HDU 2836 Traversal（线段树+离散化+DP）

来源：互联网发布：阿里云ecs无法连接ftp 编辑：程序博客网时间：2024/05/20 09:24

题目链接：点击打开链接

题意：给你n个数的序列，一个数h，求相邻数之差不超过h的子序列的个数和 % 9901。

思路：经典水题，显然用d[i]表示以a[i]结尾的满足条件的子序列个数。那么对于j < i ， | a[j] - a[i] | <= h ，等价于 a[j] <= a[i] + h && a[j] >= a[i] - h。对于这个限制用线段树下标维护，线段树用来维护d[i]的累加和。

细节参见代码：

#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)?(a):(b))#define Min(a,b) ((a)<(b)?(a):(b))using namespace std;typedef long long ll;typedef long double ld;const ld eps = 1e-9, PI = 3.1415926535897932384626433832795;const int mod = 9901;const int INF = 0x3f3f3f3f;// & 0x7FFFFFFFconst int seed = 131;const ll INF64 = ll(1e18);const int maxn = 100000 + 10;int T,n,m,sum[maxn<<2],h,d[maxn],a[maxn],b[maxn];inline void add(int& a, int b) {    a += b;    if(a >= mod) a %= mod;}void pushup(int o) {    sum[o] = sum[o<<1] + sum[o<<1|1];    if(sum[o] >= mod) sum[o] %= mod;}void build(int l, int r, int o) {    int m = (l + r) >> 1;    sum[o] = 0;    if(l == r) return ;    build(l, m, o<<1);    build(m+1, r, o<<1|1);    pushup(o);}void update(int L, int R, ll v, int l, int r, int o) {    int m = (l + r) >> 1;    if(L <= l && r <= R) {        add(sum[o], v); return ;    }    if(L <= m) update(L, R, v, l, m, o<<1);    if(m < R) update(L, R, v, m+1, r, o<<1|1);    pushup(o);}int query(int L, int R, int l, int r, int o) {    int m = (l + r) >> 1;    if(L <= l && r <= R) return sum[o];    int ans = 0;    if(L <= m) add(ans, query(L, R, l, m, o<<1));    if(m < R) add(ans, query(L, R, m+1, r, o<<1|1));    pushup(o);    return ans;}int main() {    while(~scanf("%d%d",&n,&h)) {        for(int i=1;i<=n;i++) {            scanf("%d",&a[i]);            b[i-1] = a[i];        }        sort(b, b+n);        int len = unique(b, b+n) - b;        build(1, len, 1);        int ans = 0;        for(int i=1;i<=n;i++) {            int l = lower_bound(b, b+len, a[i]-h) - b + 1;            int r = upper_bound(b, b+len, a[i]+h) - b ;            int pos = lower_bound(b, b+len, a[i]) - b + 1;            d[i] = 1;            add(d[i], query(l, r, 1, len, 1));            add(ans, d[i]-1);            update(pos, pos, d[i], 1, len, 1);        }        printf("%d\n",ans);    }    return 0;}

0 0