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线段树 (矩形面积并&&周长并 – 来自notonlysuccess)

2019年04月05日 ⁄ 综合 ⁄ 共 2927字 ⁄ 字号 评论关闭

矩形面积并

hdu1542
Atlantis

题意:矩形面积并
思路:浮点数先要离散化;然后把矩形分成两条边,上边和下边,对横轴建树,然后从下到上扫描上去,用cnt表示该区间下边比上边多几个,sum代表该区间内被覆盖的线段的长度总和
这里线段树的一个结点并非是线段的一个端点,而是该端点和下一个端点间的线段,所以题目中r+1,r-1的地方可以自己好好的琢磨一下

线段树操作:update:区间增减 query:直接取根节点的值

    #include <cstdio>  
    #include <cstring>  
    #include <cctype>  
    #include <algorithm>  
    using namespace std;  
    #define lson l , m , rt << 1  
    #define rson m + 1 , r , rt << 1 | 1  
       
    const int maxn = 2222;  
    int cnt[maxn << 2];  
    double sum[maxn << 2];  
    double X[maxn];  
    struct Seg {  
        double h , l , r;  
        int s;  
        Seg(){}  
        Seg(double a,double b,double c,int d) : l(a) , r(b) , h(c) , s(d) {}  
        bool operator < (const Seg &cmp) const {  
            return h < cmp.h;  
        }  
    }ss[maxn];  
    void PushUp(int rt,int l,int r) {  
        if (cnt[rt]) sum[rt] = X[r+1] - X[l];  
        else if (l == r) sum[rt] = 0;  
        else sum[rt] = sum[rt<<1] + sum[rt<<1|1];  
    }  
    void update(int L,int R,int c,int l,int r,int rt) {  
        if (L <= l && r <= R) {  
            cnt[rt] += c;  
            PushUp(rt , l , r);  
            return ;  
        }  
        int m = (l + r) >> 1;  
        if (L <= m) update(L , R , c , lson);  
        if (m < R) update(L , R , c , rson);  
        PushUp(rt , l , r);  
    }  
    int Bin(double key,int n,double X[]) {//离散化  
        int l = 0 , r = n - 1;  
        while (l <= r) {  
            int m = (l + r) >> 1;  
            if (X[m] == key) return m;  
            if (X[m] < key) l = m + 1;  
            else r = m - 1;  
        }  
        return -1;  
    }  
    int main() {  
        int n , cas = 1;  
        while (~scanf("%d",&n) && n) {  
            int m = 0;  
            while (n --) {  
                double a , b , c , d;  
                scanf("%lf%lf%lf%lf",&a,&b,&c,&d);//左上和右下  
                X[m] = a;  
                ss[m++] = Seg(a , c , b , 1);  
                X[m] = c;  
                ss[m++] = Seg(a , c , d , -1);  
            }  
            sort(X , X + m);  
            sort(ss , ss + m);  
            int k = 1;  
            for (int i = 1 ; i < m ; i ++) {  
                if (X[i] != X[i-1]) X[k++] = X[i];  
            }  
            memset(cnt , 0 , sizeof(cnt));  
            memset(sum , 0 , sizeof(sum));  
            double ret = 0;  
            for (int i = 0 ; i < m - 1 ; i ++) {  
                int l = Bin(ss[i].l , k , X);  
                int r = Bin(ss[i].r , k , X) - 1;  
                if (l <= r) update(l , r , ss[i].s , 0 , k - 1, 1);  
                ret += sum[1] * (ss[i+1].h - ss[i].h);  
            }  
            printf("Test case #%d\nTotal explored area: %.2lf\n\n",cas++ , ret);  
        }  
        return 0;  
    }  

矩形周长并

hdu1828
Picture

题意:矩形周长并
思路:与面积不同的地方是还要记录竖的边有几个(numseg记录),并且当边界重合的时候需要合并(用lbd和rbd表示边界来辅助)
线段树操作:update:区间增减 query:直接取根节点的值

include <cstdio>  
#include <cstring>  
#include <cctype>  
#include <algorithm>  
using namespace std;  
#define lson l , m , rt << 1  
#define rson m + 1 , r , rt << 1 | 1  
   
const int maxn = 22222;  
struct Seg{  
    int l , r , h , s;  
    Seg() {}  
    Seg(int a,int b,int c,int d):l(a) , r(b) , h(c) , s(d) {}  
    bool operator < (const Seg &cmp) const {  
        if (h == cmp.h) return s > cmp.s;  
        return h < cmp.h;  
    }  
}ss[maxn];  
bool lbd[maxn<<2] , rbd[maxn<<2];  
int numseg[maxn<<2];  
int cnt[maxn<<2];  
int len[maxn<<2];  
void PushUP(int rt,int l,int r) {  
    if (cnt[rt]) {  
        lbd[rt] = rbd[rt] = 1;  
        len[rt] = r - l + 1;  
        numseg[rt] = 2;  
    } else if (l == r) {  
        len[rt] = numseg[rt] = lbd[rt] = rbd[rt] = 0;  
    } else {  
        lbd[rt] = lbd[rt<<1];  
        rbd[rt] = rbd[rt<<1|1];  
        len[rt] = len[rt<<1] + len[rt<<1|1];  
        numseg[rt] = numseg[rt<<1] + numseg[rt<<1|1];  
        if (lbd[rt<<1|1] && rbd[rt<<1]) numseg[rt] -= 2;//两条线重合  
    }  
}  
void update(int L,int R,int c,int l,int r,int rt) {  
    if (L <= l && r <= R) {  
        cnt[rt] += c;  
        PushUP(rt , l , r);  
        return ;  
    }  
    int m = (l + r) >> 1;  
    if (L <= m) update(L , R , c , lson);  
    if (m < R) update(L , R , c , rson);  
    PushUP(rt , l , r);  
}  
int main() {  
    int n;  
    while (~scanf("%d",&n)) {  
        int m = 0;  
        int lbd = 10000, rbd = -10000;  
        for (int i = 0 ; i < n ; i ++) {  
            int a , b , c , d;  
            scanf("%d%d%d%d",&a,&b,&c,&d);//左下和右上  
            lbd = min(lbd , a);  
            rbd = max(rbd , c);  
            ss[m++] = Seg(a , c , b , 1);  
            ss[m++] = Seg(a , c , d , -1);  
        }  
        sort(ss , ss + m);  
        int ret = 0 , last = 0;  
        for (int i = 0 ; i < m ; i ++) {  
            if (ss[i].l < ss[i].r) update(ss[i].l , ss[i].r - 1 , ss[i].s , lbd , rbd - 1 , 1);  
            ret += numseg[1] * (ss[i+1].h - ss[i].h);  
            ret += abs(len[1] - last);  
            last = len[1];  
        }  
        printf("%d\n",ret);  
    }  
    return 0;  
}  

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