周中训练笔记+Just a Hook（9.14）

最近几天是真的没刷题，上课只是一个因素，在这里我的地吐槽一下， 新开的Java布置的作业太**多了，搞得我一有时间就得去敲那个，虽然题目不难，也很不错，后面有几道题目是搜索时候讲的例题，像什么全排列，整数的划分等等，但是耗时太多。。

好几天没提交过题目了，趁着今天课最少，水了一道，与单点更新不同的，线段树成段更新类型的（感觉像是照着模板打了一遍）。

还得多多趁课余时间多看看资料，加深理解。

Just a Hook

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 35386    Accepted Submission(s): 17280

Problem Description

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.

 

Input

The 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.

 

Output

For 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

11021 5 25 9 3

 

Sample Output

Case 1: The total value of the hook is 24.
#include<iostream>#include<cstdio>#include<algorithm>using namespace std;#define lson l,m,rt<<1#define rson m+1,r,rt<<1|1int col[500000];int sum[500000];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;    int a,b,c;    cin>>t;    for (int i=1;i<=t;i++)    {        scanf("%d%d",&n,&m);        build(1,n,1);        while (m--)        {            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",i,sum[1]);    }    return 0;}