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Chen Jiecao — Sun, 26 Jul 2009 04:43:00 GMT

1140. ��里的钥�?/h1>

输入文名�Q�box.in
输出文名�Q�box.out

提交讨论 �q�行状况

有N个编号�ؓ1到N的箱子和N个编号�ؓ1到N的钥匙，�W�i号钥匙只能用来打开�W�i��L��子。现在我们随机地��一把钥匙锁�q�一个箱子里�Q�即每个��子里都恰好�? 一把钥匙，保证所有的情况都等可能性地出现。现在你有M个炸弹，每个炸弹可以用来炸开一个箱子，一旦你把某个箱子打开�Q�你��可以取出其中的钥匙�Q�从而有�? 能用�q�钥匙打开更多的箱子。你的策略很��单，当没有箱子可以打开�Ӟ��随便选一个箱子，用炸弹炸开它，取出钥匙�q��l�打开��可能多的箱子，直至没有��子可以打开�Q�然后��l��用下一颗炸式V�?

现给定N�Q�你的�Q务是求出你可以取得所有钥匙的概率。这个概率必��输出成分数“A/B”的�Ş式，A和B都是正整��C��公约数必��Mؓ1�?

输入格式

输入一行，包含�I�格隔开的两个数N和M

输出格式

输出为A/B的�Ş式�?

输入样例

3 1

输出样例

1/3

数据规模与约�?/strong>

1 ≤ N ≤ 20, 1 ≤ M ≤ N
解析:
�q�个题目基本上就是一个数学题,涉及到第一�c�stirling数的求解.
所�?span style="color: red;">�W�一�c�stirling�?/span>,例如S[n,k]表示��一个大��ؓn的集合分成k个部�?每个部分的元素个��C��于1,且�Ş成环�?span style="color: red;">��L��法数.
一个元素也��作单独的环.
�Ҏ��的到
    S[1,1]=1;
    S[n,0]=0;
当n对合法的n,k,满��: S[n,k]=S[n-1,k-1]+(n-1)*S[n-1,k];
把n当作钥匙(也即��子)的个�?k为钥匙所放位�|��Ş成的"�?,每破坏一个箱�?都可以得到该��子所属环的所有钥�?k表示实际的环的个�?br>当k>m时便不可能取得到所有的钥匙.
�q�样下面的代码就很好理解�?

1 #include<iostream>
2 using namespace std;
3 const int MAXN=30;
4 template <class T>
5 T Gcd(T a,T b)
6 {
7   return (!a)? b : Gcd(b%a,a);
8 }
9
10 int main()
11 {
12   freopen("box.in","r",stdin);
13   freopen("box.out","w",stdout);
14   long long n,m,S[MAXN][MAXN];
15   cin >> n >> m;
16   memset(S,0,sizeof(S));
17   S[1][1]=1;
18   for (int i=2;i<=n;++i)
19     for (int j=1;j<=i;++j)
20       S[i][j]=S[i-1][j-1]+(i-1)*S[i-1][j];
21   long long B=1;
22   for (int i=2;i<=n;++i) B*=i;
23   long long A=B;
24   for (int i=m+1;i<=n;++i) A-=S[n][i];
25   long long G=Gcd(A,B);
26   cout << A/G << '/' << B/G << endl;
27   return 0;
28 }
29

Chen Jiecao 2009-07-26 12:43 发表评论

TJU_OI 1094 珍珠��w��

Chen Jiecao — Wed, 22 Jul 2009 07:15:00 GMT

1094. 珍珠��w��

输入文�g名：necklace.in
输出文�g名：necklace.out 提交讨论 �q�行状况

有一串由n个珍珠组成的珍珠��w��Q�珍珠的�~�号�?..n�Q�每个珍珠都有各自的价��|��我们用w[i]表示�~�号为i的珍珠的价��|��注意�Q�w[i]可以��于 �Ӟ��Q�已知这n个珍珠的价值量��d��是n-1,现要求你在项铄��某个位置断开�Q��得断开后的珍珠链满��_��于�Q意k,前k个珍珠的价值量��d��不超�q�k-1. (�Ҏ��开的一点说�? 如果在位�|�p断开, 那么得到的珍珠链��一定是p,p+1,...,n,1,2,...,p-1)

输入格式

输入文�g的第一行有一个唯一的整数n,

接下来n个用�I�格和换行符隔开的整数分别表�C�w[1],w[2],...,w[n]

输出格式

如果无解误��Z��?Impossible"�Q�不含引��P��否则输出一个整数表�C�断开后的珍珠铄��一个珍珠的�~�号

输入样例

5
1 1 1 1 0

输出样例

5

数据规模与约�?/strong>

3≤n≤200,000 -1,000,000,000≤w[i]≤1,000,000,000

40%的测试数据满��n≤1,000

题目来源 �Q?a >OIBH 信息学练习赛 #8

代码:

1 #include<iostream>
2 using namespace std;
3 const int MAXN=200000+100;
4 long long n,w[MAXN<<1];
5 int main()
6 {
7   freopen("necklace.in","r",stdin);
8   freopen("necklace.out","w",stdout);
9   cin >> n;
10   for (int i=0;i<n;++i) cin >> w[i],w[i+n]=w[i];
11   long long len=0,tot=0;;
12   for (int i=0;i<n*2;++i)
13     {
14       if (tot+w[i]<=len) tot+=w[i],++len;
15       else
16     { tot=0; len=0; }
17       if (len==n) { cout << i-len+2 << endl; return 0; }
18     }
19   cout << "Impossible\n";
20   return 0;
21 }
22
23
基本�?��是枚�D珍珠,如果某个珍珠可以作�ؓ�W�一颗珍�?那么接着把它的下一颗当作第二颗,如果合法,�l�箋下一�?如果不合�?直接把当前不合法的这一颗的下一颗当作第一颗��l�枚�?
��Z��么挡当前的这颗珍珠在�q�个位置不行,��׃��用在别的位置��试�?�q�是因�ؓ一旦某颗珍珠作为断掉之后的铄��k颗不合法的话,它也一定不可能在当作第1、第2........或第k-1颗时合法,略证如下:
若A[1],A[2],A[3]......A[k-1] 作�ؓ�?k-1)颗珍珠合�?...............................................................................(1)
而A[1]+A[2]+A[3]+......+A[k]>k-1   不合�?..........................................................................................(2)
那么我们�?br>A[t+1]+A[t+1]+......+A[k]>k-1-(A[1]+A[2]+......+A[t])       (1<=t(1)==> A[1]+A[2]+.......A[t]<=t-1
所以由(3)�?A[t+1]+A[t+2]+......A[k]>k-1-(t-1)=k-t...........................................................................(4)
而要��x��A[t+1],A[t+2],.......A[k]当作�W?,2,.........(k-t)颗珍珠且合法,必须�?br>A[t+1]+A[t+2]+.......A[k]<=k-t-1 (有k-t颗珠�?..................................................................................(5)
(4),(5)矛盾,所以红色部分得�?

Chen Jiecao 2009-07-22 15:15 发表评论

URAL 1022 Genealogical tree

Chen Jiecao — Tue, 21 Jul 2009 09:14:00 GMT
拓扑排序是最直接的算�?

1 /* 拓扑排序 */
2
3 #include<iostream>
4 #include<vector>
5 using namespace std;
6 const int MAXN=120;
7 vector<int> adj[MAXN];
8
9 int into[MAXN];
10
11 int main()
12 {
13   //freopen("data.in","r",stdin);
14   //freopen("data.out","w",stdout);
15   int n;
16   memset(into,0,sizeof(into));
17   cin >> n;
18   for (int i=1;i<=n;++i)
19     {
20       int x;
21       cin >> x;
22       while (x!=0)
23     {
24       adj[i].push_back(x);
25       ++into[x];
26       cin >> x;
27     }
28     }
29  //for (int i=1;i<=n;++i) cout << into[i] << ' '; cout << endl;
30   bool everp=0;
31   for (int k=0;k<n;++k)
32     {
33       int i=1;
34       for (;i<=n;i++) if (!into[i]) break;
35       (everp) ? cout << ' ' << i : cout << i;
36       everp=1;
37       --into[i];
38       for (int j=0;j<adj[i].size();++j) --into[ adj[i][j] ];
39     }
40   cout << endl;
41   return 0;
42 }
43

Chen Jiecao 2009-07-21 17:14 发表评论

URAL 1021 Sacrament of the sum

Chen Jiecao — Tue, 21 Jul 2009 09:12:00 GMT
水题,�׃��l�出的已�l�是单调的序�?所以有o(N)的算�?br>
1 #include<iostream>
2 using namespace std;
3 const int MAXN=50005;
4
5
6 int main()
7 {
8   freopen("data.in","r",stdin);
9   freopen("data.out","w",stdout);
10   int c,cc,num[MAXN];
11   cin >> c;
12   for (int i=0;i<c;++i) cin >> num[i];
13   num[c]=31440461;
14   cin >> cc;
15   int p=0;
16   for (int i=0;i<cc;++i)
17     {
18       int x;
19       cin >> x;
20       while (x+num[p]<10000) ++p;
21       if (x+num[p]==10000) { cout << "YES\n"; return 0; }
22     }
23   cout << "NO\n";
24   return 0;
25 }
26

Chen Jiecao 2009-07-21 17:12 发表评论

URAL 1020 Rope

Chen Jiecao — Mon, 20 Jul 2009 08:49:00 GMT
水题,没什么好说的,注意N=1的情�?

1 #include<iostream>
2 #include<math.h>
3 using namespace std;
4 const double PI=3.1415926;
5
6 template <class T>
7 T dis(T x1,T y1,T x2,T y2)
8 {
9   return sqrt( (x1-x2)*(x1-x2) + (y1-y2)*(y1-y2) );
10 }
11 int main()
12 {
13   int n;
14   double r,len=0;
15   //freopen("data.in","r",stdin);
16   //freopen("data.out","w",stdout);
17   cin >> n >> r;
18   double x0,y0,x,y;
19   cin >> x0 >> y0;
20   double px=x0,py=y0;
21 if (n==1) goto L1;
22   for (int i=1;i<n;++i)
23     {
24       cin >> x >> y;
25       len+=dis(px,py,x,y);
26       px=x;
27       py=y;
28     }
29   len+=dis(x,y,x0,y0);
30 L1:  len+= 2*PI*r ;
31   printf("%.2f\n",len);
32   return 0;
33 }
34

Chen Jiecao 2009-07-20 16:49 发表评论

URAL 1019 A line painting

Chen Jiecao — Mon, 20 Jul 2009 07:00:00 GMT

A Line painting

Time Limit: 2.0 second
Memory Limit: 16 MB

The segment of numerical axis from 0 to 10⁹ is painted into white color. After that some parts of this segment are painted into black, then some into white again and so on. In total there have been made N re-paintings (1 ≤ N ≤ 5000). You are to write a program that finds the longest white open interval after this sequence of re-paintings.

Input

The first line of input contains the only number N. Next N lines contain information about re-paintings. Each of these lines has a form:
a_i b_i c_i
where a_i and b_i are integers, c_i is symbol 'b' or 'w', a_i, b_i, c_i are separated by spaces.
This triple of parameters represents repainting of segment from a_i to b_i into color c_i ('w' �?white, 'b' �?black). You may assume that 0 < a_i < b_i < 10⁹.

Output

Output should contain two numbers x and y (x < y) divided by space(s). These numbers should define the longest white open interval. If there are more than one such an interval output should contain the one with the smallest x.

Sample

input output

4
1 999999997 b
40 300 w
300 634 w
43 47 b

47 634

�q�个题目最直接的办法是��L��?��L��化了之后��好折腾�?因�ؓ操作指��o很少,所以我选择��L��+朴素染色.
复杂度上界�ؓO(M*M),�q�里M表示��L��化后得到的区间个�?

1 #include<iostream>
2 #include<algorithm>
3 using namespace std;
4 const int MAXN=10005;
5 const int MAXL=1000000000;
6 struct re
7 {
8   int a,b;
9   char c;
10 }op[MAXN];
11
12 int n,p=0,que[MAXN];
13 bool color[MAXN<<1];
14
15 /* 二分查找 */
16 int find(int num)
17 {
18   int l=0,r=p+1,mid;
19   while (r-l>1)
20     {
21       mid=(l+r) >> 1;
22       (que[mid]<=num)? l=mid : r=mid;
23       if (que[l]==num) return l;
24     }
25   return l;
26 }
27
28 void disp(re& op)
29 {
30   /* 区间�?开始数,而不是由0开�?nbsp;*/
31   op.a=find(op.a)+1;
32   op.b=find(op.b);
33 }
34
35 void mark(int l,int r,char c)
36 {
37   for (int i=l;i<=r;++i) color[i]=(c=='b');
38 }
39
40 int main()
41 {
42  // freopen("data.in","r",stdin);
43  // freopen("data.out","w",stdout);
44   cin >> n;
45   for (int i=0;i<n;++i)
46     {
47       cin >> op[i].a >> op[i].b >> op[i].c;
48       que[++p]=op[i].a;
49       que[++p]=op[i].b;
50     }
51   que[++p]=MAXL;
52   /* 删除重复出现的数�?nbsp;*/
53   int rp=0;
54   for (int i=1;i<=p;++i)
55     if (que[rp]!=que[i]) que[++rp]=que[i];
56   p=rp;
57   /* 排序,便于定位和离散化 */
58   sort(que,que+p+1);
59   que[p+1]=MAXL+1;
60   /* �Ҏ��作进行离散处�?nbsp;*/
61   for (int i=0;i<n;++i) disp(op[i]);
62   /* 处理操作序列 */
63   memset(color,0,sizeof(color));
64   for (int i=0;i<n;++i) mark(op[i].a,op[i].b,op[i].c);
65   int t=1,a,b,mlen=0;
66   for (int i=2;i<=p+1;++i)
67     if (color[i]!=color[t] || i>p)
68       {
69     if ((!color[t]) && (que[i-1]-que[t-1]>mlen)) { a=que[t-1];b=que[i-1];mlen=que[i-1]-que[t-1];}
70     t=i;
71       }
72   cout << a << ' ' << b << endl;
73   //system("pause");
74   return 0;
75 }
76
个别目标�q�大的同学可能�ƈ不仅仅是惌��册��个题�?�q�时候徏议你使用��L��?�U�段�?复杂度会大大地降�?br>

Chen Jiecao 2009-07-20 15:00 发表评论

URAL 1018 A Binary Apple Tree

Chen Jiecao — Sun, 19 Jul 2009 15:02:00 GMT

A Binary Apple Tree

Time Limit: 1.0 second
Memory Limit: 16 MB

Let's imagine how apple tree looks in binary computer world. You're right, it looks just like a binary tree, i.e. any biparous branch splits up to exactly two new branches. We will enumerate by natural numbers the root of binary apple tree, points of branching and the ends of twigs. This way we may distinguish different branches by their ending points. We will assume that root of tree always is numbered by 1 and all numbers used for enumerating are numbered in range from 1 to N, where N is the total number of all enumerated points. For instance in the picture below N is equal to 5. Here is an example of an enumerated tree with four branches:

2 5
\ /
3 4
\ /
1

As you may know it's not convenient to pick an apples from a tree when there are too much of branches. That's why some of them should be removed from a tree. But you are interested in removing branches in the way of minimal loss of apples. So your are given amounts of apples on a branches and amount of branches that should be preserved. Your task is to determine how many apples can remain on a tree after removing of excessive branches.

Input

First line of input contains two numbers: N and Q (1 ≤ Q ≤ N; 1 < N ≤ 100). N denotes the number of enumerated points in a tree. Q denotes amount of branches that should be preserved. Next N−1 lines contains descriptions of branches. Each description consists of a three integer numbers divided by spaces. The first two of them define branch by it's ending points. The third number defines the number of apples on this branch. You may assume that no branch contains more than 30000 apples.

Output

Output should contain the only number �?amount of apples that can be preserved. And don't forget to preserve tree's root ;-)

Sample

input output

5 2
1 3 1
1 4 10
2 3 20
3 5 20

21

��?
      �q�是一个简单的树�Ş动态规划问�?大概可以拿来当这�c�题目的入门训练�?虽然�q�是URAL上的�W�一个树形DP,但是我奇怪的是它的通过率�ƈ不很�?
      对于原题目的囑�Ş,用数�l�value[a][b]表示a,b炚w��Ҏ��的个�?用nd[p].L,nd[p].R分别表示节点p的左叛_��?通过build_tree(1)获得数组nd[],从而获得整��|��的信�?
接着,用ans[p][i]表示以节点p为祖宗的子树,保留的枝条不��过i条时所能保留的最多的�Ҏ��,状态�{�U�L��一下几�U�情�?
在除��d��余枝条的后的图中,
1. p只与一个儿子相�q?
    ans[p][i]=max(ans[left_son][i-1]+value[left_son][p],ans[right_son][i-1]+value[right_son][p]);
2. p与两个儿子相�q?
    for (int j=0;j<=i-2;++j)
      ans[p][i]=max(ans[p][i],ans[left_son][j]+ans[right_son][i-j-2]+d);
    �q�里,d=value[left_son][p]+value[right_son][p];

��法在o(N*Q*Q)�U�别

1 #include<iostream>
2 using namespace std;
3 const int MAXN=102;
4 int n,q,value[MAXN][MAXN],ans[MAXN][MAXN];
5 struct node
6 {
7   int l,r;
8 }nd[MAXN];
9
10 void build_tree(int p)
11 {
12   int flg=0;
13   for (int i=1;i<=n;++i)
14     if (value[p][i] && (!nd[i].l))
15       {
16     flg=1;
17     if (nd[p].l==0) nd[p].l=i;
18     else
19       {nd[p].r=i; break;}
20       }
21   if (!flg) return;
22   if (nd[p].l) build_tree(nd[p].l);
23   if (nd[p].r) build_tree(nd[p].r);
24 }
25
26 void calc(int p)
27 {
28   if (!nd[p].l) return;
29   int l=nd[p].l,r=nd[p].r;
30   calc(l);
31   calc(r);
32   ans[p][1]=max(value[l][p],value[r][p]);
33
34   int d=value[l][p]+value[r][p];
35   for (int i=2;i<=q;++i)
36   {
37     ans[p][i]=max(ans[l][i-1]+value[l][p],ans[r][i-1]+value[r][p]);
38     for (int j=0;j<=i-2;++j)
39       ans[p][i]=max(ans[p][i],ans[l][j]+ans[r][i-j-2]+d);
40   }
41 }
42
43
44 int main()
45 {
46   //freopen("data.in","r",stdin);
47   //freopen("data.out","w",stdout);
48   cin >> n >> q;
49   memset(value,0,sizeof(value));
50   for (int i=1;i<n;++i)
51     {
52       int a,b,c;
53       cin >> a >> b >> c;
54       value[a][b]=c;
55       value[b][a]=c;
56     }
57   memset(nd,0,sizeof(nd));
58   build_tree(1);
59   calc(1);
60   cout << ans[1][q] << endl;
61   return 0;
62 }
63

Chen Jiecao 2009-07-19 23:02 发表评论

Chen Jiecao — Sat, 18 Jul 2009 08:38:00 GMT
       C++,写写OI的东西现在是够了,也看了一些书,比如 Essential C++, Accelerate C++,我的目的倒也不是惌��C��么高�?像Template,基本能用��p��,�c�M��只要有点概念,能看懂就够了.而在USACO上做�q�些已经做过的题目多��显得乏�?再�?�׃��脑子里留着对原来算法的印象,我的潜意识��L��引导我直接默�?�q�样的coding�l�对��不上一�U��n�?
      所以我打算挑几个题目做�?而不必机械式地重�?也可以去USACO上找些历�ơ月赛的题目,金牌�l�的那些�q�是有点意思的,很多要求寚w��U�数据结构进行改�?我以前做了几个题�?感觉收获极丰�?银牌�l�的也还比较有趣.
      我的耐性真的不怎么�?Vijos上做�?9�?TJU的网站上做了几十�?URAL上做了几十题,PKU做过一�?SPOJ也玩�q�一�?�q�郭佛_��同志的OJ上也切了一些题�?花样很多,可是没有一个题库上一��N��?耐性不�?也是�E�序员的大忌.
     所谓USACO,重走长征�?看来是要当逃兵�?或�?��是�t�高�?Accelerate?

Chen Jiecao 2009-07-18 16:38 发表评论

USACO 1.5.1 Number Triangles

Chen Jiecao — Fri, 17 Jul 2009 03:18:00 GMT
�q�个题目,或者说�q�样的题目是动态规划的入门例题.�q�里的最优结构很明显,��法上没有什么好讲的.实现�?��利用滚动数组以节省空�?虽然你开个二�l�的也未必会挤爆�I�间.

1 /*
2 ID: 31440461
3 LANG: C++
4 TASK: numtri
5 */
6 #include<iostream>
7 using namespace std;
8 const int MAXN=1000+10;
9 int data[2][MAXN];
10
11 int main()
12 {
13   freopen("numtri.in","r",stdin);
14   freopen("numtri.out","w",stdout);
15   int n,flg=0;
16   cin >> n;
17   for (int i=1;i<=n;++i)
18     {
19       flg=1-flg;
20       for (int j=1;j<=i;++j)
21     {
22       int x;
23       cin >> x;
24       data[flg][j]=max(data[!flg][j-1],data[!flg][j])+x;
25     }
26     }
27   int m=0;
28   for (int i=1;i<=n;++i) m=max(m,data[flg][i]);
29   cout << m << endl;
30   return 0;
31 }
32

Chen Jiecao 2009-07-17 11:18 发表评论