Bzoj1109 [poi2007] stacked wood klo

Dynamic Planning is the first thing we can see.

Then write the equation: F [I] indicates the maximum number of I blocks that meet the requirements.

Then f [I] = max (F [J] + 1)

J meets the following three conditions:

(1) j <I

(2) A [J] <A [I]

(3) A [I]-A [J] <= I-j

Deformation (3): A [I]-I <= A [J]-J... (4)

At this time, the zyh God told me that (2) + (3) can launch (1), that is, only two conditions (2) and (4) can be met.

 

The obtained method is as follows: first sort (first keyword A [I]-I in ascending order, second keyword A [I] in descending order), and then directly obtain the lis of the new series.

In fact, when lis is obtained here, the maximum value of [1, I] is required each time. Line tree maintenance is not required, as long as the tree array is enough.

 

 1 /************************************************************** 2     Problem: 1109 3     User: rausen 4     Language: C++ 5     Result: Accepted 6     Time:332 ms 7     Memory:3148 kb 8 ****************************************************************/ 9  10 #include <cstdio>11 #include <cmath>12 #include <algorithm>13  14 using namespace std;15  16 struct Rec{17     int num, x;18 } a[150000];19 int ans, n, f[150000], Bit[150000];20  21 bool cmp(Rec a, Rec b){22     return a.x == b.x ? a.num < b.num : a.x > b.x;23 } 24  25 inline int lowbit(int x){26     return x & (-x);27 }28  29 int query(int x){30     if (!x) return 0;31     int res = 0;32     while (x){33         res = max(res, Bit[x]);34         x -= lowbit(x);35     }36     return res;37 }38  39 void add(int x, int y){40     while (x <= n){41         Bit[x] = max(Bit[x], y);42         x += lowbit(x);43     }44 }45  46 int main(){47     scanf("%d", &n);48     for (int i = 1; i <= n; ++i){49         scanf("%d", &a[i].num);50         a[i].x = a[i].num - i;51     }52     sort(a + 1, a + n + 1, cmp);53      54     for (int i = 1; i <= n; ++i){55         if (a[i].x > 0) continue;56         int X = query(a[i].num - 1);57         f[i] = X + 1;58         ans = max(ans, f[i]);59         add(a[i].num, f[i]);60     }       61     printf("%d\n", ans);62     return 0;63 }

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Bzoj1109 [poi2007] stacked wood klo