[POI2011]Meteors
阿新 • • 發佈:2018-12-19
嘟嘟嘟
做了幾道題之後,對整體二分有點感覺了。
整體二分的本質就是二分答案。所以這道題二分的就是次數。
然後就是套路了,把小於\(mid\)的操作都新增減去,然後查詢,如果查詢的值\(x\)比給定值大,就把這個詢問放到左區間,否則減去\(x\),放到右區間。
具體的操作,要支援區間加和單點查,第一反應是線段樹,結果有兩個點TLE的很慘。所以改成了樹狀陣列差分維護字首和,竟然過了,而且還跑的特別塊。常數真是致命啊……
#include<cstdio> #include<iostream> #include<cmath> #include<algorithm> #include<cstring> #include<cstdlib> #include<cctype> #include<vector> #include<stack> #include<queue> using namespace std; #define enter puts("") #define space putchar(' ') #define Mem(a, x) memset(a, x, sizeof(a)) #define rg register typedef long long ll; typedef double db; const int INF = 0x3f3f3f3f; const db eps = 1e-8; const int maxn = 3e5 + 5; inline ll read() { ll ans = 0; char ch = getchar(), last = ' '; while(!isdigit(ch)) last = ch, ch = getchar(); while(isdigit(ch)) ans = (ans << 1) + (ans << 3) + ch - '0', ch = getchar(); if(last == '-') ans = -ans; return ans; } inline void write(ll x) { if(x < 0) x = -x, putchar('-'); if(x >= 10) write(x / 10); putchar(x % 10 + '0'); } int n, m, k, val[maxn]; vector<int> con[maxn]; struct Node { int L, R, w, id; }t[maxn << 1], tl[maxn << 1], tr[maxn << 1]; int ans[maxn]; ll c[maxn]; int lowbit(int x) {return x & -x;} inline void clear(int pos) { for(; pos <= m; pos += lowbit(pos)) if(c[pos]) c[pos] = 0; else break; } inline void add(int pos, int d) { if(!pos) return; for(; pos <= m; pos += lowbit(pos)) c[pos] += d; } inline ll query(int pos) { ll ret = 0; for(; pos; pos -= lowbit(pos)) ret += c[pos]; return ret; } inline void Change(Node q) { if(q.L > q.R) add(q.L, q.w), add(1, q.w), add(q.R + 1, -q.w); else add(q.L, q.w), add(q.R + 1, -q.w); } inline void Clear(Node q) { if(q.L > q.R) clear(q.L), clear(1), clear(q.R + 1); else clear(q.L), clear(q.R + 1); } inline void solve(int kl, int kr, int ql, int qr) { if(ql > qr) return; if(kl == kr) { for(int i = ql; i <= qr; ++i) if(t[i].id > k + 1) ans[t[i].id - k - 1] = kl; return; } int mid = (kl + kr) >> 1, id1 = 0, id2 = 0; for(int i = ql; i <= qr; ++i) { if(t[i].id <= k + 1) { if(t[i].id <= mid) Change(t[i]), tl[++id1] = t[i]; else tr[++id2] = t[i]; } else { ll tot = 0; int x = t[i].id - k - 1; for(int j = 0; j < (int)con[x].size(); ++j) if((tot += query(con[x][j])) >= t[i].w) break; if(tot >= t[i].w) tl[++id1] = t[i]; else t[i].w -= tot, tr[++id2] = t[i]; } } for(int i = 1; i <= id1; ++i) if(tl[i].id <= k + 1 && tl[i].id <= mid) Clear(tl[i]); for(int i = 1; i <= id1; ++i) t[ql + i - 1] = tl[i]; for(int i = 1; i <= id2; ++i) t[ql + id1 + i - 1] = tr[i]; solve(kl, mid, ql, ql + id1 - 1); solve(mid + 1, kr, ql + id1, qr); } int main() { n = read(); m = read(); for(int i = 1, x; i <= m; ++i) x = read(), con[x].push_back(i); for(int i = 1; i <= n; ++i) val[i] = read(); k = read(); for(int i = 1; i <= k; ++i) t[i].L = read(), t[i].R = read(), t[i].w = read(), t[i].id = i; t[k + 1] = (Node){1, m, INF, k + 1}; for(int i = 1; i <= n; ++i) t[k + 1 + i] = (Node){0, 0, val[i], k + 1 + i}; solve(1, k + 1, 1, k + n + 1); for(int i = 1; i <= n; ++i) if(ans[i] > k) puts("NIE"); else write(ans[i]), enter; return 0; }