JXOI2017颜色

题面

loj

分析

这道题非常妙啊
对于可保留区间[l, r]
枚举右端点r
考虑l的取值范围有两重约数
记颜色i出现的最右侧位置是\(max_i\) 最左侧位置是\(min_i\)
r前最后一次出现的位置是pre[i]
1.若max[i] > r 则 l > pre[i]
2.若max[i] <= r 则 l 不能取(min[i], max[i] ]
限制一维护一下单调栈就好啦 限制二线段树维护一下

#include <cstdio>
#include <algorithm>
#include <cmath>
#include <cstring>
using namespace std;
const int N = 3e5 + 5;
const int inf = 0x3f3f3f3f;
int n, a[N], mx[N], mn[N];
long long ans;
int stk[N], top;

struct Seg{
    int w[N << 2]; bool flag[N << 2];
    void update(int rt){
        w[rt] = w[rt << 1] + w[rt << 1 | 1];
    }
    void pushdown(int rt, int l, int r){
        if(!flag[rt]) return ;
        int mid = l + ((r - l) >> 1);
        w[rt << 1] = mid - l + 1;
        w[rt << 1 | 1] = r - mid;
        flag[rt << 1] = flag[rt << 1 | 1] = 1;
        flag[rt] = 0;
    }
    void clr(int rt, int l, int r){
        w[rt] = flag[rt] = 0;
        if(l == r) return ;
        int mid = l + ((r - l) >> 1);
        clr(rt << 1, l, mid);
        clr(rt << 1 | 1, mid + 1, r);
    }
    void mdf(int rt, int l, int r, int x, int y){
        if(l > r) return ;
        if(l >= x && r <= y){
            w[rt] = r - l + 1; flag[rt] = 1;
            return ;
        }
        pushdown(rt, l, r);
        int mid = l + ((r - l) >> 1);
        if(x <= mid) mdf(rt << 1, l, mid, x, y);
        if(y > mid) mdf(rt << 1 | 1, mid + 1, r, x, y);
        update(rt);
    }
    int qry(int rt, int l, int r, int x, int y){
        if(l > r) return 0;
        if(l >= x && r <= y) return w[rt];
        pushdown(rt, l, r);
        int mid = l + ((r - l) >> 1), ret = 0;
        if(x <= mid) ret += qry(rt << 1, l, mid, x, y);
        if(y > mid) ret += qry(rt << 1 | 1, mid + 1, r, x, y);
        return ret;
    }
}seg; 

int main(){
    int T; scanf("%d", &T);
    while(T--){
        scanf("%d", &n);
        
        seg.clr(1, 1, n);
        for(int i = 0; i <= n; ++i) mx[i] = 0, mn[i] = inf;
        
        for(int i = 1; i <= n; ++i){
            scanf("%d", &a[i]);
            mx[a[i]] = max(mx[a[i]], i);
            mn[a[i]] = min(mn[a[i]], i);
        }
        ans = 0;
        top = 0;
        mn[0] = mx[0] = 0;
        for(int r = 1; r <= n; ++r){
            if(r == mx[a[r]] && r > mn[a[r]]){
                seg.mdf(1, 1, n, mn[a[r]] + 1, r);
            //  printf("%d %d\n", mn[a[r]] + 1, r);
            }
            stk[++top] = r;
            while(top && mx[a[stk[top]]] <= r) --top;
            //这里维护颜色调了好久 蠢了蠢了 注意那个pre不是单调递增的哦 
            
            ans += (r - stk[top] - seg.qry(1, 1, n, stk[top] + 1, r));
            //printf("ans = %d %d %d %d %d %d\n", ans, r, stk[top], stk[top] + 1, r, seg.qry(1, 1, n, stk[top] + 1, r));
        }
        printf("%lld\n", ans);
    }
    return 0;   
}

来源：https://www.cnblogs.com/hjmmm/p/10624563.html