2016北京集训测试赛（十七）Problem C: 数组

Solution

线段树好题.
我们考虑用last[i]表示(i)这个位置的颜色的上一个出现位置. 考虑以一个位置(R)为右端点的区间最远能向左延伸到什么位置: (L = max_{i le j} last[j]).
而我们的答案就等于

[sum_{i = 1}^n (i - (max_{1 le j le i} last[j])) = sum_{i = 1}^n i - sum_{i = 1}^n max_{1 le j le i} last[j] ]

第一项可以直接计算, 考虑如何维护第二项.
我们开一颗线段树, 假设一个节点所维护的区间是([L, R]), 则节点维护(sum_{i = L}^R max_{L le j le i} last[j]). 具体怎么实现, 看代码即可.

#include <cstdio>
#include <cctype>
#include <set>
#include <algorithm>
#define iter set<int>::iterator
 
using namespace std;
namespace Zeonfai
{
    inline int getInt()
    {
        int a = 0, sgn = 1; char c;
        while(! isdigit(c = getchar())) if(c == '-') sgn *= -1;
        while(isdigit(c)) a = a * 10 + c - '0', c = getchar();
        return a * sgn;
    }
}
const int N = (int)1e5;
int n;
int col[N + 1], pre[N + 1];
set<int> st[N + 1];
inline iter getPrevious(iter p) {return -- p;}
inline iter getNext(iter p) {return ++ p;}
struct segmentTree
{
    struct node
    {
        int mx;
        long long sum;
    }nd[N << 2];
    long long work(int u, int L, int R, int val)
    {
        int mid = L + R >> 1;
        if(L == R) {return max(val, nd[u].mx);}
        else if(nd[u].mx <= val) {return (long long)(R - L + 1) * val;}
        else if(nd[u << 1].mx <= val) return (long long)val * (mid - L + 1) + work(u << 1 | 1, mid + 1, R, val);
        else return work(u << 1, L, mid, val) + nd[u].sum - nd[u << 1].sum;
    }
    inline void pushUp(int u, int L, int R)
    {
        int mid = L + R >> 1;
        nd[u].mx = max(nd[u << 1].mx, nd[u << 1 | 1].mx);
        nd[u].sum = (long long)nd[u << 1].sum + work(u << 1 | 1, mid + 1, R, nd[u << 1].mx);
    }
    void build(int u, int L, int R)
    {
        if(L == R) {nd[u].mx = nd[u].sum = pre[L]; return;}
        int mid = L + R >> 1;
        build(u << 1, L, mid); build(u << 1 | 1, mid + 1, R);
        pushUp(u, L, R);
    }
    inline void build() {build(1, 1, n);}
    void modify(int u, int L, int R, int pos, int val)
    {
        if(L == R) {nd[u].sum = nd[u].mx = val; return;}
        int mid = L + R >> 1;
        if(pos <= mid) modify(u << 1, L, mid, pos, val); else modify(u << 1 | 1, mid + 1, R, pos, val);
        pushUp(u, L, R);
    }
    inline void modify(int pos, int val) {modify(1, 1, n, pos, val);}
}seg;
int main()
{
 
    #ifndef ONLINE_JUDGE
 
    freopen("array.in", "r", stdin);
    freopen("array.out", "w", stdout);
 
    #endif
 
    using namespace Zeonfai;
    n = getInt();
    for(int i = 1; i <= n; ++ i) st[i].clear();
    for(int i = 1; i <= n; ++ i)
    {
        col[i] = getInt();
        if(st[col[i]].empty()) pre[i] = 0; else pre[i] = *getPrevious(st[col[i]].find(i));
        st[col[i]].insert(i);
    }
    seg.build();
    int m = getInt();
    for(int i = 0; i < m; ++ i)
    {
        int opt = getInt();
        if(opt)
        {
            int pos = getInt(), x = getInt();
            iter p = getNext(st[col[pos]].find(pos));
            st[col[pos]].erase(st[col[pos]].find(pos));
            if(p != st[col[pos]].end()) pre[*p] = p == st[col[pos]].begin() ? 0 : *getPrevious(p);
            if(p != st[x].end()) seg.modify(*p, pre[*p]);
            col[pos] = x; st[x].insert(pos);
            pre[pos] = st[x].find(pos) == st[x].begin() ? 0 : *getPrevious(st[x].find(pos));
            p = getNext(st[x].find(pos)); if(p != st[x].end()) pre[*p] = pos;
            if(p != st[x].end()) seg.modify(*p, pre[*p]);
            seg.modify(pos, pre[pos]);
        }
        else printf("%lld
", (long long)n * (n + 1) / 2 - seg.nd[1].sum);
    }
}