kuangbin专题七 HDU1698 Just a Hook （set线段树）

发布时间：2020-12-14 04:17:09 所属栏目：大数据来源：网络整理

导读：In the game of DotA,Pudge’s meat hook is actually the most horrible thing for most of the heroes. The hook is made up of several consecutive metallic sticks which are of the same length. Now Pudge wants to do some operations on the hook.

In the game of DotA,Pudge’s meat hook is actually the most horrible thing for most of the heroes. The hook is made up of several consecutive metallic sticks which are of the same length.

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.

InputThe input consists of several test cases. The first line of the input is the number of the cases. There are no more than 10 cases.
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,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.
OutputFor each case,print a number in a line representing the total value of the hook after the operations. Use the format in the example.
Sample Input

Sample Output

Case 1: The total value of the hook is 24.


将一段区间设置为同一个数，然后区间求和。自己敲得时候，还是发现了一点细节问题。这个和修改有区别，pushdown的左右孩子的sum是= 而不是 += 可以理解 
还有最重要的一点!!! 以后的lazy还是不要直接写了，应该先判断一下，就因为这个，找了好长时间bug


还有一个问题是 区间求和时的 Pushdown和Pushup 不是必要的？？？不是太懂。

  1 #include <iostream>
  2 #include <stdio.h>
  3 #include <math.h>
  4 #include <string.h>
  5 #include <stdlib.h>
  6 #include <string>
  7 #include <vector>
  8 #include <set>
  9 #include <map>
 10 #include <queue>
 11 #include <algorithm>
 12 #include <sstream>
 13 #include <stack>
 14 using namespace std;
 15 #define FO freopen("in.txt","r",stdin);
 16 #define rep(i,a,n) for (int i=a;i<n;i++)
 17 #define per(i,n) for (int i=n-1;i>=a;i--)
 18 #define pb push_back
 19 #define mp make_pair
 20 #define all(x) (x).begin(),(x).end()
 21 #define fi first
 22 #define se second
 23 #define SZ(x) ((int)(x).size())
 24 #define debug(x) cout << "&&" << x << "&&" << endl;
 25 #define lowbit(x) (x&-x)
 26 #define mem(a,b) memset(a,b,sizeof(a));
 27 typedef vector<int> VI;
 28 typedef long long ll;
 29 typedef pair<int,int> PII;
 30 const ll mod=1000000007;
 31 const int inf = 0x3f3f3f3f;
 32 ll powmod(ll a,ll b) {ll res=1;a%=mod;for(;b;b>>=1){if(b&1)res=res*a%mod;a=a*a%mod;}return res;}
 33 ll gcd(ll a,ll b) { return b?gcd(b,a%b):a;}
 34 //head
 35 
 36 const int maxn=100010;
 37 int sum[maxn<<2],lazy[maxn<<2],_,n,q,l,r,val;
 38 
 39 void Pushup(int rt) {
 40     sum[rt]=sum[rt<<1]+sum[rt<<1|1];
 41 }
 42 
 43 void Pushdown(int rt,int x) {
 44     if(lazy[rt]) {//!!!
 45         lazy[rt<<1]=lazy[rt<<1|1]=lazy[rt];
 46         sum[rt<<1]=(x-(x>>1))*lazy[rt];//!!! =
 47         sum[rt<<1|1]=(x>>1)*lazy[rt];
 48         lazy[rt]=0;
 49     }
 50 }
 51 
 52 void build(int rt,int L,int R) {
 53     lazy[rt]=0;
 54     if(L==R) {
 55         sum[rt]=1;
 56         return;
 57     }
 58     int mid=(L+R)>>1;
 59     build(rt<<1,L,mid);
 60     build(rt<<1|1,mid+1,R);
 61     Pushup(rt);
 62 }
 63 
 64 void Updata(int rt,int R,int l,int r) {
 65     if(L>=l&&R<=r) {//核心
 66         lazy[rt]=val;
 67         sum[rt]=(R-L+1)*val;
 68         return;
 69     }
 70     Pushdown(rt,R-L+1);
 71     int mid=(L+R)>>1;
 72     if(l<=mid) Updata(rt<<1,mid,r);
 73     if(r>mid) Updata(rt<<1|1,R,r);
 74     Pushup(rt);
 75 }
 76 
 77 int Query(int rt,int r) {
 78     if(L>=l&&R<=r) return sum[rt];
 79     Pushdown(rt,R-L+1);
 80     int ans=0;
 81     int mid=(L+R)>>1;
 82     if(l<=mid) ans+=Query(rt<<1,r);
 83     if(r>mid) ans+=Query(rt<<1|1,r);
 84     Pushup(rt);
 85     return ans;
 86 
 87 }
 88 
 89 int main() {
 90     int cur=1;
 91     for(scanf("%d",&_);_;_--) {
 92         scanf("%d",&n);
 93         build(1,1,n);
 94         scanf("%d",&q);
 95         while(q--) {
 96             scanf("%d%d%d",&l,&r,&val);
 97             Updata(1,r);
 98         }
 99         printf("Case %d: The total value of the hook is %d.n",cur++,Query(1,1,n));
100     }
101 }

（编辑：李大同）

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