BZOJ 3747: [POI2015]Kinoman

考虑枚举起点，找一个最远的右端点使得以这个点为起点的答案最大

用线段树维护，需要设计一种标记，使得含有多个数的前缀中减去这个数的贡献

考虑这样设计，对于第一次出现的数，对其赋值$w[i]$

第二次出现 赋值$-w[i]$

第三次出现 赋值为0

这样前缀和就满足多次出现的数不会计算贡献

再维护前驱节点和后驱节点，每次删去开头的数，维护一下后驱节点

#include <bits/stdc++.h>
using namespace std;

#define ll long long
#define N 1000010
#define INF 0x3f3f3f3f3f3f3f3f
int n, m, f[N], pre[N], nx[N];
ll w[N], v[N];
map <int, int> mp;

namespace SEG
{
    ll Max[N << 2], lazy[N << 2];
    void init()
       { 
        memset(Max, 0, sizeof Max);
        memset(lazy, 0, sizeof lazy);   
    }
    void pushdown(int id)
    {
        if (!lazy[id]) return;
        lazy[id << 1] += lazy[id];
        Max[id << 1] += lazy[id];
        lazy[id << 1 | 1] += lazy[id];
        Max[id << 1 | 1] += lazy[id]; 
        lazy[id] = 0;
    }
    void update(int id, int l, int r, int ql, int qr, ll val)
    {
        if (l >= ql && r <= qr) 
        {
            lazy[id] += val;
            Max[id] += val;
            return;
        }
        pushdown(id); 
        int mid = (l + r) >> 1;
        if (ql <= mid) update(id << 1, l, mid, ql, qr, val);
        if (qr > mid) update(id << 1 | 1, mid + 1, r, ql, qr, val);
        Max[id] = max(Max[id << 1], Max[id << 1 | 1]); 
    }
}

int main()
{
    while (scanf("%d%d", &n, &m) != EOF)
    {
        for (int i = 1; i <= n; ++i) scanf("%d", f + i);
        for (int i = 1; i <= m; ++i) scanf("%lld", w + i);
        mp.clear();
        for (int i = 1; i <= n; ++i)
        {
            if (mp.find(f[i]) != mp.end())
                pre[i] = mp[f[i]];
            else
                pre[i] = 0;
            mp[f[i]] = i;
        }
        mp.clear(); 
        for (int i = n; i >= 1; --i)
        {
            if (mp.find(f[i]) != mp.end())
                nx[i] = mp[f[i]];
            else
                nx[i] = n + 1;
            mp[f[i]] = i;
        }
        for (int i = 1; i <= n; ++i)
        {
            if (pre[i] == 0) v[i] = w[f[i]];
            else if (v[pre[i]] == w[f[pre[i]]]) v[i] = -w[f[i]]; 
            else v[i] = 0;
        }
        SEG::init();
        for (int i = 1; i <= n; ++i)
            SEG::update(1, 1, n, i, n, v[i]);
        ll res = 0;
        for (int i = 1; i <= n; ++i)
        {
            res = max(res, SEG::Max[1]);
            SEG::update(1, 1, n, i, n, -v[i]);
            if (nx[i] != n + 1)
            {
                SEG::update(1, 1, n, nx[i], n, -v[nx[i]]);
                v[nx[i]] = w[f[nx[i]]];
                SEG::update(1, 1, n, nx[i], n, v[nx[i]]);
                if (nx[nx[i]] != n + 1)
                {
                    v[nx[nx[i]]] = -v[nx[i]];
                    SEG::update(1, 1, n, nx[nx[i]], n, -v[nx[i]]);
                }
            }
        }
        printf("%lld
", res);
    }
    return 0;
}