目要求输出以某个字�W�串为前�~��?font face="Times New Roman">word的数目，建立字典�?w��i)之后就是个��单的查询了，��Z��性能采用了静态字典树(w��i)�Q�由于不知道�?x��)有多少个单词就猜了下感�?/font>10w应该够了吧，提交上去access violation�Q�明昄��界讉K��Q�修改�ؓ(f��)20W一样出错，后来火了�Q�直接开�?/font>50w�q�了�Q�测试数据相当狠呀�?/font>

不多说了�Q�参考代码如下�?/span>

#include <stdio.h>
#include <stdlib.h>
struct node
{
    int cnt;
    int childs[26];
};
int avail = 1;
int cur = 0;
struct node tree[500000];
char buf[15];
int main(void)
{
    int i;
    int root;
    int index;
    int flag;
    /*freopen("in.txt", "r", stdin);*/
    while (fgets(buf, 15, stdin) != NULL && buf[0] != '\n')
    {
        i = 0;
        root = 0;
        while (buf[i] != '\n')
        {
            index = buf[i]-'a';
            if (tree[root].childs[index] == 0)
            {
                tree[root].childs[index] = avail++;
            }
            ++tree[tree[root].childs[index]].cnt;
            root = tree[root].childs[index];
            ++i;
        }
    }

    while (fgets(buf, 15, stdin) != NULL && buf[0] != '\n')
    {
        i = 0;
        root = 0;
        flag = 1;
        while (buf[i] != '\n')
        {
            index = buf[i] - 'a';
            if (tree[root].childs[index] == 0)
            {
                flag = 0;
                break;
            }
            root = tree[root].childs[index];
            ++i;
        }
        printf("%d\n", flag == 1 ? tree[root].cnt : 0);
    }
    return 0;
}

李东�?/a> 2010-12-07 16:34 发表评论

ZOJ 1808 Immediate Decodability

Fri, 05 Nov 2010 09:12:00 GMT

ZOJ 1808 Immediate Decodability

�q�道题给�?/span>n个有1�?/span>0�l�成的字�W�串集合�Q�然后要求判断是否有某一个字�W�串是另一个字�W�串的前�~�。是字典�?w��i)的典型应用�?/span>

字典�?w��i)有静态和动态之分，动态字典树(w��i)��是在插入时�Ҏ(gu��)��需要动�?/span>malloc节点�Q�而静态字典树(w��i)则是事先开辟一个较大的数组�Q�然后设�|�一个变�?/span>index指向当前数组中未被占用的节点下标的最��|��即下一个可用节点的下标。跟C语言中实现静态链表类伹{��这两种�Ҏ(gu��)��各有优劣�Q�动态字典树(w��i)理论上可以插入�Q意多个节点，但是每次�?/span>malloc�?qi��ng)最后的free�?x��)消耗很多时间。而静态字典树(w��i)省去了内存的动态申请和释放�Q�节省了旉��Q�但是可以插入节�Ҏ(gu��)��目受��C��先开辟的数组大小限制�Q�可扩展性较差。具体采用哪�U�实现方式根据需求而定。就本题而言旉��要求1s�Q�可以初步需要插入的判断节点数目不会(x��)太多�Q�因此�ؓ(f��)了提高运行速度采用了静态字典树(w��i)�?/span>

参考代码如下：(x��)

#include <stdio.h>
#include <stdlib.h>
#include <string.h>
struct dick
{
    /*左右孩子指针�Q�指向左叛_��子在数组中的下标�Q�做孩子�?�Q�右孩子�?*/
    int child[2];
    /*是否是字�W�串的最后一个字�W?/span>*/
    int leaf;
};
/*从该数组中分配节�?/span>*/
struct dick d[1000];
/*指向下个可用节点的下�?/span>*/
int index;
int main(void)
{
    char buf[100];
    int no = 0;
    int flag = 1;
    int i;
    index = 0;
    int start;
    int tmp;
    int test;
    memset(d, 0, sizeof(d));
    freopen("in.txt", "r", stdin);
    while (gets(buf) != NULL)
    {
        if (buf[0] == '9' && buf[1]  == '\0')
        {
            ++no;
            if (flag == 1)
            {
                printf("Set %d is immediately decodable\n", no);
            }
            else
            {
                printf("Set %d is not immediately decodable\n", no);
            }
            /*清理和初始化数据*/
            flag = 1;
            memset(d, 0, sizeof(d));
            index = 0;
        }
        else if (flag == 1)
        {
            i = 0;
            start = 0;
            test = 1;
            /*逐字�W�插入数据到字典�?w��i)�?/span>*/
            while (buf[i] != '\0')
            {
                if (d[start].child[buf[i]-'0'] == 0)
                {
                    if (d[start].leaf == 1)
                    {
                        break;/*发现已插入字�W�串是本字符串的前缀*/
                    }
                    /*分配新的节点*/
                    d[start].child[buf[i]-'0'] = ++index;
                    test = 0;
                }
                tmp = d[start].child[buf[i]-'0'];
                if (buf[i+1] == '\0')
                {
                    d[tmp].leaf = 1;
                }
                start = tmp;
                ++i;
            }
            if (test == 1)
            {
                flag = 0;
            }
        }
    }
    return 0;
}

李东�?/a> 2010-11-05 17:12 发表评论

ZOJ1058_Currency Exchange

Tue, 02 Nov 2010 03:30:00 GMT

ZOJ1058 Currency Exchange

水题一道，唯一需要注意的是题目中说只能取到货币的癑ֈ�之一�Q�因此在每次�q�行货币汇率转换之后都要�q�行处理�Q?/span>WA了一�ơ就是因为到最后输出的时候才四舍五入�Q�这个操作应该在每次转换汇率后都�q�行�?/span>

参考代码如下：(x��)

#include <stdio.h>
#include <stdlib.h>
double rates[5][5];
int indx[10];
int main(void)
{
    int n;
    int t;
    double money;
    int i, j;
    int prev;
    /*freopen("in.txt", "r", stdin);*/
    scanf("%d", &t);
    while (t--)
    {
        for (i = 0; i < 5; ++i)
        {
            for (j = 0; j < 5; ++j)
            {
                scanf("%lf", &rates[i][j]);
            }
        }
        while (scanf("%d", &n) == 1 && n != 0)
        {
            prev = 0;
            for (i = 0; i < n; ++i)
            {
                scanf("%d", &indx[i]);
            }
            scanf("%lf", &money);
            money *= 100;
            for (i = 0; i < n; ++i)
            {
                money *= rates[prev][indx[i]-1];
                prev = indx[i]-1;
                if (money - (int)money >= 0.5)
                    money = (int)money+1;
                else
                    money = (int)money;
            }
            money *= rates[prev][0];
            if (money - (int)money >= 0.5)
                money = (int)money+1;
            else
                money = (int)money;
            printf("%.2f\n", money/100);
        }
        if (t != 0)
        {
            printf("\n");
        }
    }
    return 0;
}

李东�?/a> 2010-11-02 11:30 发表评论

ZOJ_1334_Basically Speaking

Tue, 19 Oct 2010 12:02:00 GMT

ZOJ 1334 Basically Speaking

�q�是一道简单的�q�制转换题，也是一道让我无语的题�?/span>

题目大意较�ؓ(f��)��单，但是提交�?/span>n�ơ，一�?/span>PE�Q�检查了好多地方�Q�实在感觉没头��A了，��活马当死马医，�?/span>printf(“%*c”, len, ‘ ’)换成了��@环，因�ؓ(f��)要右寚w��Q�所以输��Z��I�格�Q�竟�?/span>AC了，竟然是对printf的这�U�输出格式理解有误，无语呀�?/span>

参考代码如下：(x��)

#include <stdio.h>
#include <stdlib.h>
#include <ctype.h>
#include <string.h>
#include <math.h>

char a[50];
char ch[] = "0123456789ABCDEF";
int main(void)
{
    int to, from;
    unsigned sum;
    int len;
    int i;
    unsigned t;
    freopen("in.txt", "r", stdin);
    while (scanf("%s %d %d", a, &from, &to) != EOF)
    {
        sum = 0;
        t = 1;
        len = strlen(a);
        for (i = len-1; i >= 0; --i)
        {
            if (isdigit(a[i]))
            {
                sum +=  (a[i] - '0')*t;
            }
            else
            {
                sum += (a[i] - 'A' + 10)*t;
            }
            t *= from;
        }
        if (to == 10)
        {
            len = (int)log10(sum)+1;
            if (len > 7)
            {
                printf("  ERROR\n");
            }
            else
            {
                printf("%7d\n", sum);
            }
        }
        else
        {
            i = 0;
            while (sum > 0)
            {
                a[i++] = ch[sum%to];
                sum /= to;
            }

            if (i > 7)
            {
                printf("  ERROR\n");
            }
            else
            {
                len = 7-i;
                printf("%*c", len, ' ');
                --i;
                while (i >= 0)
                {
                    putchar(a[i--]);
                }
                printf("\n");
            }
        }
    }
    return 0;
}

李东�?/a> 2010-10-19 20:02 发表评论