Codeforces Beta Round #99 (Div. 1) C Mushroom Gnomes - 2(单点查询)
2013-03-30 00:54
477 查看
题意:有n棵树,m个蘑菇,每棵树有坐标a,高度h,向左边倒的概率,向右的概率(概率用0-100表示),向左倒范围[x-h,x)内的蘑菇被破坏,向右倒范围(x,x+h]范围内的蘑菇被破坏。每个蘑菇有坐标b,及它的魔力值z。
问树倒下后,所以蘑菇的魔力值的和的期望。
离散化之后,成段更新,即树倒下范围内的区间全部乘以倒向这边的概率。最后遍历每个蘑菇的存活的概率乘以它的魔力值就可以了。
问树倒下后,所以蘑菇的魔力值的和的期望。
离散化之后,成段更新,即树倒下范围内的区间全部乘以倒向这边的概率。最后遍历每个蘑菇的存活的概率乘以它的魔力值就可以了。
#include <iostream> #include <cstdio> #include <cstring> #include <vector> #include <map> #include <algorithm> using namespace std; #define LL(x) (x<<1) #define RR(x) (x<<1|1) #define MID(a,b) (a+((b-a)>>1)) const int N=1e5+5; struct OP { double flag; int st,ed,lft; OP(){} OP(int a,int b,double c,int d){st=a;ed=b;flag=c;lft=d;} }; struct node { double P; int lft,rht,flag; int mid(){return MID(lft,rht);} void fun(double tmp) { P*=tmp; flag=1; } }; int B ,Z ; vector<OP> op; vector<int> Y; map<int,int> H; struct Segtree { node tree[N*4*4]; void PushDown(int ind) { if(tree[ind].flag) { tree[LL(ind)].fun(tree[ind].P); tree[RR(ind)].fun(tree[ind].P); tree[ind].flag=0; tree[ind].P=1; } } void build(int lft,int rht,int ind) { tree[ind].lft=lft; tree[ind].rht=rht; tree[ind].P=1; tree[ind].flag=0; if(lft!=rht) { int mid=tree[ind].mid(); build(lft,mid,LL(ind)); build(mid+1,rht,RR(ind)); } } void updata(int st,int ed,int ind,double flag) { int lft=tree[ind].lft,rht=tree[ind].rht; if(st<=lft&&rht<=ed) tree[ind].fun(flag); else { PushDown(ind); int mid=tree[ind].mid(); if(st<=mid) updata(st,ed,LL(ind),flag); if(ed> mid) updata(st,ed,RR(ind),flag); } } double query(int pos,int ind) { if(tree[ind].lft==tree[ind].rht) return tree[ind].P; else { PushDown(ind); int mid=tree[ind].mid(); if(pos<=mid) return query(pos,LL(ind)); else return query(pos,RR(ind)); } } }seg; int main() { int n,m; while(scanf("%d%d",&n,&m)!=EOF) { Y.clear(); H.clear(); op.clear(); for(int i=0;i<n;i++) { int a,b,c,d; scanf("%d%d%d%d",&a,&b,&c,&d); Y.push_back(a-b); Y.push_back(a+b); Y.push_back(a); op.push_back(OP(a-b,a,(100-c)/100.0,1)); op.push_back(OP(a,a+b,(100-d)/100.0,0)); } for(int i=0;i<m;i++) { scanf("%d%d",&B[i],&Z[i]); Y.push_back(B[i]); } sort(Y.begin(),Y.end()); Y.erase(unique(Y.begin(),Y.end()),Y.end()); for(int i=0;i<(int)Y.size();i++) H[Y[i]]=i; seg.build(0,(int)Y.size(),1); for(int i=0;i<(int)op.size();i++) { if(op[i].lft) seg.updata(H[op[i].st],H[op[i].ed]-1,1,op[i].flag); else seg.updata(H[op[i].st]+1,H[op[i].ed],1,op[i].flag); } double res=0; for(int i=0;i<m;i++) { res+=Z[i]*seg.query(H[B[i]],1); } printf("%.4lf\n",res); } return 0; }
相关文章推荐
- Codeforces Beta Round #73 (Div. 2 Only)——A,B,C
- Codeforces Beta Round #75 (Div. 1 Only)---B.Queue
- Codeforces Beta Round #57 (Div. 2) Enemy is weak
- Codeforces Beta Round #69 (Div. 1 Only) C. Beavermuncher-0xFF 树上贪心
- Codeforces Beta Round #27 (Codeforces format, Div. 2)
- Codeforces Beta Round #32 (Div. 2) C (math+思维)
- Codeforces Beta Round #92 (Div. 2 Only) B. Permutations 模拟
- Codeforces Beta Round #6 (Div. 2 Only) D. Lizards and Basements 2 dp
- 【Codeforces Round 263 (Div 2)E】【坐标映射 脑洞】Appleman and a Sheet of Paper 折纸游戏 区间查询
- Codeforces Beta Round #6 (Div. 2 Only) A. Triangle 水题
- Codeforces Beta Round #34 (Div. 2) E. Collisions
- Codeforces Beta Round #6 (Div. 2 Only)
- Codeforces Beta Round #98 (Div. 2) / 137A Postcards and photos (模拟)
- Codeforces Beta Round #69 (Div. 2 Only) E题
- Codeforces Beta Round #75 (Div. 2 Only) A题
- Codeforces Beta Round #85 (Div. 1 Only) C (状态压缩或是数学?)
- Codeforces Beta Round #80 (Div. 2 Only)
- Codeforces Beta Round #92 (Div. 1 Only)
- Codeforces Beta Round #44 (Div. 2)——B
- Codeforces Beta Round #65 (Div. 2)