HDU 5773 The All-Purpose Zero (DP)

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Ideas:

First of all: we have to use all 0 is sure not to lose.

Next: We will find the remaining non-0 number of LIS, but these 0 may not be all inserted in the LIS, it is bound to be in the LIS out some number. A very simple way, we subtract each number from the number of 0 in front of him, so that the equivalent of 0 to leave the space, because it is all reduced, the relative size is constant.

See the code for details:

#include <cstdio> #include <cstring> #include <algorithm> #include <iostream> #include < string> #include <vector> #include <stack> #include <bitset> #include <cstdlib> #include < cmath> #include <set> #include <list> #include <deque> #include <map> #include <queue># Define MAX (a) > (b)? ( A):(B) #define MIN (a) < (b) ( A):(B)) using namespace Std;typedef long long ll;typedef long double ld;const double eps = 1e-6;const double PI = ACOs (-1); const INT mod = 1e9 + 7;const int INF = 0x3f3f3f3f;const int seed = 131;const ll INF64 = ll (1e18); const int MAXN = 1e5 + 1    0;int N, M, D[MAXN], G[MAXN], a[maxn], Kase = 0;int main () {int T; scanf ("%d", &t);        while (t--) {scanf ("%d", &n);            for (int i = 1; I <= n; i++) {g[i] = INF;        scanf ("%d", &a[i]);        } int num = 0, ans = 0;           for (int i = 1; I <= n; i++) {if (a[i] = = 0) {     num++;            Continue            } a[i]-= num;            int k = Lower_bound (g+1, g+n+1, A[i])-G;            D[i] = k;            G[K] = A[i];        ans = max (ans, d[i]);    } printf ("Case #%d:%d\n", ++kase, Ans+num); } return 0;}

HDU 5773 The All-Purpose Zero (DP)