BZOJ3688 折线统计【dp + BIT】

最新推荐文章于 2019-02-27 19:43:17 发布

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最新推荐文章于 2019-02-27 19:43:17 发布

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原文链接：http://www.cnblogs.com/Mychael/p/9260492.html

本文提供了一种解决BZOJ3688问题的有效算法，通过将点排序并定义状态f[i][j][0|1]来求解方案数，采用BIT优化实现O(knlogn)的时间复杂度。

题目链接

BZOJ3688

题解

将点排序
设\(f[i][j][0|1]\)表示以第\(i\)点结尾，有\(j\)段，最后一段上升或者下降的方案数
以上升为例
\[f[i][j][0] = \sum\limits_{k = 1}^{i - 1}\sum\limits_{y_k < y_i}f[k][j][0] + \sum\limits_{k = 1}^{i - 1}\sum\limits_{y_k < y_i}f[k][j - 1][1]\]
\(bit\)优化成\(O(knlogn)\)

#include<algorithm>
#include<iostream>
#include<cstdlib>
#include<cstring>
#include<cstdio>
#include<vector>
#include<queue>
#include<cmath>
#include<map>
#define LL long long int
#define REP(i,n) for (int i = 1; i <= (n); i++)
#define Redge(u) for (int k = h[u],to; k; k = ed[k].nxt)
#define cls(s,v) memset(s,v,sizeof(s))
#define mp(a,b) make_pair<int,int>(a,b)
#define cp pair<int,int>
#define lbt(x) (x & -x)
using namespace std;
const int maxn = 100005,maxm = 100005,INF = 0x3f3f3f3f,P = 100007;
inline int read(){
    int out = 0,flag = 1; char c = getchar();
    while (c < 48 || c > 57){if (c == '-') flag = 0; c = getchar();}
    while (c >= 48 && c <= 57){out = (out << 1) + (out << 3) + c - 48; c = getchar();}
    return flag ? out : -out;
}
int f[maxn][12][2],N = 100000;
int S[2][2][maxn],n,x[maxn],y[maxn],id[maxn],K;
void add(int* s,int u,int v){while (u <= N) s[u] = (s[u] + v) % P,u += lbt(u);}
int query(int* s,int u){int re = 0; while (u) re = (re + s[u]) % P,u -= lbt(u); return re;}
int sum(int* s,int l,int r){return query(s,r) - query(s,l - 1);}
inline bool cmp(const int& a,const int& b){return x[a] < x[b];}
int main(){
    n = read(); K = read();
    REP(i,n) x[i] = read(),y[i] = read(),id[i] = i;
    sort(id + 1,id + 1 + n,cmp);
    REP(i,n) f[i][0][0] = f[i][0][1] = 1;
    for (int k = 1; k <= K; k++){
        cls(S,0);
        for (int i = 1; i <= n; i++){
            int u = id[i];
            f[u][k][0] = (sum(S[1][0],1,y[u]) + sum(S[0][1],1,y[u])) % P;
            f[u][k][1] = (sum(S[1][1],y[u],N) + sum(S[0][0],y[u],N)) % P;
            add(S[0][0],y[u],f[u][k - 1][0]);
            add(S[1][0],y[u],f[u][k][0]);
            add(S[0][1],y[u],f[u][k - 1][1]);
            add(S[1][1],y[u],f[u][k][1]);
        }
    }
    int ans = 0;
    for (int i = 1; i <= n; i++) ans = (ans + f[i][K][0] + f[i][K][1]) % P;
    printf("%d\n",(ans + P) % P);
    return 0;
}

转载于:https://www.cnblogs.com/Mychael/p/9260492.html