• 【BZOJ 2120】 数颜色


    【题目链接】

             https://www.lydsy.com/JudgeOnline/problem.php?id=2120

    【算法】

                   带修改的莫队算法

    【代码】

               

    #include<bits/stdc++.h>
    using namespace std;
    #define MAXN 10010
    const int MAXC = 1e6 + 5;
    
    int i,len,x,y,sum,l,r,n,now,qnum,m,cnt;
    int a[MAXN],block[MAXN],s[MAXC],ans[MAXN];
    char op[5];
    
    struct info
    {
        int x,y,last;
    } c[MAXN];
    struct Query
    {
        int x,y,id,cur;
    } q[MAXN];
    
    inline bool cmp(Query a,Query b)
    {
        if (block[a.x] == block[b.x])
        {
            if (block[a.y] == block[b.y]) return a.cur < b.cur;
            else return block[a.y] < block[b.y];
        } else return block[a.x] < block[b.x];        
    }
    inline void add(int x)
    {
        s[a[x]]++;
        if (s[a[x]] == 1) sum++;    
    }
    inline void dec(int x)
    {
        s[a[x]]--;
        if (s[a[x]] == 0) sum--;
    }
    inline void worka(int now)
    {
        int x = c[now].x,y = c[now].y;
        if (x >= l && x <= r) dec(x);
        c[now].last = a[x];
        a[x] = y;
        if (x >= l && x <= r) add(x);    
    }
    inline void workb(int now)
    {
        int x = c[now].x,y = c[now].y;
        if (x >= l && x <= r) dec(x);
        a[x] = c[now].last;
        if (x >= l && x <= r) add(x);
    }
    
    int main()
    {
        
        scanf("%d%d",&n,&m);
        len = sqrt(n);
        for (i = 1; i <= n; i++) block[i] = (i - 1) / len + 1;
        for (i = 1; i <= n; i++) scanf("%d",&a[i]); 
        for (i = 1; i <= m; i++)
        {
            scanf("%s%d%d",&op,&x,&y);
            if (op[0] == 'Q') q[++qnum] = (Query){x,y,qnum,cnt};
            else c[++cnt] = (info){x,y,0};    
        }
        sort(q+1,q+qnum+1,cmp);
        l = 1; r = 0; now = 0; sum = 0;
        for (i = 1; i <= qnum; i++)
        {
            for (; r < q[i].y; r++) add(r+1);        
            for (; r > q[i].y; r--) dec(r);
            for (; l < q[i].x; l++) dec(l);
            for (; l > q[i].x; l--) add(l-1);
            for (now; now < q[i].cur; now++) worka(now+1);
            for (now; now > q[i].cur; now--) workb(now);
            ans[q[i].id] = sum;
        }
        for (i = 1; i <= qnum; i++) printf("%d
    ",ans[i]);
    
        return 0;
    }
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  • 原文地址:https://www.cnblogs.com/evenbao/p/9312997.html
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