HDU 3340 rain in acstar line segment tree + simple ry?

All the "Rain points" can be finally divided into a trapezoid (a triangle is one of the bottom is 0), and then the lazy mark is set to the upper bottom and lower bottom of the trapezoid. The number is large. Of course, discretization is required. Pay attention to the relationship between the processing points and the line segments between points, and note that the labels may change during the update process when there are downward update labels.

#include <cstdio>#include <cstring>#include <iostream>#include <map>#include <set>#include <vector>#include <string>#include <queue>#include <deque>#include <bitset>#include <list>#include <cstdlib>#include <climits>#include <cmath>#include <ctime>#include <algorithm>#include <stack>#include <sstream>#include <numeric>#include <fstream>#include <functional>using namespace std;#define MP make_pair#define PB push_back#define lson rt << 1,l,mid#define rson rt << 1 | 1,mid + 1,rtypedef long long LL;typedef unsigned long long ULL;typedef vector<int> VI;typedef pair<int,int> pii;const int INF = INT_MAX / 3;const double eps = 1e-8;const LL LINF = 1e17;const double DINF = 1e60;const int maxn = 25000 + 10;const int maxv = maxn * 10;VI numx;vector<pii> rpoint[maxn];char cmd[maxn];int ql[maxn],qr[maxn],n,cnt[maxn];double lbd[maxv << 2], rbd[maxv << 2], sum[maxv << 2];int len[maxv << 2],knumx;int getID(int Val) {    return lower_bound(numx.begin(),numx.end(),Val) - numx.begin();}void input() {    int nx,ny;    numx.clear();    scanf("%d",&n);    for(int i = 1;i <= n;i++) {        scanf(" %c",&cmd[i]);        if(cmd[i] == ‘Q‘) {            scanf("%d%d",&ql[i],&qr[i]);            numx.PB(ql[i]); numx.PB(qr[i]);        }        else {            scanf("%d",&cnt[i]);            rpoint[i].clear();            for(int j = 0;j < cnt[i];j++) {                scanf("%d%d",&nx,&ny);                rpoint[i].PB(MP(nx,ny));                numx.PB(nx);            }        }    }    sort(numx.begin(),numx.end());    numx.erase(unique(numx.begin(),numx.end()),numx.end());}void pushup(int rt) {    sum[rt] = sum[rt << 1] + sum[rt << 1 | 1];}void pushdown(int rt) {    int lc = rt << 1, rc = rt << 1 | 1;    double mid = lbd[rt] + (rbd[rt] - lbd[rt]) * len[lc] / len[rt];    lbd[lc] += lbd[rt]; rbd[lc] += mid;    lbd[rc] += mid; rbd[rc] += rbd[rt];    sum[lc] += (lbd[rt] + mid) * len[lc] / 2;    sum[rc] += (mid + rbd[rt]) * len[rc] / 2;    lbd[rt] = rbd[rt] = 0;}void build(int rt,int l,int r) {    len[rt] = numx[r + 1] - numx[l];    if(l == r) return;    int mid = (l + r) >> 1;    build(lson); build(rson);}void update(int rt,int l,int r,int ql,int qr,double ld,double rd) {    if(ql == l && qr == r) {        lbd[rt] += ld; rbd[rt] += rd; sum[rt] += (ld + rd) * len[rt] / 2;    }    else {        pushdown(rt);        int mid = (l + r) >> 1;        if(qr <= mid) update(lson,ql,qr,ld,rd);        else if(ql > mid) update(rson,ql,qr,ld,rd);        else {            double midv = ld + (rd - ld) * (numx[mid + 1] - numx[ql]) / (numx[qr + 1] - numx[ql]);            update(lson,ql,mid,ld,midv);            update(rson,mid + 1,qr,midv,rd);        }        pushup(rt);    }}double query(int rt,int l,int r,int ql,int qr) {    if(ql <= l && qr >= r) return sum[rt];    else {        int mid = (l + r) >> 1; double ret = 0;        pushdown(rt);        if(ql <= mid) ret += query(lson,ql,qr);        if(qr > mid) ret += query(rson,ql,qr);        return ret;    }}void Handle(vector<pii> &sp,int cnt) {    sp.PB(sp[0]);    for(int i = 0;i < cnt;i++) {        pii now = sp[i], nxt = sp[i + 1];        if(now.first < nxt.first) {            int x1 = getID(now.first), x2 = getID(nxt.first) - 1;            update(1,0,knumx - 1,x1,x2,-now.second,-nxt.second);        }        if(now.first > nxt.first) {            int x1 = getID(nxt.first), x2 = getID(now.first) - 1;            update(1,0,knumx - 1,x1,x2,nxt.second,now.second);        }    }}void solve() {    knumx = numx.size() - 1;    build(1,0,knumx - 1);    memset(lbd,0,sizeof(lbd));    memset(rbd,0,sizeof(rbd));    memset(sum,0,sizeof(sum));    for(int i = 1;i <= n;i++) {        if(cmd[i] == ‘Q‘) {            int x1 = getID(ql[i]), x2 = getID(qr[i]) - 1;            printf("%.3f\n",query(1,0,knumx - 1,x1,x2));        }        else Handle(rpoint[i],cnt[i]);    }}int main() {    int T; scanf("%d",&T);    while(T--) {        input();        solve();    }    return 0;}

 

HDU 3340 rain in acstar line segment tree + simple ry?