HDU - 1698 Just a Hook

Posted 2020-12-31 wpbing

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<=100,000, 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. 
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

1
10
2
1 5 2
5 9 3

Sample Output

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

一段线段由n条小线段组成，每次操作把一个区间的小线段变成金银铜之一（金的价值为3，银为2，铜为1），最初可当做全为铜；最后求这条线段的总价值。

#include<iostream>
#include<cstdio>
#include<cstring>
#include<cmath>
using namespace std;
typedef long long ll;
const int maxn=100005;
int t,n,q;
ll sum[maxn*4],add[maxn*4];
void pushdown(int rt,int len){
    if(add[rt]){
        add[rt<<1]=add[rt];//注意这里是都变成这个数，而不是求和 
        add[rt<<1|1]=add[rt];
        sum[rt<<1]=add[rt]*(len-(len>>1));//sum依然为区间和 
        sum[rt<<1|1]=add[rt]*(len>>1);
        add[rt]=0;
    }
}
void build(int l,int r,int rt){
    add[rt]=0;
    if(l==r){
        sum[rt]=1;//初始为一 
        return;
    }
    int m=(l+r)>>1;
    build(l,m,rt<<1);
    build(m+1,r,rt<<1|1);
    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){
        add[rt]=c;
        sum[rt]=(ll)c*(r-l+1);
        return;
    }
    pushdown(rt,r-l+1);//下推标记 
    int m=(l+r)>>1;
    if(L<=m) update(L,R,c,l,m,rt<<1);
    if(R>m) update(L,R,c,m+1,r,rt<<1|1);
    sum[rt]=sum[rt<<1]+sum[rt<<1|1];
}
ll querry(int L,int R,int l,int r,int rt){
    if(L<=l&&R>=r) return sum[rt];
    pushdown(rt,r-l+1);//下推标记 
    int m=(l+r)>>1;
    ll ans=0;
    if(L<=m) ans+=querry(L,R,l,m,rt<<1);
    if(R>m) ans+=querry(L,R,m+1,r,rt<<1|1);
    return ans;
}
int main()
{
    int x,y,z;
    scanf("%d",&t);
    for(int cas=1;cas<=t;cas++){
        scanf("%d%d",&n,&q);
        build(1,n,1);
        while(q--){
            scanf("%d%d%d",&x,&y,&z);
            update(x,y,z,1,n,1);
        }
        printf("Case %d: The total value of the hook is %I64d.
",cas,querry(1,n,1,n,1));
    }
}