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1221.�?

Sat, 07 Apr 2012 02:21:00 GMT

此题是一个排列组合加递归的题目。具体解释看代码里面的注释：

1 #include
  2 int x,y,k;
  3 int array[36][36];
  4 int result[31];
  5 //初始化杨辉三�?nbsp;
  6 void init_array()
  7 {
  8     array[0][0] = 1;
  9     for(int i = 1;i < 31;++i)
10     {
11         for(int j = 0;j <= i;++j)
12         {
13             if(j == 0 ||j == i)
14             {
15                 array[i][j] = 1;
16             }
17             else
18             {
19                 array[i][j] = array[i - 1][j - 1] + array[i- 1][j];
20             }
21         }
22     }
23 }
24
25 int get_num_count(int num)
26 {
27     int total = 0;
28     while(num > 0)
29     {
30         ++total;
31         num /= 2;
32     }
33     return total;
34 }
35
36 //��L��首位�?
37 int delete_highest(int num)
38 {
39     int count = get_num_count(num);
40     int total = 1<<(count - 1);
41     total = ~total;
42     return num & total;
43 }
44
45 //计算跟一个数2�q�制位数相同的最大�?nbsp;
46 int getMin(int total_nums,int one_nums)
47 {
48     int total = 0;
49     if(total_nums < one_nums)
50         return -1;
51     for(int i = 0;i < one_nums;++i)
52     {
53         total = total * 2 + 1;
54     }
55     for(int j = one_nums;j < total_nums;++j)
56     {
57         total *= 2;
58     }
59     return total;
60 }
61 //计算跟一个数2�q�制位数相同的最��?nbsp;
62 int getMax(int total_nums,int one_nums)
63 {
64     int total = 1;
65     if(total_nums < one_nums)
66         return 1000000001;
67     for(int i = 0;i < total_nums - one_nums;++i)
68     {
69         total = total * 2;
70     }
71     for(int j = 1;j < one_nums;++j)
72     {
73         total = total * 2 + 1;
74     }
75     return total;
76 }
77
78
79 int recursive_deal(int start,int end,int one_nums)
80 {
81     if(start > end)
82         return 0;
83     int start_count = get_num_count(start);
84     int end_count = get_num_count(end);
85     if(one_nums == 0)
86     {
87         if(start == 0 || end == 0)
88             return 1;
89         else
90             return 0;
91     }
92     else
93     //如果始末两个数的位数相同的话�Q�就��L��最高位
94     if(start_count == end_count)
95     {
96             return recursive_deal(delete_highest(start),delete_highest(end),one_nums - 1);
97     }
98     else    //如果始末两个数的位数不同的话�Q�分别计��和起始�Q�终止的的数�?�q�制位数�Q�中间的数的数量可以通过排列�l�合计算出来
99     {
100         int total = 0;
101         for(int i = start_count + 1;i < end_count;++i)
102         {
103             total += array[i - 1][one_nums - 1];
104         }
105         return total + recursive_deal(start,getMin(start_count,one_nums),one_nums) + recursive_deal(getMax(end_count,one_nums),end,one_nums);
106     }
107 }
108
109 int main()
110 {
111     int tests;
112     init_array();
113     scanf("%d",&tests);
114     while(--tests >= 0)
115     {
116         scanf("%d%d%d",&x,&y,&k);
117         printf("%d\n",recursive_deal(x,y,k));
118     }
119     return 0;
120 }

崔佳�?/a> 2012-04-07 10:21 发表评论

1025.The Dragon of Loowater

Fri, 06 Apr 2012 12:43:00 GMT

此题属于比较��单的题目�Q�读完题(前提是你能读完，因�ؓ它真的很�?��发现其实是贪心题目�?br />所以，直接�Q�排序，贪心�Q�AC.下面是代�?

1 #include
2 #include
3 int n,m;
4 int heads[20010];
5 int knights[20010];
6
7
8 int cmp(const void *first,const void * second)
9 {
10     return *((int*)first) - *((int*)second);
11 }
12
13 //�q�回-1表示不可�?nbsp;
14 int deal()
15 {
16     if(n > m)
17         return -1;
18     qsort(heads,n,sizeof(int),cmp);
19     qsort(knights,m,sizeof(int),cmp);
20     int i = 0,j = 0;
21     int total = 0;
22     for(i = 0;i < n;++i)
23     {
24         while(knights[j] < heads[i] && j < m)
25         {
26             ++ j;
27         }
28         if(j == m)
29             break;
30         else
31         {
32             total += knights[j];
33             ++j;
34         }
35     }
36     if(i == n)
37         return total;
38     else
39         return -1;
40
41 }
42
43 int main()
44 {
45     int i;
46     int result;
47     while(scanf("%d%d",&n,&m))
48     {
49         if(m == 0 &&n == 0)
50             break;
51         for(i = 0;i < n;++i)
52         {
53             scanf("%d",&heads[i]);
54         }
55         for(i = 0;i < m;++i)
56         {
57             scanf("%d",&knights[i]);
58         }
59         result = deal();
60         if(result == -1)
61         {
62             printf("Loowater is doomed!\n");
63         }
64         else
65         {
66             printf("%d\n",result);
67         }
68     }
69     return 0;
70 }

崔佳�?/a> 2012-04-06 20:43 发表评论

1077.安全�|�络 ver.4

Fri, 06 Apr 2012 11:59:00 GMT

此题大意为求从顶�?到顶点n的最短�\径�?br />所以采用最短�\径的��法�?br />想省事的话，可以使用弗洛伊�d�Q�但是n的��gؓ10^5,�q�样肯定��时�?br />所以本题采用SPFA��法。见下面代码

#include
#include
#include
#include<string.h>
using namespace std;
typedef struct node
{
    int id;
    int distance;
    struct node *next;
}Node,*LNode;
int n,m;
LNode array[100010];//��L��链表
int in_queue[100010];//记录一个点是否在队列中
int between_distance[100010];//记录起始点到两个点之间的距离
queue<int> q;//建立一个队�?/span>

void init_array()
{
    for(int i = 0;i <= n;++i)
    {
        array[i] = (LNode)malloc(sizeof(Node));
        array[i]->id = i;
        array[i]->distance = 0;
        array[i]->next = NULL;
    }
}

void insert_node(LNode header,LNode &insert_node)
{
    while(header->next != NULL)
        header = header->next;
    insert_node->next = header->next;
    header->next = insert_node;
}

//��M��两点之间的距��设�|��ؓ-1�Q�表�C�正无穷
void init_distance()
{
    for(int i = 0;i <= n;++i)
    {
        between_distance[i] = -1;
    }
    //�W�一个结点到自��n�?
    between_distance[1] = 0;
}

//村ּ�
void relax(int index)
{
    if(between_distance[index] == -1)
        return ;
    LNode header = array[index]->next;
    while(header != NULL)
    {
        if(between_distance[header->id] > between_distance[index] + header->distance || between_distance[header->id] == -1)
        {
            between_distance[header->id] = between_distance[index] + header->distance;
            if(!in_queue[header->id])
            {
                q.push(header->id);
                in_queue[header->id] = 1;
            }
        }
        header = header->next;
    }
}

void SPFA()
{
    //�W�一个点入队�?/span>
    q.push(1);
    in_queue[1] = 1;
    int index = 0;
    while(!q.empty())
    {
        index = q.front();
        q.pop();
        in_queue[index] = 0;
        relax(index);
    }
}

int main()
{
    int x,y,distance;
    scanf("%d%d",&n,&m);
    init_array();
    for(int i = 0;i < m;++i)
    {
        scanf("%d%d%d",&x,&y,&distance);
        LNode node = (LNode)malloc(sizeof(Node));
        node->id = y;
        node->distance = distance;
        insert_node(array[x],node);
        node = (LNode)malloc(sizeof(Node));
        node->id = x;
        node->distance = distance;
        insert_node(array[y],node);
    }
    init_distance();
    memset(in_queue,0,sizeof(in_queue));
    SPFA();
    if(between_distance[n] == -1)
    {
        printf("-1\n");
    }
    else
        printf("%d\n",between_distance[n]);
    return 0;
}

崔佳�?/a> 2012-04-06 19:59 发表评论

田教�ȝ��卡片

Fri, 06 Apr 2012 02:35:00 GMT

acm.xmu.edu.cn上的1303题�?br />此题为厦门大学信息科学与技术学院暑期ACM培训的内部测试题�?br />直接看题�Q�明显�ؓ一�l�区间的动态规划�?br />

1 #include
2 char array[110];
3 int zero,one;
4 int i;
5 int dp(int n)
6 {
7     if(array[0] == '0')
8     {
9         zero = 0;
10         one = 1;
11     }
12     else
13     {
14         zero = 1;
15         one = 0;
16     }
17     for(i = 1;i < n;++i)
18     {
19         if(array[i] == '0')
20         {
21             if(array[i] != array[i - 1])
22                 ++one;
23         }
24         else
25         {
26             if(array[i] != array[i - 1])
27                 ++ zero;
28         }
29     }
30     return zero < one?zero:one;
31 }
32
33 int main()
34 {
35     int t,n;
36     scanf("%d",&t);
37     while(t-- != 0)
38     {
39         scanf("%d",&n);
40         scanf("%s",array);
41         printf("%d\n",dp(n));
42     }
43     return 0;
44 }

崔佳�?/a> 2012-04-06 10:35 发表评论