洛谷P4559 [JSOI2018]列队 【70分二分 + 主席树】
2018-06-16 19:52
375 查看
题目链接
题解
只会做\(70\)分的\(O(nlog^2n)\)
如果本来就在区间内的人是不用动的,区间右边的人往区间最右的那些空位跑,区间左边的人往区间最左的那些空位跑
找到这些空位就用二分 + 主席树
理应可以在主席树上的区间二分而做到\(O(nlogn)\),但是写不出来,先留着坑
#include<algorithm> #include<iostream> #include<cstring> #include<cstdio> #include<cmath> #include<map> #define REP(i,n) for (register int i = 1; i <= (n); i++) #define mp(a,b) make_pair<int,long long int>(a,b) #define cp pair<int,long long int> #define LL long long int using namespace std; const int maxn = 500005,maxm = 11000005,INF = 1000000000; inline int read(){ int out = 0,flag = 1; char c = getchar(); while (c < 48 || c > 57){if (c == '-') flag = -1; c = getchar();} while (c >= 48 && c <= 57){out = (out << 3) + (out << 1) + c - 48; c = getchar();} return out * flag; } int N,n,m,rt[maxn]; int ls[maxm],rs[maxm],num[maxm],cnt; LL sum[maxm]; void modify(int& u,int pre,int l,int r,int pos){ u = ++cnt; sum[u] = sum[pre] + pos; num[u] = num[pre] + 1; ls[u] = ls[pre]; rs[u] = rs[pre]; if (l == r) return; int mid = l + r >> 1; if (mid >= pos) modify(ls[u],ls[pre],l,mid,pos); else modify(rs[u],rs[pre],mid + 1,r,pos); } int q_num(int u,int v,int l,int r,int L,int R){ if (l >= L && r <= R) return num[u] - num[v]; int mid = l + r >> 1; if (mid >= R) return q_num(ls[u],ls[v],l,mid,L,R); if (mid < L) return q_num(rs[u],rs[v],mid + 1,r,L,R); return q_num(ls[u],ls[v],l,mid,L,R) + q_num(rs[u],rs[v],mid + 1,r,L,R); } LL q_sum(int u,int v,int l,int r,int L,int R){ if (l >= L && r <= R) return sum[u] - sum[v]; int mid = l + r >> 1; if (mid >= R) return q_sum(ls[u],ls[v],l,mid,L,R); if (mid < L) return q_sum(rs[u],rs[v],mid + 1,r,L,R); return q_sum(ls[u],ls[v],l,mid,L,R) + q_sum(rs[u],rs[v],mid + 1,r,L,R); } inline LL S(int l,int r){ return 1ll * (l + r) * (r - l + 1) / 2; } inline LL q_pre(int u,int v,int L,int R,int k){ int ll = L,rr = R,mid; LL a; while (ll < rr){ mid = ll + rr >> 1; a = q_num(u,v,1,N,L,mid); if ((mid - L + 1) - a >= k) rr = mid; else ll = mid + 1; } a = q_sum(u,v,1,N,L,ll); return S(L,ll) - a; } inline LL q_post(int u,int v,int L,int R,int k){ int ll = L,rr = R,mid,a; while (ll < rr){ mid = ll + rr + 1 >> 1; a = q_num(u,v,1,N,mid,R); if ((R - mid + 1) - a >= k) ll = mid; else rr = mid - 1; } a = q_sum(u,v,1,N,mid,R); return S(ll,R) - a; } void work(){ int l,r,L,R,a,s; LL ans,b; while (m--){ l = read(); r = read(); L = read(); R = L + r - l; ans = 0; if (L > 1){ a = q_num(rt[r],rt[l - 1],1,N,1,L - 1); if (a){ s = q_sum(rt[r],rt[l - 1],1,N,1,L - 1); b = q_pre(rt[r],rt[l - 1],L,R,a); ans += b - s; } } a = q_num(rt[r],rt[l - 1],1,N,R + 1,N); if (a){ s = q_sum(rt[r],rt[l - 1],1,N,R + 1,N); b = q_post(rt[r],rt[l - 1],L,R,a); ans += s - b; } printf("%lld\n",ans); } } int main(){ n = read(); m = read(); N = 1000000 + n + 1; int x; REP(i,n){ x = read(),modify(rt[i],rt[i - 1],1,N,x); } work(); return 0; }
相关文章推荐
- [CTSC2018]混合果汁(二分答案+主席树)
- [JSOI2018]军训列队
- BZOJ5319 & 洛谷4559 & LOJ2551:[JSOI2018]军训列队——题解
- BZOJ_5343_[Ctsc2018]混合果汁_二分答案+主席树
- BZOJ5343 [Ctsc2018]混合果汁 【二分 + 主席树】
- bzoj 5319: [Jsoi2018]军训列队
- BZOJ5343: [Ctsc2018]混合果汁 二分答案+主席树
- POJ-2018 Best Cow Fences(二分加DP)
- JSOI2018简要题解
- Codeforces Hello 2018 D. Too Easy Problems (二分)
- bzoj 1567: [JSOI2008]Blue Mary的战役地图【二分+hash】
- [BZOJ2653]middle(主席树+二分)
- bzoj1012: [JSOI2008]最大数maxnumber[单调队列+二分]
- 【BZOJ1822】[JSOI2010]Frozen Nova 冷冻波 几何+二分+网络流
- bzoj4012 [HNOI2015]开店(动态点分治+二分+STL/树链剖分+主席树)
- BZOJ 1567: [JSOI2008]Blue Mary的战役地图( 二分答案 + hash )
- POJ-2018 Best Cow Fences(二分加DP)
- 主席树+CDQ分治+整体二分
- BZOJ 1014: [JSOI2008]火星人prefix Splay+二分
- BZOJ5315 [JSOI2018]防御网络 【仙人掌 + dp】