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HDU 6047 Maximum Sequence (贪心+单调队列)
题意:给定一个序列,让你构造出一个序列,满足条件,且最大。条件是 选取一个ai <= max{a[b[j], j]-j}
析:贪心,贪心策略就是先尽量产生大的,所以就是对于B序列尽量从头开始,由于数据比较大,采用桶排序,然后维护一个单调队列,使得最头上最大。
代码如下:
#pragma comment(linker, "/STACK:1024000000,1024000000") #include <cstdio> #include <string> #include <cstdlib> #include <cmath> #include <iostream> #include <cstring> #include <set> #include <queue> #include <algorithm> #include <vector> #include <map> #include <cctype> #include <cmath> #include <stack> #include <sstream> #include <list> #define debug() puts("++++"); #define gcd(a, b) __gcd(a, b) #define lson l,m,rt<<1 #define rson m+1,r,rt<<1|1 #define freopenr freopen("in.txt", "r", stdin) #define freopenw freopen("out.txt", "w", stdout) using namespace std; typedef long long LL; typedef unsigned long long ULL; typedef pair<int, int> P; const int INF = 0x3f3f3f3f; const double inf = 0x3f3f3f3f3f3f; const double PI = acos(-1.0); const double eps = 1e-8; const int maxn = 250000 + 10; const LL mod = 1e9 + 7; const int dr[] = {-1, 0, 1, 0}; const int dc[] = {0, 1, 0, -1}; const char *de[] = {"0000", "0001", "0010", "0011", "0100", "0101", "0110", "0111", "1000", "1001", "1010", "1011", "1100", "1101", "1110", "1111"}; int n, m; const int mon[] = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31}; const int monn[] = {0, 31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31}; inline bool is_in(int r, int c){ return r > 0 && r <= n && c > 0 && c <= m; } int num[maxn]; int a[maxn<<1]; int q[maxn<<1]; int main(){ while(scanf("%d", &n) == 1){ memset(num, 0, sizeof num); for(int i = 1; i <= n; ++i) scanf("%d", a+i); for(int i = 0; i < n; ++i){ scanf("%d", &m); ++num[m]; } int fro = 0, rear = 0; q[++rear] = 1; for(int i = 2; i <= n; ++i){ while(rear > fro && a[i] - i >= a[q[rear]] - q[rear]) --rear; q[++rear] = i; } int cnt = 1; for(int i = n+1; i <= n+n; ++i){ while(!num[cnt]) ++cnt; --num[cnt]; while(q[fro+1] < cnt) ++fro; a[i] = a[q[fro+1]] - q[fro+1]; while(rear > fro && a[i] - i >= a[q[rear]] - q[rear]) --rear; q[++rear] = i; } int ans = 0; for(int i = n+1; i <= n+n; ++i) ans = (ans + a[i]) % mod; printf("%d\n", ans); } return 0; }
HDU 6047 Maximum Sequence (贪心+单调队列)
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