[POI 2015] Kinoman

[题目链接]

          https://www.lydsy.com/JudgeOnline/problem.php?id=3747

[算法]

        首先 ， 预处理nxt[i]表示下一个和第i天放映同样电影的是哪一天

        枚举左端点 ， 不妨计算以每个点为右端点所能获得“好看值”的总和 ， 当左端点右移一位时 ， [i , nxt[i] - 1]的答案减少了w[f[i]] ， [nxt[i] , nxt[nxt[i]] - 1]的答案增加了w[f[i]]

        维护一棵支持区间修改 ， 维护最值的线段树即可

        时间复杂度 : O(NlogN)

[代码]

       

#include<bits/stdc++.h>
using namespace std;
#define MAXN 1000010
typedef long long LL;

int n , m;
LL f[MAXN] , w[MAXN];
int last[MAXN] , nxt[MAXN];

struct Segment_Tree
{
        struct Node
        {
                int l , r;
                LL mx , tag;
        } Tree[MAXN << 2];
        inline void build(int index , int l , int r)
        {
                Tree[index] = (Node){l , r , 0 , 0};
                if (l == r) return;
                int mid = (l + r) >> 1;
                build(index << 1 , l , mid);
                build(index << 1 | 1 , mid + 1 , r);
        }
        inline void pushdown(int index)
        {
                Tree[index << 1].mx += Tree[index].tag;
                Tree[index << 1 | 1].mx += Tree[index].tag;
                Tree[index << 1].tag += Tree[index].tag;
                Tree[index << 1 | 1].tag += Tree[index].tag;
                Tree[index].tag = 0;
        }
        inline void update(int index)
        {
                Tree[index].mx = max(Tree[index << 1].mx , Tree[index << 1 | 1].mx);
        }
        inline void modify(int index , int l , int r , LL value)
        {
                if (Tree[index].l == l && Tree[index].r == r)
                {
                        Tree[index].mx += value;
                        Tree[index].tag += value;
                        return;
                }
                pushdown(index);
                int mid = (Tree[index].l + Tree[index].r) >> 1;
                if (mid >= r) modify(index << 1 , l , r , value);
                else if (mid + 1 <= l) modify(index << 1 | 1 , l , r , value);
                else
                {
                        modify(index << 1 , l , mid , value);
                        modify(index << 1 | 1 , mid + 1 , r , value);
                }
                update(index);
        }
        inline LL query()
        {
                return Tree[1].mx;
        }
} SGT;

template <typename T> inline void chkmax(T &x,T y) { x = max(x,y); }
template <typename T> inline void chkmin(T &x,T y) { x = min(x,y); }
template <typename T> inline void read(T &x)
{
    T f = 1; x = 0;
    char c = getchar();
    for (; !isdigit(c); c = getchar()) if (c == '-') f = -f;
    for (; isdigit(c); c = getchar()) x = (x << 3) + (x << 1) + c - '0';
    x *= f;
}

int main()
{
        
        read(n); read(m);
        for (int i = 1; i <= n; i++) read(f[i]);
        for (int i = 1; i <= m; i++) read(w[i]);
        for (int i = n; i >= 1; i--)
        {
                nxt[i] = last[f[i]];
                last[f[i]] = i;        
        }
        SGT.build(1 , 1 , n);
        for (int i = 1; i <= m; i++)
        {
                if (last[i])
                {
                        if (!nxt[last[i]]) SGT.modify(1 , last[i] , n , w[i]);
                        else SGT.modify(1 , last[i] , nxt[last[i]] - 1 , w[i]);
                }
        }
        LL ans = 0;
        for (int i = 1; i <= n; i++)
        {
                chkmax(ans , SGT.query());
                if (nxt[i])
                {
                        SGT.modify(1 , i , nxt[i] - 1 , -w[f[i]]);
                        if (nxt[nxt[i]]) SGT.modify(1 , nxt[i] , nxt[nxt[i]] - 1 , w[f[i]]);
                        else SGT.modify(1 , nxt[i] , n , w[f[i]]);
                } else SGT.modify(1 , i , n , -w[f[i]]);
        }
        printf("%lld
" , ans);
        
        return 0;
    
}