hdu5542 The Battle of Chibi【树状数组】【离散化】

The Battle of Chibi

Time Limit: 6000/4000 MS (Java/Others)    Memory Limit: 65535/65535 K (Java/Others)
Total Submission(s): 2899    Accepted Submission(s): 1043

Problem Description

Cao Cao made up a big army and was going to invade the whole South China. Yu Zhou was worried about it. He thought the only way to beat Cao Cao is to have a spy in Cao Cao's army. But all generals and soldiers of Cao Cao were loyal, it's impossible to convince any of them to betray Cao Cao.

So there is only one way left for Yu Zhou, send someone to fake surrender Cao Cao. Gai Huang was selected for this important mission. However, Cao Cao was not easy to believe others, so Gai Huang must leak some important information to Cao Cao before surrendering.

Yu Zhou discussed with Gai Huang and worked out N information to be leaked, in happening order. Each of the information was estimated to has ai value in Cao Cao's opinion.

Actually, if you leak information with strict increasing value could accelerate making Cao Cao believe you. So Gai Huang decided to leak exact M information with strict increasing value in happening order. In other words, Gai Huang will not change the order of the N information and just select M of them. Find out how many ways Gai Huang could do this.

 

Input

The first line of the input gives the number of test cases, T(1≤100). T test cases follow.

Each test case begins with two numbers N(1≤N≤103) and M(1≤M≤N), indicating the number of information and number of information Gai Huang will select. Then N numbers in a line, the ith number ai(1≤ai≤109) indicates the value in Cao Cao's opinion of the ith information in happening order.

 

Output

For each test case, output one line containing Case #x: y, where x is the test case number (starting from 1) and y is the ways Gai Huang can select the information.

The result is too large, and you need to output the result mod by 1000000007(109+7).

 

Sample Input

2
3 2
1 2 3
3 2
3 2 1

 

Sample Output

Case #1: 3
Case #2: 0

Hint

In the first cases, Gai Huang need to leak 2 information out of 3. He could leak any 2 information as all the information value are in increasing order.
In the second cases, Gai Huang has no choice as selecting any 2 information is not in increasing order.

 

Source

The 2015 China Collegiate Programming Contest

 

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题意：

问一个序列之中有多少个长度为M的严格递增子序列。

思路：

我们用dp[i][j]表示前j个数中，以Aj为结尾的长度为i的严格递增子序列的个数。

那么对于dp[i][j]，我们只需要枚举所有小于j的k，并且Ak < Aj，将所有的dp[i-1][k]求和，可以得到dp[i][j]

很容易想到O(n^3)的算法，但是显然会超时，所以我们需要进行一些优化。

当我们枚举内层循环j,k时，外层循环i就可以被当成是定值。当j增加1，k的取值只是多了k = j这个新的决策。

因此我们用树状数组维护一个前缀和，表示1~j区间，长度为i-1时的方案数。

由于add时第一个参数放的是a[j]，也就是说现在直接用了他的权值作为了下标，所以只要下标比他小的，权值一定比他小，也就不需要进行比较了。直接去前缀和就可以了。【日常感谢家庭教师】

最开始离散化用的set和map TLE了

后来改用了另一个数组，先排序然后lower_bound，并且直接把原数组的值改掉

 #include <iostream>
 #include <set>
 #include <cmath>
 #include <stdio.h>
 #include <cstring>
 #include <algorithm>
 #include <map>
 using namespace std;
 typedef long long LL;
 #define inf 0x7f7f7f7f

 int t, n, m, cnt;
 const int maxn = ;
 const LL mod = 1e9 + ;
 int a[maxn], b[maxn];
 LL sum[maxn], dp[maxn][maxn];
 map<LL, int>discrete;
 set<LL>sss;
 set<LL>::iterator iter;

 void add(int pos, LL x)
 {
     while(pos <= n + ){
         sum[pos] = (sum[pos] + x) % mod;
         pos += (pos & -pos);
     }
 }

 LL ask(int pos)
 {
     LL ans = ;
     while(pos){
         ans = (ans + sum[pos]) % mod;;
         pos -= (pos & -pos);
     }
     return ans;
 }

 void init()
 {
     discrete.clear();
     sss.clear();
     //memset(dp, 0, sizeof(dp));
     for(int i = ; i <= m; i++){
         for(int j = ; j <= n; j++){
             dp[i][j] = ;
         }
     }
     cnt = ;
 }

 int main()
 {
     scanf("%d", &t);
     for(int cas = ; cas <= t; cas++){
         init();
         scanf("%d%d", &n, &m);
         a[] = b[] = -inf;
         dp[][] = ;
         //sss.insert(a[0]);
         for(int i = ; i <= n; i++){
             scanf("%d", &a[i]);
             b[i] = a[i];
             //sss.insert(a[i]);
         }
         sort(b, b + n + );
         for(int i = ; i <= n; i++){
             a[i] = lower_bound(b, b +  + n, a[i]) - b + ;
             //cout<<a[i]<<endl;
         }
         /*for(iter = sss.begin(); iter != sss.end(); iter++){
             discrete[*iter] = ++cnt;
         }*/
         //cout<<a[0]<<endl;
         for(int i = ; i <= m; i++){
             //memset(sum, 0, sizeof(sum));
             for(int j = ; j <= n + ; j++){
                 sum[j] = ;
             }
             add(a[], dp[i - ][]);

             for(int j = ; j <= n; j++){
                 dp[i][j] = ask(a[j] - );
                 //if(discrete[a[j]] < discrete[a[j + 1]])
                 add(a[j], dp[i - ][j]);
             }
         }

         int ans = ;
         for(int i = ; i <= n; i++){
             ans = (ans + dp[m][i]) % mod;
         }
         printf("Case #%d: %d\n", cas, ans);
     }
 }