ACM_大牛总结的线段树专辑
【完全版】线段树
这是从大牛那里粘过来的总结，对于刚训练线段树的我来说帮助很大。希望这种清新的代码风格同样能让你受益.
在代码前先介绍一些我的线段树风格:
· maxn是题目给的最大区间,而节点数要开4倍,确切的来说节点数要开大于maxn的最小2x的两倍
· lson和rson分辨表示结点的左儿子和右儿子,由于每次传参数的时候都固定是这几个变量,所以可以用预定于比较方便的表示
· 以前的写法是另外开两个个数组记录每个结点所表示的区间,其实这个区间不必保存,一边算一边传下去就行,只需要写函数的时候多两个参数,结合lson和rson的预定义可以很方便
· PushUP(int rt)是把当前结点的信息更新到父结点
· PushDown(int rt)是把当前结点的信息更新给儿子结点
· rt表示当前子树的根(root),也就是当前所在的结点
整理这些题目后我觉得线段树的题目整体上可以分成以下四个部分:
· 单点更新:最最基础的线段树,只更新叶子节点,然后把信息用PushUP(int r)这个函数更新上来
o hdu1166 敌兵布阵
题意:O(-1)
思路:O(-1)
线段树功能:update:单点增减 query:区间求和
?View Code CPP
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#include
#define lson l , m , rt << 1 #define rson m + 1 , r , rt << 1 | 1 const int maxn = 55555; int sum[maxn<<2]; void PushUP(int rt) { sum[rt] = sum[rt<<1] + sum[rt<<1|1]; } void build(int l,int r,int rt) { if (l == r) { scanf(“%d”,&sum[rt]); return ; } int m = (l + r) >> 1; build(lson); build(rson); PushUP(rt); } void update(int p,int add,int l,int r,int rt) { if (l == r) { sum[rt] += add; return ; } int m = (l + r) >> 1; if (p <= m) update(p , add , lson); else update(p , add , rson); PushUP(rt); } int query(int L,int R,int l,int r,int rt) { if (L <= l && r <= R) { return sum[rt]; } int m = (l + r) >> 1; int ret = 0; if (L <= m) ret += query(L , R , lson); if (R > m) ret += query(L , R , rson); return ret; } int main() { int T , n; scanf(“%d”,&T); for (int cas = 1 ; cas <= T ; cas ++) { printf(“Case %d:\n”,cas); scanf(“%d”,&n); build(1 , n , 1); char op[10]; while (scanf(“%s”,op)) { if (op[0] == ‘E’) break; int a , b; scanf(“%d%d”,&a,&b); if (op[0] == ‘Q’) printf(“%d\n”,query(a , b , 1 , n , 1)); else if (op[0] == ‘S’) update(a , -b , 1 , n , 1); else update(a , b , 1 , n , 1); } } return 0; } |
o hdu1754 I Hate It
题意:O(-1)
思路:O(-1)
线段树功能:update:单点替换 query:区间最值
?View Code CPP
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#include #include using namespace std;
#define lson l , m , rt << 1 #define rson m + 1 , r , rt << 1 | 1 const int maxn = 222222; int MAX[maxn<<2]; void PushUP(int rt) { MAX[rt] = max(MAX[rt<<1] , MAX[rt<<1|1]); } void build(int l,int r,int rt) { if (l == r) { scanf(“%d”,&MAX[rt]); return ; } int m = (l + r) >> 1; build(lson); build(rson); PushUP(rt); } void update(int p,int sc,int l,int r,int rt) { if (l == r) { MAX[rt] = sc; return ; } int m = (l + r) >> 1; if (p <= m) update(p , sc , lson); else update(p , sc , rson); PushUP(rt); } int query(int L,int R,int l,int r,int rt) { if (L <= l && r <= R) { return MAX[rt]; } int m = (l + r) >> 1; int ret = 0; if (L <= m) ret = max(ret , query(L , R , lson)); if (R > m) ret = max(ret , query(L , R , rson)); return ret; } int main() { int n , m; while (~scanf(“%d%d”,&n,&m)) { build(1 , n , 1); while (m –) { char op[2]; int a , b; scanf(“%s%d%d”,op,&a,&b); if (op[0] == ‘Q’) printf(“%d\n”,query(a , b , 1 , n , 1)); else update(a , b , 1 , n , 1); } } return 0; } |
o hdu1394 Minimum Inversion Number
题意:求Inversion后的最小逆序数
思路:用O(nlogn)复杂度求出最初逆序数后,就可以用O(1)的复杂度分别递推出其他解
线段树功能:update:单点增减 query:区间求和
?View Code CPP
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#include #include using namespace std;
#define lson l , m , rt << 1 #define rson m + 1 , r , rt << 1 | 1 const int maxn = 5555; int sum[maxn<<2]; void PushUP(int rt) { sum[rt] = sum[rt<<1] + sum[rt<<1|1]; } void build(int l,int r,int rt) { sum[rt] = 0; if (l == r) return ; int m = (l + r) >> 1; build(lson); build(rson); } void update(int p,int l,int r,int rt) { if (l == r) { sum[rt] ++; return ; } int m = (l + r) >> 1; if (p <= m) update(p , lson); else update(p , rson); PushUP(rt); } int query(int L,int R,int l,int r,int rt) { if (L <= l && r <= R) { return sum[rt]; } int m = (l + r) >> 1; int ret = 0; if (L <= m) ret += query(L , R , lson); if (R > m) ret += query(L , R , rson); return ret; } int x[maxn]; int main() { int n; while (~scanf(“%d”,&n)) { build(0 , n - 1 , 1); int sum = 0; for (int i = 0 ; i < n ; i ++) { scanf(“%d”,&x[i]); sum += query(x[i] , n - 1 , 0 , n - 1 , 1); update(x[i] , 0 , n - 1 , 1); } int ret = sum; for (int i = 0 ; i < n ; i ++) { sum += n - x[i] - x[i] - 1; ret = min(ret , sum); } printf(“%d\n”,ret); } return 0; } |
o hdu2795 Billboard
题意:h*w的木板,放进一些1*L的物品,求每次放空间能容纳且最上边的位子
思路:每次找到最大值的位子,然后减去L
线段树功能:query:区间求最大值的位子(直接把update的操作在query里做了)
?View Code CPP
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#include #include using namespace std;
#define lson l , m , rt << 1 #define rson m + 1 , r , rt << 1 | 1 const int maxn = 222222; int h , w , n; int MAX[maxn<<2]; void PushUP(int rt) { MAX[rt] = max(MAX[rt<<1] , MAX[rt<<1|1]); } void build(int l,int r,int rt) { MAX[rt] = w; if (l == r) return ; int m = (l + r) >> 1; build(lson); build(rson); } int query(int x,int l,int r,int rt) { if (l == r) { MAX[rt] -= x; return l; } int m = (l + r) >> 1; int ret = (MAX[rt<<1] >= x) ? query(x , lson) : query(x , rson); PushUP(rt); return ret; } int main() { while (~scanf(“%d%d%d”,&h,&w,&n)) { if (h > n) h = n; build(1 , h , 1); while (n –) { int x; scanf(“%d”,&x); if (MAX[1] < x) puts(“-1”); else printf(“%d\n”,query(x , 1 , h , 1)); } } return 0; } |
· 练习:
o poj2828 Buy Tickets
o poj2886 Who Gets the Most Candies?
· 成段更新(通常这对初学者来说是一道坎),需要用到延迟标记(或者说懒惰标记),简单来说就是每次更新的时候不要更新到底,用延迟标记使得更新延迟到下次需要更新or询问到的时候
o hdu1698 Just a Hook
题意:O(-1)
思路:O(-1)
线段树功能:update:成段替换 (由于只query一次总区间,所以可以直接输出1结点的信息)
?View Code CPP
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#include #include using namespace std;
#define lson l , m , rt << 1 #define rson m + 1 , r , rt << 1 | 1 const int maxn = 111111; int h , w , n; int col[maxn<<2]; int sum[maxn<<2]; void PushUp(int rt) { sum[rt] = sum[rt<<1] + sum[rt<<1|1]; } void PushDown(int rt,int m) { if (col[rt]) { col[rt<<1] = col[rt<<1|1] = col[rt]; sum[rt<<1] = (m - (m >> 1)) * col[rt]; sum[rt<<1|1] = (m >> 1) * col[rt]; col[rt] = 0; } } void build(int l,int r,int rt) { col[rt] = 0; sum[rt] = 1; if (l == r) return ; int m = (l + r) >> 1; build(lson); build(rson); PushUp(rt); } void update(int L,int R,int c,int l,int r,int rt) { if (L <= l && r <= R) { col[rt] = c; sum[rt] = c * (r - l + 1); return ; } PushDown(rt , r - l + 1); int m = (l + r) >> 1; if (L <= m) update(L , R , c , lson); if (R > m) update(L , R , c , rson); PushUp(rt); } int main() { int T , n , m; scanf(“%d”,&T); for (int cas = 1 ; cas <= T ; cas ++) { scanf(“%d%d”,&n,&m); build(1 , n , 1); while (m –) { int a , b , c; scanf(“%d%d%d”,&a,&b,&c); update(a , b , c , 1 , n , 1); } printf(“Case %d: The total value of the hook is %d.\n”,cas , sum[1]); } return 0; } |
o poj3468 A Simple Problem with Integers
题意:O(-1)
思路:O(-1)
线段树功能:update:成段增减 query:区间求和
?View Code CPP
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#include #include using namespace std;
#define lson l , m , rt << 1 #define rson m + 1 , r , rt << 1 | 1 #define LL long long const int maxn = 111111; LL add[maxn<<2]; LL sum[maxn<<2]; void PushUp(int rt) { sum[rt] = sum[rt<<1] + sum[rt<<1|1]; } void PushDown(int rt,int m) { if (add[rt]) { add[rt<<1] += add[rt]; add[rt<<1|1] += add[rt]; sum[rt<<1] += add[rt] * (m - (m >> 1)); sum[rt<<1|1] += add[rt] * (m >> 1); add[rt] = 0; } } void build(int l,int r,int rt) { add[rt] = 0; if (l == r) { scanf(“%lld”,&sum[rt]); return ; } int m = (l + r) >> 1; build(lson); build(rson); PushUp(rt); } void update(int L,int R,int c,int l,int r,int rt) { if (L <= l && r <= R) { add[rt] += c; sum[rt] += (LL)c * (r - l + 1); return ; } PushDown(rt , r - l + 1); int m = (l + r) >> 1; if (L <= m) update(L , R , c , lson); if (m < R) update(L , R , c , rson); PushUp(rt); } LL query(int L,int R,int l,int r,int rt) { if (L <= l && r <= R) { return sum[rt]; } PushDown(rt , r - l + 1); int m = (l + r) >> 1; LL ret = 0; if (L <= m) ret += query(L , R , lson); if (m < R) ret += query(L , R , rson); return ret; } int main() { int N , Q; scanf(“%d%d”,&N,&Q); build(1 , N , 1); while (Q –) { char op[2]; int a , b , c; scanf(“%s”,op); if (op[0] == ‘Q’) { scanf(“%d%d”,&a,&b); printf(“%lld\n”,query(a , b , 1 , N , 1)); } else { scanf(“%d%d%d”,&a,&b,&c); update(a , b , c , 1 , N , 1); } } return 0; } |
o poj2528 Mayor’s posters
题意:在墙上贴海报,海报可以互相覆盖,问最后可以看见几张海报
思路:这题数据范围很大,直接搞超时+超内存,需要离散化:
离散化简单的来说就是只取我们需要的值来用,比如说区间[1000,2000],[1990,2012]我们用不到[-∞,999][1001,1989][1991,1999][2001,2011][2013,+∞]这些值,所以我只需要1000,1990,2000,2012就够了,将其分别映射到0,1,2,3,在于复杂度就大大的降下来了
所以离散化要保存所有需要用到的值,排序后,分别映射到1~n,这样复杂度就会小很多很多
而这题的难点在于每个数字其实表示的是一个单位长度(并且一个点),这样普通的离散化会造成许多错误(包括我以前的代码,poj这题数据奇弱)
给出下面两个简单的例子应该能体现普通离散化的缺陷:
1-10 1-4 5-10
1-10 1-4 6-10
为了解决这种缺陷,我们可以在排序后的数组上加些处理,比如说[1,2,6,10]
如果相邻数字间距大于1的话,在其中加上任意一个数字,比如加成[1,2,3,6,7,10],然后再做线段树就好了.
线段树功能:update:成段替换 query:简单hash
?View Code CPP
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#include #include #include using namespace std; #define lson l , m , rt << 1 #define rson m + 1 , r , rt << 1 | 1
const int maxn = 11111; bool hash[maxn]; int li[maxn] , ri[maxn]; int X[maxn*3]; int col[maxn<<4]; int cnt;
void PushDown(int rt) { if (col[rt] != -1) { col[rt<<1] = col[rt<<1|1] = col[rt]; col[rt] = -1; } } void update(int L,int R,int c,int l,int r,int rt) { if (L <= l && r <= R) { col[rt] = c; return ; } PushDown(rt); int m = (l + r) >> 1; if (L <= m) update(L , R , c , lson); if (m < R) update(L , R , c , rson); } void query(int l,int r,int rt) { if (col[rt] != -1) { if (!hash[col[rt]]) cnt ++; hash[ col[rt] ] = true; return ; } if (l == r) return ; int m = (l + r) >> 1; query(lson); query(rson); } int Bin(int key,int n,int X[]) { 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 T , n; scanf(“%d”,&T); while (T –) { scanf(“%d”,&n); int nn = 0; for (int i = 0 ; i < n ; i ++) { scanf(“%d%d”,&li[i] , &ri[i]); X[nn++] = li[i]; X[nn++] = ri[i]; } sort(X , X + nn); int m = 1; for (int i = 1 ; i < nn; i ++) { if (X[i] != X[i-1]) X[m ++] = X[i]; } for (int i = m - 1 ; i > 0 ; i –) { if (X[i] != X[i-1] + 1) X[m ++] = X[i-1] + 1; } sort(X , X + m); memset(col , -1 , sizeof(col)); for (int i = 0 ; i < n ; i ++) { int l = Bin(li[i] , m , X); int r = Bin(ri[i] , m , X); update(l , r , i , 0 , m , 1); } cnt = 0; memset(hash , false , sizeof(hash)); query(0 , m , 1); printf(“%d\n”,cnt); } return 0; } |
o poj3225 Help with Intervals
题意:区间操作,交,并,补等
思路:
我们一个一个操作来分析:(用0和1表示是否包含区间,-1表示该区间内既有包含又有不包含)
U:把区间[l,r]覆盖成1
I:把[-∞,l)(r,∞]覆盖成0
D:把区间[l,r]覆盖成0
C:把[-∞,l)(r,∞]覆盖成0 , 且[l,r]区间0/1互换
S:[l,r]区间0/1互换
成段覆盖的操作很简单,比较特殊的就是区间0/1互换这个操作,我们可以称之为异或操作
很明显我们可以知道这个性质:当一个区间被覆盖后,不管之前有没有异或标记都没有意义了
所以当一个节点得到覆盖标记时把异或标记清空
而当一个节点得到异或标记的时候,先判断覆盖标记,如果是0或1,直接改变一下覆盖标记,不然的话改变异或标记
开区间闭区间只要数字乘以2就可以处理(偶数表示端点,奇数表示两端点间的区间)
线段树功能:update:成段替换,区间异或 query:简单hash
?View Code CPP
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#include #include #include #include using namespace std; #define lson l , m , rt << 1 #define rson m + 1 , r , rt << 1 | 1
const int maxn = 131072; bool hash[maxn]; int cover[maxn<<2]; int XOR[maxn<<2]; void FXOR(int rt) { if (cover[rt] != -1) cover[rt] ^= 1; else XOR[rt] ^= 1; } void PushDown(int rt) { if (cover[rt] != -1) { cover[rt<<1] = cover[rt<<1|1] = cover[rt]; XOR[rt<<1] = XOR[rt<<1|1] = 0; cover[rt] = -1; } if (XOR[rt]) { FXOR(rt<<1); FXOR(rt<<1|1); XOR[rt] = 0; } } void update(char op,int L,int R,int l,int r,int rt) { if (L <= l && r <= R) { if (op == ‘U’) { cover[rt] = 1; XOR[rt] = 0; } else if (op == ‘D’) { cover[rt] = 0; XOR[rt] = 0; } else if (op == ‘C’ || op == ‘S’) { FXOR(rt); } return ; } PushDown(rt); int m = (l + r) >> 1; if (L <= m) update(op , L , R , lson); else if (op == ‘I’ || op == ‘C’) { XOR[rt<<1] = cover[rt<<1] = 0; } if (m < R) update(op , L , R , rson); else if (op == ‘I’ || op == ‘C’) { XOR[rt<<1|1] = cover[rt<<1|1] = 0; } } void query(int l,int r,int rt) { if (cover[rt] == 1) { for (int it = l ; it <= r ; it ++) { hash[it] = true; } return ; } else if (cover[rt] == 0) return ; if (l == r) return ; PushDown(rt); int m = (l + r) >> 1; query(lson); query(rson); } int main() { cover[1] = XOR[1] = 0; char op , l , r; int a , b; while ( ~scanf(“%c %c%d,%d%c\n”,&op , &l , &a , &b , &r) ) { a <<= 1 , b <<= 1; if (l == ‘(‘) a ++; if (r == ‘)’) b –; if (a > b) { if (op == ‘C’ || op == ‘I’) { cover[1] = XOR[1] = 0; } } else update(op , a , b , 0 , maxn , 1); } query(0 , maxn , 1); bool flag = false; int s = -1 , e; for (int i = 0 ; i <= maxn ; i ++) { if (hash[i]) { if (s == -1) s = i; e = i; } else { if (s != -1) { if (flag) printf(” “); flag = true; printf(“%c%d,%d%c”,s&1?‘(‘:‘[‘ , s>>1 , (e+1)>>1 , e&1?‘)’:‘]’); s = -1; } } } if (!flag) printf(“empty set”); puts(“”); return 0; } |
· 练习:
o poj1436 Horizontally Visible Segments
o poj2991 Crane
o Another LCIS
o Bracket Sequence
· 区间合并
这类题目会询问区间中满足条件的连续最长区间,所以PushUp的时候需要对左右儿子的区间进行合并
o poj3667 Hotel
题意:1 a:询问是不是有连续长度为a的空房间,有的话住进最左边
2 a b:将[a,a+b-1]的房间清空
思路:记录区间中最长的空房间
线段树操作:update:区间替换 query:询问满足条件的最左断点
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#include #include #include #include using namespace std; #define lson l , m , rt << 1 #define rson m + 1 , r , rt << 1 | 1
const int maxn = 55555; int lsum[maxn<<2] , rsum[maxn<<2] , msum[maxn<<2]; int cover[maxn<<2];
void PushDown(int rt,int m) { if (cover[rt] != -1) { cover[rt<<1] = cover[rt<<1|1] = cover[rt]; msum[rt<<1] = lsum[rt<<1] = rsum[rt<<1] = cover[rt] ? 0 : m - (m >> 1); msum[rt<<1|1] = lsum[rt<<1|1] = rsum[rt<<1|1] = cover[rt] ? 0 : (m >> 1); cover[rt] = -1; } } void PushUp(int rt,int m) { lsum[rt] = lsum[rt<<1]; rsum[rt] = rsum[rt<<1|1]; if (lsum[rt] == m - (m >> 1)) lsum[rt] += lsum[rt<<1|1]; if (rsum[rt] == (m >> 1)) rsum[rt] += rsum[rt<<1]; msum[rt] = max(lsum[rt<<1|1] + rsum[rt<<1] , max(msum[rt<<1] , msum[rt<<1|1])); } void build(int l,int r,int rt) { msum[rt] = lsum[rt] = rsum[rt] = r - l + 1; cover[rt] = -1; if (l == r) return ; int m = (l + r) >> 1; build(lson); build(rson); } void update(int L,int R,int c,int l,int r,int rt) { if (L <= l && r <= R) { msum[rt] = lsum[rt] = rsum[rt] = c ? 0 : r - l + 1; cover[rt] = c; return ; } PushDown(rt , r - l + 1); int m = (l + r) >> 1; if (L <= m) update(L , R , c , lson); if (m < R) update(L , R , c , rson); PushUp(rt , r - l + 1); } int query(int w,int l,int r,int rt) { if (l == r) return l; PushDown(rt , r - l + 1); int m = (l + r) >> 1; if (msum[rt<<1] >= w) return query(w , lson); else if (rsum[rt<<1] + lsum[rt<<1|1] >= w) return m - rsum[rt<<1] + 1; return query(w , rson); } int main() { int n , m; scanf(“%d%d”,&n,&m); build(1 , n , 1); while (m –) { int op , a , b; scanf(“%d”,&op); if (op == 1) { scanf(“%d”,&a); if (msum[1] < a) puts(“0”); else { int p = query(a , 1 , n , 1); printf(“%d\n”,p); update(p , p + a - 1 , 1 , 1 , n , 1); } } else { scanf(“%d%d”,&a,&b); update(a , a + b - 1 , 0 , 1 , n , 1); } } return 0; } |
· 练习:
o hdu3308 LCIS
o hdu3397 Sequence operation
o hdu2871 Memory Control
o hdu1540 Tunnel Warfare
o CF46-D Parking Lot
· 扫描线
这类题目需要将一些操作排序,然后从左到右用一根扫描线(当然是在我们脑子里)扫过去
最典型的就是矩形面积并,周长并等题
o 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
线段树操作:update:区间增减 query:直接取根节点的值
?View Code CPP
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#include #include #include #include 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[]) { 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); 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; } |
o 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
?View Code CPP
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#include #include #include #include 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); 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; } |
· 练习
o hdu3265 Posters
hdu3642 Get The Treasury
poj2482 Stars in Your Window
poj2464 Brownie Points II
hdu3255 Farming
ural1707 Hypnotoad’s Secret
uva11983 Weird Advertisement
线段树与其他结合练习(欢迎大家补充):
· hdu3954 Level up
· hdu4027 Can you answer these queries?
· hdu3333 Turing Tree
· hdu3874 Necklace
· hdu3016 Man Down
· hdu3340 Rain in ACStar
· zju3511 Cake Robbery
· UESTC1558 Charitable Exchange
· CF85-D Sum of Medians
· spojGSS2 Can you answer these queries II