• codechef QCHEF(不删除莫队)


    题意

    题目链接

    给出长度为(n)的序列,每次询问区间([l, r]),要求最大化

    (max |x − y| : L_i ≤ x, y ≤ R_i and A_x = A_y)

    Sol

    标算神仙的一批看不懂。

    维护好每个数出现的左右位置之后直接上不删除莫队就行了

    #include<bits/stdc++.h>
    
    const int MAXN = 1e5 + 10, INF = 1e9 + 7;
    using namespace std;
    template<typename A, typename B> inline bool chmax(A &x, B y) {
    	if(y > x) {x = y; return 1;}
    	else return 0;
    }
    template<typename A, typename B> inline bool chmin(A &x, B y) {
    	if(y < x) {x = y; return 1;}
    	else return 0;
    }
    inline int read() {
    	char c = getchar(); int x = 0, f = 1;
    	while(c < '0' || c > '9') {if(c == '-') f = -1; c = getchar();}
    	while(c >= '0' && c <= '9') x = x * 10 + c - '0', c = getchar();
    	return x * f;
    }
    int N, K, M, a[MAXN], l[MAXN], r[MAXN], tl[MAXN], tr[MAXN], belong[MAXN], block, cnt, ans[MAXN];
    struct query {
    	int l, r, id;
    	bool operator < (const query &rhs) const {
    		return belong[l] == belong[rhs.l] ? r < rhs.r : belong[l] < belong[rhs.l];
    	}
    };
    vector<query> q[MAXN];
    int solve(int l, int r) {
    	for(int i = l; i <= r; i++) tl[a[i]] = INF, tr[a[i]] = 0;
    	int ans = 0;
    	for(int i = l; i <= r; i++) {
    		int x = a[i];
    		chmin(tl[x], i), chmax(tr[x], i);
    		chmax(ans, tr[x] - tl[x]);
    	}
    	return ans;
    }
    int update(int x) {
    	chmax(tr[a[x]], x);
    	chmin(tl[a[x]], x);
    	return tr[a[x]] - tl[a[x]];
    }
    int update2(int x) {
    	int v = a[x];
    	chmax(r[v], x);
    	chmin(l[v], x);
    	return max(tr[v] - l[v], r[v] - l[v]);
    }
    void LxlDuLiu(vector<query> v, int id) {
    	int base = id * block, ll = base, rr = ll - 1, pre = 0, now = 0;
    	memset(tr, 0, sizeof(tr)); memset(r, 0, sizeof(r));
    	memset(tl, 0x3f, sizeof(tl)); memset(l, 0x3f, sizeof(l));
    
    	for(auto &x : v) {
    		//memset(l, 0x3f, sizeof(l)); memset(r, 0, sizeof(r));
    		while(rr < x.r) chmax(now, update(++rr));
    		pre = now;
    		while(ll > x.l) chmax(now, update2(--ll));
    		chmax(ans[x.id], now);
    		while(ll < base) r[a[ll]] = 0, l[a[ll]] = INF, ll++;
    		now = pre;
    	}
    }
    int main() {
    //	freopen("a.in", "r", stdin);
    	//freopen("b.out", "w", stdout);
    	
    	N = read(); K = read(); M = read(); block = sqrt(N); int mx = 0;
    	for(int i = 1; i <= N; i++) a[i] = read(), belong[i] = (i - 1) / block + 1, chmax(mx, belong[i]);
    	for(int i = 1; i <= M; i++) {
    		int l = read(), r = read();
    		if(belong[l] == belong[r]) ans[i] = solve(l, r);
    		else q[belong[l]].push_back({l, r, i});
    	}
    	for(int i = 1; i <= mx; i++) sort(q[i].begin(), q[i].end());
    	for(int i = 1; i <= mx; i++)
    		LxlDuLiu(q[i], i);
    	for(int i = 1; i <= M; i++) printf("%d
    ", ans[i]);
    	
    	return 0;
    }
    
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  • 原文地址:https://www.cnblogs.com/zwfymqz/p/10357461.html
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