//HDOJ 1823 Luck and Love 二维线段树 单点更新 成段求最值/*题意：当操作符为‘I’时，表示有一个MM报名，后面接着一个整数，H表示身高，两个浮点数，A表示活泼度，L表示缘分值。 （100<=H<=200， 0.0<=A，L<=100.0）当操作符为‘Q’时，后面接着四个浮点数，H1，H2表示身高区间，A1，A2表示活泼度区间，输出符合身高和活泼度要求的MM中的缘分最高值。 （100<=H1，H2<=200， 0.0<=A1，A2<=100.0）所有输入的浮点数，均只有一位小数。思路：求二维区间内的最大值 二维线段树 树套树实现*/#include<stdio.h>#include<string.h>#include<stdlib.h>#include<algorithm>using namespace std;#define N 105#define M 1005#define lson rt<<1,l,mid#define rson rt<<1|1,mid+1,rint sum[N<<2][M<<2];int Max(int x,int y){return x>y?x:y;}void SubBuild(int rt,int l,int r,int t){sum[t][rt] = -1;if(l != r){int mid = (l + r) >> 1;SubBuild(lson,t);SubBuild(rson,t);}}void Build(int rt,int l,int r){SubBuild(1,0,1000,rt);if(l!=r){int mid = (l + r) >> 1;Build(lson);Build(rson);}}void SubUpdate(int rt,int l,int r,int y,int val,int t){if(l == r)sum[t][rt] = Max(sum[t][rt],val);else{int mid = (l + r) >> 1;if(y <= mid) SubUpdate(lson,y,val,t);else SubUpdate(rson,y,val,t);sum[t][rt] = Max(sum[t][rt<<1],sum[t][rt<<1|1]);}}void Update(int rt,int l,int r,int x,int y,int val){SubUpdate(1,0,1000,y,val,rt);if(l!=r){int mid = (l + r) >> 1;if(x <= mid) Update(lson,x,y,val);else Update(rson,x,y,val);}}int SubQuery(int rt,int l,int r,int LY,int RY,int t){if(LY <= l && RY >= r){return sum[t][rt];}int mid = (l + r) >> 1;int ans = -1;if(LY <= mid) ans = Max(ans,SubQuery(lson,LY,RY,t));if(RY > mid ) ans = Max(ans,SubQuery(rson,LY,RY,t));return ans;}int Query(int rt,int l,int r,int LX,int RX,int LY,int RY){if(LX <= l && RX >= r)return SubQuery(1,0,1000,LY,RY,rt);int mid = (l + r) >> 1;int ans = -1;if(LX <= mid) ans = Max(ans,Query(lson,LX,RX,LY,RY));if(RX > mid ) ans = Max(ans,Query(rson,LX,RX,LY,RY));return ans;}int main(){int n,i;int H;double A,L,H1,H2,A1,A2;char op[5];while(scanf("%d",&n),n){Build(1,100,200);for(i = 0; i < n; ++i){scanf("%s",op);if(op[0] == 'I'){scanf("%d %lf %lf",&H,&A,&L);Update(1,100,200,H,int(A*10),int(L*10));}else{scanf("%lf %lf %lf %lf",&H1,&H2,&A1,&A2);if(H1 > H2)swap(H1,H2);if(A1 > A2) swap(A1,A2);double ans = double(Query(1,100,200,(int)H1,(int)H2,int(A1*10),int(A2*10)))/10;if(ans < 0) printf("-1\n");else printf("%.1lf\n",ans);}}}return 0;}
