[2019杭电多校第三场][hdu6606]Distribution of books(线段树&&dp)

题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=6606

题意为在n个数中选m(自选)个数，然后把m个数分成k块，使得每块数字之和最大的最小。

求数字和最大的最小一般都是二分，二分后可以dp来判断合法，dp[i]表示第i个数字最大可以在的块数。则$dp[i]=max(dp[j])+1,{sum[i]-sum[j]<=x}$，sum为前缀和，x为二分的值。

但是这样的复杂度O(n²logn)，显然不行。

则可以优化一下dp，dp的合法转移条件是sum[i]-sum[j]<=x，移项为sum[j]>=sum[i]-x。所以将前缀和离散化，然后每次在权值线段树上查询大于等于sum[i]-x的位置的最大值，即是每次转移的最优位置。

然后用答案更新sum[i]位置的权值即可。

 #include<iostream>
 #include<algorithm>
 #include<cstring>
 #include<string>
 #include<cmath>
 #include<vector>
 #define lson l,mid,i<<1
 #define rson mid+1,r,i<<1|1
 using namespace std;
 typedef long long ll;
 const ll maxn = 2e5 + ;
 const ll inf = 1e18;
 ll Max[maxn * ];
 ll a[maxn], b[maxn];
 void up(ll i) {
     Max[i] = max(Max[i << ], Max[i <<  | ]);
 }
 void build(ll l, ll r, ll i) {
     Max[i] = -inf;
     if (l == r)
         return;
     ll mid = l + r >> ;
     build(lson);
     build(rson);
 }
 void update(ll pos, ll val, ll l, ll r, ll i) {
     if (l == r) {
         Max[i] = val;
         return;
     }
     ll mid = l + r >> ;
     if (pos <= mid)update(pos, val, lson);
     else update(pos, val, rson);
     up(i);
 }
 ll query(ll L, ll R, ll l, ll r, ll i) {
     if (L > R)return -inf;
     if (L <= l && r <= R)
         return Max[i];
     ll ans = -inf;
     ll mid = l + r >> ;
     if (L <= mid)ans = max(ans, query(L, R, lson));
     if (R > mid)ans = max(ans, query(L, R, rson));
     return ans;
 }
 ll n, k;
 ll dp[maxn];
 bool check(ll x, ll len) {
     build(, len, );
     for (ll i = ; i <= n; i++) {
         ll pos = lower_bound(b + , b +  + len, a[i] - x) - b;
         ll pos2 = lower_bound(b + , b +  + len, a[i]) - b;
         dp[i] = query(pos, len, , len, ) + ;
         if (a[i] <= x)
             dp[i] = max(dp[i], 1ll);
         if (dp[i] >= k)
             return true;
         if (dp[i] > )
             update(pos2, dp[i], , len, );
     }
     return false;
 }
 int main() {
     int t;
     scanf("%d", &t);
     while (t--) {
         ll x;
         scanf("%lld%lld", &n, &k);
         for (ll i = ; i <= n; i++)
             scanf("%lld", &x), a[i] = a[i - ] + x, b[i] = a[i];
         sort(b + , b +  + n);
         ll len = unique(b + , b +  + n) - b - ;
         ll l = -inf, r = inf, ans;
         while (l <= r) {
             ll mid = l + r >> ;
             if (check(mid, len)) {
                 ans = mid;
                 r = mid - ;
             }
             else
                 l = mid + ;
         }
         printf("%lld\n", ans);
     }
 }