hdu 4719 Oh My Holy FFF(dp线段树优化)

Oh My Holy FFF

Time Limit: 5000/2500 MS (Java/Others)    Memory Limit: 65535/65535 K (Java/Others)
Total Submission(s): 848    Accepted Submission(s): 219

Problem Description

N soldiers from the famous "*FFF* army" is standing in a line, from left to right.

 o   o   o   o   o   o   o   o   o   o   o   o   o   o   o   o   o   o
/F\ /F\ /F\ /F\ /F\ /F\ /F\ /F\ /F\ /F\ /F\ /F\ /F\ /F\ /F\ /F\ /F\ /F\
/ \ / \ / \ / \ / \ / \ / \ / \ / \ / \ / \ / \ / \ / \ / \ / \ / \ / \

You, as the captain of *FFF*, want to divide them into smaller groups, but each group should still be continous in the original line. Like this:

 o   o   o  |  o   o   o   o  |  o   o   o   o   o   o  |  o   o   o   o   o
/F\ /F\ /F\ | /F\ /F\ /F\ /F\ | /F\ /F\ /F\ /F\ /F\ /F\ | /F\ /F\ /F\ /F\ /F\
/ \ / \ / \ | / \ / \ / \ / \ | / \ / \ / \ / \ / \ / \ | / \ / \ / \ / \ / \

In your opinion, the number of soldiers in each group should be no more than L.
Meanwhile, you want your division be "holy". Since the soldier may have different heights, you decide that for each group except the first one, its last soldier(which is the rightmost one) should be strictly taller than the previous group's last soldier. That is, if we set bi as the height of the last soldier in group i. Then for i >= 2, there should be b_i > b_i-1.
You give your division a score, which is calculated as , b₀ = 0 and 1 <= k <= M, if there are M groups in total. Note that M can equal to 1.
Given the heights of all soldiers, please tell us the best score you can get, or declare the division as impossible.

 

Input

The first line has a number T (T <= 10) , indicating the number of test cases.
For each test case, first line has two numbers N and L (1 <= L <= N <= 10⁵), as described above.
Then comes a single line with N numbers, from H1 to Hn, they are the height of each soldier in the line, from left to right. (1 <= H_i <= 10⁵)

 

Output

For test case X, output "Case #X: " first, then output the best score.

 

Sample Input

2
5 2
1 4 3 2 5
5 2
5 4 3 2 1

 

Sample Output

Case #1: 31
Case #2: No solution

 

Source

2013 ACM/ICPC Asia Regional Online —— Warmup2

 

Recommend

zhuyuanchen520

 

题意：有n个数，划分为多个部分，如果M份，每份不能多于L个。每一个数有一个h[i]，每份最右边的那个数要大于前一份最右边的那个数。

设每份最右         边的数为b[i]。求最大的sum{b[i]² - b[i - 1]}，1≤i≤M，当中b[0] = 0。

题解：dp[i]表示前i个数能达到的最大值。dp[i]=max(dp[i],dp[j]+h[i]*h[i]-h[j]),i-L<=j<i.

         线段树优化：先排个序。按h从小到大排。相等的pos大的在前。这是为了保证严格升序。

         然后逐个插入，更新a[i].pos位置的单点值，再维护区间最大值，具体见代码。

#include<cstring>
#include<cstdio>
#include<iostream>
#include<algorithm>
#include<cmath>
#include<queue>
#define lson l,mid,idx<<1
#define rson mid+1,r,idx<<1|1
#define lc idx<<1
#define rc idx<<1|1
#define N 100010
#define ll long long

using namespace std;

int n,m;

struct node {
    int pos;
    ll num;
} a[N];

ll tree[N<<2];

void build(int l,int r,int idx) {
    tree[idx]=-1;
    if(r==l)return;
    int mid=(l+r)>>1;
    build(lson);
    build(rson);
}

void update(int l,int r,int idx,int x,ll k) {///把x位置的值更新
    if(l==r) {
        tree[idx]=max(tree[idx],k);
        return;
    }
    int mid=(l+r)>>1;
    if(x<=mid)update(lson,x,k);
    else      update(rson,x,k);
    tree[idx]=max(tree[lc],tree[rc]);
}

ll query(int l,int r,int idx,int x,int y) {
    if(l>=x&&y>=r) {
        return tree[idx];
    }
    int mid=(l+r)>>1;
    ll ans=-1;
    if(x<=mid)ans=max(ans,query(lson,x,y));
    if(y>mid) ans=max(ans,query(rson,x,y));
    return ans;
}

bool cmp(node a,node b) {
    if(a.num==b.num)return a.pos>b.pos;
    return a.num<b.num;
}

int main() {
    //freopen("test.in","r",stdin);
    int t;
    cin>>t;
    int ca=1;
    while(t--) {
        scanf("%d%d",&n,&m);
        for(int i=1; i<=n; i++) {
            scanf("%I64d",&a[i].num);
            a[i].pos=i+1;
        }
        sort(a+1,a+n+1,cmp);
        n++;
        build(1,n,1);
        update(1,n,1,1,0);
        ll ans=-1;
        for(int i=1; i<n; i++) {
            int l=(a[i].pos-m)>0?
(a[i].pos-m):1,r=a[i].pos-1;///能一组的最左区间和最右区间-1
            ll it=query(1,n,1,l,r);
            if(it==-1&&a[i].pos==n) {///-1表示分不了组
                break;
            }
            if(it==-1) continue;
            it+=a[i].num*a[i].num;
            if(a[i].pos==n) {//已经更新到n了。直接跳出
                ans=it;
                break;
            }
            update(1,n,1,a[i].pos,it-a[i].num);//减去当前a[i].num,非常巧妙
        }
        printf("Case #%d: ",ca++);
        if(ans==-1)printf("%s\n","No solution" );
        else         printf("%I64d\n",ans );
    }
    return 0;
}