hdu 1823 Luck and Love ，二维线段树

Luck and Love

Time Limit: 10000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 5282    Accepted Submission(s): 1324

Input
本题有多个测试数据，第一个数字M，表示接下来有连续的M个操作，当M＝0时处理中止。
接下来是一个操作符C。
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。
 

Output
对于每一次询问操作，在一行里面输出缘分最高值，保留一位小数。
对查找不到的询问，输出-1。
 

Sample Input
8
I 160 50.5 60.0
I 165 30.0 80.5
I 166 10.0 50.0
I 170 80.5 77.5
Q 150 166 10.0 60.0
Q 166 177 10.0 50.0
I 166 40.0 99.9
Q 166 177 10.0 50.0
0

Sample Output
80.5
50.0
99.9

 

二维线段树

两种板子（都是C++ AC, G++ WA 。。。。。）

#include <cstdio>
#include <algorithm>

using namespace std;

#define lson (o<<1)
#define rson ((o<<1)|1)

const int maxh = 100 + 10;
const int maxa = 1000 + 10;
int mmax[maxh<<2][maxa<<2];

void build_A(int O, int o, int L, int R) {      //活泼度建树
    mmax[O][o] = -1;
    if(L == R) return;
    int M = (L + R) >> 1;
    build_A(O, lson, L, M);
    build_A(O, rson, M+1, R);
}

void build_H(int o, int L, int R) {     //身高建树
    build_A(o, 1, 0, 1000);
    if(L == R) return;
    int M = (L + R) >> 1;
    build_H(lson, L, M);
    build_H(rson, M+1, R);
}

void Insert_A(int O, int o, int L, int R, int A, int D) {       //根据活泼度建树
    if(L == R) {
        mmax[O][o] = max(mmax[O][o], D);
        return;
    }
    int M = (L + R) >> 1;
    if(A <= M) Insert_A(O, lson, L, M, A, D);
    else Insert_A(O, rson, M+1, R, A, D);
    mmax[O][o] = max(mmax[O][lson], mmax[O][rson]);
}

void Insert(int o, int L, int R, int H, int A, int D) {
    Insert_A(o, 1, 0, 1000, A, D);
    if(L == R) return;
    int M = (L + R) >> 1;
    if(H <= M) Insert(lson, L, M, H, A, D);
    else Insert(rson, M+1, R, H, A, D);
}

int query_A(int O, int o, int L, int R, int A1, int A2) {       //根据活泼度查询
    if(A1 <= L && R <= A2) return mmax[O][o];
    int M = (L + R) >> 1;
    int Max1 = -1, Max2 = -1;
    if(A1 <= M) Max1 = query_A(O, lson, L, M, A1, A2);
    if(A2 > M) Max2 = query_A(O, rson, M+1, R, A1, A2);
    return (Max1 > Max2) ? Max1 : Max2;
}

int query(int o, int L, int R, int H1, int H2, int A1, int A2) {
    if(H1 <= L && R <= H2) return query_A(o, 1, 0, 1000, A1, A2);
    int M = (L + R) >> 1;
    int Max1 = -1, Max2 = -1;
    if(H1 <= M) Max1 = query(lson, L, M, H1, H2, A1, A2);
    if(H2 > M) Max2 = query(rson, M+1, R, H1, H2, A1, A2);
    return (Max1 > Max2) ? Max1 : Max2;
}

int main() {
    int M;
    char op;
    while(scanf("%d", &M) == 1 && M) {
        build_H(1, 100, 200);
        for(int i = 0; i < M; i++) {
            getchar();
            op = getchar();
            if(op == 'I') {
                int H, A, D;
                double AA, DD;
                scanf("%d%lf%lf", &H, &AA, &DD);
                A = (int)(AA * 10);
                D = (int)(DD * 10);
                Insert(1, 100, 200, H, A, D);
            } else {
                int H1, H2, A1, A2;
                double AA, BB;
                scanf("%d%d%lf%lf", &H1, &H2, &AA, &BB);
                A1 = (int)(AA * 10);
                A2 = (int)(BB * 10);
                if(A1 > A2) swap(A1, A2);
                if(H1 > H2) swap(H1, H2);
                int ret = query(1, 100, 200, H1, H2, A1, A2);
                ret != -1 ? printf("%.1f\n", ret/10.0) : puts("-1");
            }
        }
    }
    return 0;
}

#include<cstdio>
#include<cstring>
#include<algorithm>

#define lson rt<<1
#define rson lson|1

using namespace std;
const int inf = 0x3f;
const int N = 200;
const int M = 1000;

double ans;

struct IntervalTree2D {
    double Max[N*4+10][M*4+10], v;
    int n, m, hleaf, Hrt, h, h1, h2, a, a1, a2;

    void query1D(int rt, int A1, int A2) {
        if(a1<=A1 && A2<=a2) {
            ans = max(ans, Max[Hrt][rt]);
            return ;
        }
        int mid = (A1 + A2)/2;
        if(mid>=a1)
            query1D(lson, A1, mid);
        if(mid<a2)
            query1D(rson, mid+1, A2);
    }

    void query2D(int rt, int H1, int H2) {
        if(h1<=H1 && H2<=h2) {
            Hrt = rt;
            query1D(1, 0, m);
            return ;
        }
        int mid = (H1 + H2)/2;
        if(mid>=h1)
            query2D(lson, H1, mid);
        if(mid<h2)
            query2D(rson, mid+1, H2);
    }

    void modify1D(int rt, int A1, int A2) {
        if(A1==A2) {
            if(hleaf==1) {
                Max[Hrt][rt] = max(v, Max[Hrt][rt]);
                return ;
            }
            Max[Hrt][rt] = max(Max[Hrt<<1][rt], Max[Hrt<<1|1][rt]);
            return ;
        }
        int mid = (A1 + A2)/2;
        if(mid>= a)
            modify1D(lson, A1, mid);
        else
            modify1D(rson, mid+1, A2);
        Max[Hrt][rt] = max(Max[Hrt][lson], Max[Hrt][rson]);
    }

    void modify2D(int rt, int H1, int H2) {
        if(H1 == H2) {
            Hrt = rt;
            hleaf = 1;
            modify1D(1, 0, m);
            return ;
        }
        int mid = (H1+H2)/2;
        if(mid>=h)
            modify2D(lson, H1, mid);
        else
            modify2D(rson, mid+1, H2);
        Hrt = rt;
        hleaf = 0;
        modify1D(1, 0, m);
    }

    void query() {
        ans = -inf;
        query2D(1, 100, n);
    }

    void modify() {
        modify2D(1, 100, n);
    }

    void clear() {
        for(int i=0; i<=N*4+5; ++i)
            for(int j=0; j<=1000*4+5; ++j)
                Max[i][j] = -inf;
    }
};

IntervalTree2D t;


int main() {
    int n;
    while(~scanf("%d",&n)&&n) {
        t.clear();
        t.n = 200;
        t.m = 1000;
        while(n--) {
            char ch;
            scanf(" %c", &ch);
            int h, hh;
            double a, aa;
            if(ch=='I') {
                scanf("%d%lf%lf", &h, &a, &aa);
                t.h = h;
                t.a = int(a*10);
                t.v = aa;
                t.modify();
            } else {
                scanf("%d%d%lf%lf", &h, &hh, &a, &aa);
                if(h>hh) swap(h, hh);
                if(a>aa) swap(a, aa);
                t.h1 = h;
                t.h2 = hh;
                t.a1 = int(a*10);
                t.a2 = int(aa*10);
                t.query();
                if(ans < 0)
                    printf("-1\n");
                else
                    printf("%.1f\n", ans);
            }
        }
    }
    return 0;
}