題目鏈接

洛谷P4559

題解

只會做\(70\)分的\(O(nlog^2n)\)

如果本來就在區間內的人是不用動的，區間右邊的人往區間最右的那些空位跑，區間左邊的人往區間最左的那些空位跑
找到這些空位就用二分 + 主席樹
理應可以在主席樹上的區間二分而做到\(O(nlogn)\)，但是寫不出來，先留著坑

#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; }

轉載于:https://www.cnblogs.com/Mychael/p/9191209.html

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