HDU 3340 Rain in ACStar 线段树+简单几何？

所有的“雨点”最后都可以分解成梯形（三角形就是其中一个底为0），然后lazy标记设置成梯形的上底和下底就好。数目很大，当然要离散化，注意处理点和点之间线段的关系，而且向下更新还有下放标记的时候要注意更新的过程中标记是有可能会变的。

#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,r
typedef 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
",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;
}