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;

}