魔兽世界---屠夫(Just a Hook)
Just a Hook
Time Limit: 4000/2000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 23450 Accepted Submission(s): 11742
Now Pudge wants to do some operations on the hook.
Let us number the consecutive metallic sticks of the hook from 1 to N. For each operation, Pudge can change the consecutive metallic sticks, numbered from X to Y, into cupreous sticks, silver sticks or golden sticks.
The total value of the hook is calculated as the sum of values of N metallic sticks. More precisely, the value for each kind of stick is calculated as follows:
For each cupreous stick, the value is 1.
For each silver stick, the value is 2.
For each golden stick, the value is 3.
Pudge wants to know the total value of the hook after performing the operations.
You may consider the original hook is made up of cupreous sticks.
For each case, the first line contains an integer N, 1<=N<=100,000, which is the number of the sticks of Pudge’s meat hook and the second line contains an integer Q, 0<=Q<=100,000, which is the number of the operations.
Next Q lines, each line contains three integers X, Y, 1<=X<=Y<=N, Z, 1<=Z<=3, which defines an operation: change the sticks numbered from X to Y into the metal kind Z, where Z=1 represents the cupreous kind, Z=2 represents the silver kind and Z=3 represents the golden kind.
#include<stdio.h>
const int MAXN=;
struct Node{
int l,r;
int sum,lazy,val;//val不能全局变量。。。
};
Node tree[MAXN<<];
#define NOW tree[root].sum=tree[root<<1].sum+tree[root<<1|1].sum
#define lson root<<1,tree[root].l,mid
#define rson root<<1|1,mid+1,tree[root].r
void build(int root,int l,int r){
tree[root].l=l;
tree[root].r=r;
tree[root].lazy=;
tree[root].val=;
if(l==r)tree[root].sum=;
else{
int mid=(l+r)>>;
build(lson);
build(rson);
NOW;
}
}
void update(int root,int l,int r,int v){
if(l==tree[root].l&&r==tree[root].r){
tree[root].lazy=;
tree[root].val=v;
tree[root].sum=(r-l+)*v;
}
else{
int mid=(tree[root].l+tree[root].r)>>;
if(tree[root].lazy==){
tree[root].lazy=;
update(lson,tree[root].val);
update(rson,tree[root].val);
tree[root].val=;
}
if(r<=mid)update(root<<,l,r,v);
else if(l>mid)update(root<<|,l,r,v);
else{//这个不能少了,代表l,r在tree的两个孩子节点内,也就是找到了l,r的区间;
update(root<<,l,mid,v);
update(root<<|,mid+,r,v);
}
NOW;
}
}
int main(){
int T,N,Q,flot=;
scanf("%d",&T);
while(T--){
scanf("%d",&N);
scanf("%d",&Q);
build(,,N);
while(Q--){
int x,y,z;
scanf("%d%d%d",&x,&y,&z);
update(,x,y,z);
}
printf("Case %d: The total value of the hook is %d.\n",++flot,tree[].sum);
}
return ;
}
过了一段时间又写了一遍,增强了自己的理解:
代码:
#include<iostream>
#include<cstdio>
#include<cstring>
#include<cmath>
#include<vector>
#include<algorithm>
#include<set>
using namespace std;
#define mem(x,y) memset(x,y,sizeof(x))
const int MAXN=100010;
struct Node{
int lazy,v;
};
Node tree[MAXN<<2];
#define ll root<<1
#define rr root<<1|1
#define lson ll,l,mid
#define rson rr,mid+1,r
#define LAZY(x) tree[x].lazy
#define V(x) tree[x].v
int ans;
void pushdown(int root,int x){
if(LAZY(root)){
LAZY(ll)=LAZY(root);
LAZY(rr)=LAZY(root);
V(ll)=LAZY(root)*(x-(x>>1));//这里错了半天。。。。。右节点有时候要多一。。。
V(rr)=LAZY(root)*(x>>1);
LAZY(root)=0;
}
}
void pushup(int root){
V(root)=V(ll)+V(rr);
}
void build(int root,int l,int r){
LAZY(root)=0;
int mid=(l+r)>>1;
if(l==r){
V(root)=1;
return;
}
build(lson);
build(rson);
pushup(root);
}
void update(int root,int l,int r,int L,int R,int v){
if(l>=L&&r<=R){
LAZY(root)=v;
V(root)=(r-l+1)*v;
return;
}
pushdown(root,r-l+1);
int mid=(l+r)>>1;
/*if(mid>=R)update(lson,L,R,v);
else if(mid<L)update(rson,L,R,v);
else{
update(lson,L,mid,v);
update(rson,mid+1,R,v);
}*///这样写也可以。。。
if(mid>=L)update(lson,L,R,v);
if(mid<R)update(rson,L,R,v);
pushup(root);
}
void query(int root,int l,int r,int L,int R){
int mid=(l+r)>>1;
if(l>=L&&r<=R){
ans+=V(root);
return;
}
pushdown(root,r-l+1);
if(mid>=L)query(lson,L,R);
if(mid<R)query(rson,L,R);
}
int main(){
int T,N,kase=0;
scanf("%d",&T);
while(T--){
scanf("%d",&N);
build(1,1,N);
int q,a,b,c;
scanf("%d",&q);
while(q--){
scanf("%d%d%d",&a,&b,&c);
update(1,1,N,a,b,c);
}
ans=0;
query(1,1,N,1,N);//也可以不询问直接输出tree[1].v
printf("Case %d: The total value of the hook is %d.\n",++kase,ans);
}
return 0;
}
map超时;
代码:
#include<iostream>
#include<cstdio>
#include<map>
#include<algorithm>
using namespace std;
map<int,int>mp;
map<int,int>::iterator it1;
const int MAXN=100010;
struct Node{
int s,e,v;
};
Node dt[MAXN];
int main(){
int T,N,q,kase=0;
scanf("%d",&T);
while(T--){
scanf("%d",&N);
for(int i=1;i<=N;i++)mp[i]=1;
scanf("%d",&q);
for(int i=0;i<q;i++)scanf("%d%d%d",&dt[i].s,&dt[i].e,&dt[i].v);
for(int i=0;i<q;i++){
for(it1=mp.lower_bound(dt[i].s);it1!=mp.end();){
if(it1->first<=dt[i].e)it1->second=dt[i].v,it1++;
else break;
}
}
int ans=0;
for(int i=1;i<=N;i++)ans+=mp[i];
printf("Case %d: The total value of the hook is %d.\n",++kase,ans);
}
return 0;
}
过了一段时间又写了遍,本想着一遍a的,但是错了几个小时,实在找不出,问了群里面的大神,原来我是x-(x>>1)没有加括号,这就是代码风格的问题了,我的代码风格存在很大的问题,以至于错误了很难找出来,大神指出了几点问题:
1:关于define 的括号问题;
2:关于字母与运算符的缩进问题;
3:关于宏定义的名称问题;都存在一定问题;以后要注意了,参照谷歌的吧
另一种线段树写法:
extern "C++"{
#include<iostream>
#include<cstdio>
#include<cstring>
#include<cmath>
#include<algorithm>
using namespace std;
typedef long long LL;
typedef unsigned u;
typedef unsigned long long ull;
#define mem(x,y) memset(x,y,sizeof(x))
void SI(double &x){scanf("%lf",&x);}
void SI(int &x){scanf("%d",&x);}
void SI(LL &x){scanf("%lld",&x);}
void SI(u &x){scanf("%u",&x);}
void SI(ull &x){scanf("%llu",&x);}
void SI(char *s){scanf("%s",s);} void PI(int x){printf("%d",x);}
void PI(double x){printf("%lf",x);}
void PI(LL x){printf("%lld",x);}
void PI(u x){printf("%u",x);}
void PI(ull x){printf("%llu",x);}
void PI(char *s){printf("%s",s);} #define NL puts("");
#define ll root<<1
#define rr root<<1|1
#define lson ll,l,mid
#define rson rr,mid+1,r
const int INF=0x3f3f3f3f;
const int MAXN=;
int tree[MAXN<<];
int lazy[MAXN<<];
}
/*
void pushup(int root){
tree[root]=tree[ll]+tree[rr];
}
void pushdown(int root,int x){
if(lazy[root]){
lazy[ll]=lazy[root];
lazy[rr]=lazy[root];
// tree[ll]=lazy[root]*(mid-l+1);
// tree[rr]=lazy[root]*(r-mid);
tree[ll]=lazy[root]*(x-(x>>1));
tree[rr]=lazy[root]*(x>>1); lazy[root]=0;
}
}
void build(int root,int l,int r){
int mid=(l+r)>>1;
lazy[root]=0;
if(l==r){
tree[root]=1;return ;
}
build(lson);
build(rson);
pushup(root);
}
void update(int root,int l,int r,int L,int R,int C){
if(l>=L&&r<=R){
tree[root]=(r-l+1)*C;
lazy[root]=C;
return ;
}
int mid=(l+r)>>1;
pushdown(root,r-l+1);
if(mid>=L)update(lson,L,R,C);
if(mid<R)update(rson,L,R,C);
pushup(root);
}
*/
void pushdown(int root){
lazy[ll] = lazy[rr] = lazy[root];
lazy[root] = ;
}
void update(int root,int l,int r,int L,int R,int C){
if(l >= L && r <= R){
lazy[root] = C;
return;
}
if(lazy[root])
pushdown(root);
int mid = (l + r) >> ;
if(mid >= L)
update(lson,L,R,C);
if(mid < R)
update(rson,L,R,C);
}
int sum(int root,int l,int r){
int mid = (l + r) >> ;
if(lazy[root])
return (r - l + ) * lazy[root];
return sum(lson) + sum(rson);
}
int main(){
//assert(false);
int T,kase=;
int N,M;
SI(T);
while(T--){
SI(N);
lazy[]=;
SI(M);
int a,b,c;
while(M--){
scanf("%d%d%d",&a,&b,&c);
update(,,N,a,b,c);
}
printf("Case %d: The total value of the hook is %d.\n",++kase,sum(,,N));
}
return ;
}
java超时了。。。
注意:左右查找都是mid判断的。。。注意lazy要把左右支lazy
代码:
import java.util.Scanner; public class hdoj1698{ public static void main(String[] argv){
Scanner cin = new Scanner(System.in);
SegmentTree atree = new SegmentTree(100010 << 2);
int T, N, q;
T = cin.nextInt();
int kase = 0;
while(T-- > 0){
N = cin.nextInt();
atree.build(1, 1, N);
q = cin.nextInt();
while(q-- > 0){
int a, b, c;
a = cin.nextInt();
b = cin.nextInt();
c = cin.nextInt();
atree.update(1, 1, N, a, b, c);
}
// System.out.println(atree.tree[1]); System.out.println("Case " + (++kase) +
": The total value of the hook is " +
atree.query(1, 1, N, 1, N) + ".");
}
}
} class SegmentTree{
private int[] tree;
private int[] lazy;
public SegmentTree(int size) {
tree = new int[size];
lazy = new int[size];
}
private void pushup(int root){
tree[root] = tree[root << 1] + tree[root<<1 | 1];
}
private void pushdown(int root, int x){
if(lazy[root] != 0){
lazy[root << 1] = lazy[root << 1|1] = lazy[root];
tree[root << 1] = (x - (x >> 1))*lazy[root];
tree[root << 1|1] = (x >> 1)*lazy[root];
lazy[root] = 0;
}
}
public void build(int root, int l, int r){
lazy[root] = 0;
if(l == r){
tree[root] = 1;
return ;
}
int mid = (l + r) >> 1;
build(root << 1, l, mid);
build(root << 1|1, mid + 1, r);
pushup(root);
}
public void update(int root, int l, int r, int L, int R, int v){
if(l >= L && r <= R){
tree[root] = (r - l + 1)*v;
lazy[root] = v;
return;
}
pushdown(root, r - l + 1);
int mid = (r + l) >> 1;
if(mid >= L){
update(root << 1, l, mid, L, R, v);
}
if(mid < R){
update(root << 1|1, mid + 1, r, L, R, v);
//日了狗了。。。。。。。。。。
}
pushup(root);
}
public int query(int root, int l, int r, int L, int R){
if(l >= L && r <= R){
return tree[root];
}
pushdown(root, r - l + 1);
int sum = 0;
int mid = (l + r) >> 1;
if(mid >= L){
sum += query(root << 1, l, mid, L, R);
}
if(mid < R){
sum += query(root << 1|1, mid +1, r, L, R);
}
return sum;
} }
魔兽世界---屠夫(Just a Hook)的更多相关文章
- HDU-1698 JUST A HOOK 线段树
最近刚学线段树,做了些经典题目来练手 Just a Hook Time Limit: 4000/2000 MS (Java/Others) Memory Limit: 32768/32768 K (J ...
- HDU 1698 Just a Hook(线段树
Just a Hook Time Limit: 4000/2000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total ...
- svnserver hook python
在使用中可能会遇到的错误排除 :1.Error: svn: 解析"D:\www\test"出错,或svn: E020024: Error resolving case of 'D: ...
- Android Hook技术
原文:http://blog.csdn.net/u011068702/article/details/53208825 附:Android Hook 全面入侵监听器 第一步.先爆项目demo照片,代码 ...
- Frida HOOK微信实现骰子作弊
由于微信摇骰子的功能在本地进行随机后在发送,所以存在可以hook掉判断骰子数的方法进行修改作弊. 1.frida实现hook java层函数1)写个用来测试的demo,当我们点击按钮的时候会弹出窗口显 ...
- java的关闭钩子(Shutdown Hook)
Runtime.getRuntime().addShutdownHook(shutdownHook); 这个方法的含义说明: 这个方法的意思就是在jvm中增加一个关闭的钩子,当jv ...
- IDT HOOK思路整理
IDT(中断描述符表)分为IRQ(真正的硬件中断)和软件中断(又叫异常). HOOK的思路为,替换键盘中断处理的函数地址为自己的函数地址.这样在键盘驱动和过滤驱动之前就可以截获键盘输入. 思路确定之后 ...
- Android Hook 借助Xposed
主要就是使用到了Xposed中的两个比较重要的方法,handleLoadPackage获取包加载时候的回调并拿到其对应的classLoader:findAndHookMethod对指定类的方法进行Ho ...
- iOS App 无代码入侵的方法hook
继续Objective-C runtime的研究 最近公司项目在做用户行为分析 于是App端在某些页面切换,交互操作的时候需要给统计系统发送一条消息 在几十个Controller 的项目里,一个一个地 ...
随机推荐
- class属性多个样式的用法
今天看到一个非常好用的样式用法,给已经在睡梦中苏醒的你们来一段代码头脑风暴.大家都知道现在div+css布局的使用已经不是可以用潮流来概括的了,换个词应该是:普及.以前的表格布局现在是少之极少,因为表 ...
- HCE:Host-based Card Emulation基于Android设备的卡片模拟器
HCE技术支持提供了一个软实现SE的通路,Service实现的方式很多,可以使用文件,使用网络,甚至连接真正的SE.支持HCE的测试手机:目前可以确定使用了NXP PN547作为CLF的NFC手机已经 ...
- ListView.setOnItemClickListener 点击无效
如果ListView中的单个Item的view中存在checkbox,button等view,会导致ListView.setOnItemClickListener无效, 事件会被子View捕获到,Li ...
- ByteBuffer用法总结
转自:http://blog.csdn.net/mars5337/article/details/6576417 在NIO中,数据的读写操作始终是与缓冲区相关联的.读取时信道(SocketChanne ...
- Hibernate 1、Hello Hibernate
所使用到的jar 包: 1.创建实体类 public class User { private Integer id; private String name; private String addr ...
- linux学习之(六)-主机名、网络IP的配置与查看
设置Linux 本机IP有一个非常好用的命令就是setup命令,在Linux终端打入setup命令就会打开Linux配置窗口,如下图: . 在打开的窗口中通过上下键选择 Network config ...
- wamp 虚拟目录的设置(转载)
现在先来配置虚拟主机:1.先打开apache的配置文件httpd.conf,并去掉#Include conf/extra/httpd-vhosts.conf前面的#!!2.打开apache的apach ...
- SqlServer2008 数据库同步的两种方式(Sql JOB)
尊重原著作:本文转载自http://www.cnblogs.com/tyb1222/archive/2011/05/27/2060075.html 数据库同步是一种比较常用的功能.下面介绍的就是数据库 ...
- Linux学习之停止进程
首先,用ps查看进程,方法如下: ps -ef ……smx 1822 1 0 11:38 ? 00:00:49 gnome-terminalsmx 18 ...
- mysql 数据库热备份
https://www.percona.com/doc/percona-xtrabackup/2.2/index.html