Just a Hook

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 34542    Accepted Submission(s): 16868

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

1
10
2
1 5 2
5 9 3

Sample Output

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

Source

2008 “Sunline Cup” National Invitational Contest

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wangye

题目大意：对于每组数据，给定钩子段数n，默认每段长度开始为1，接着q次变换，每次变换将x~y的钩子长度变为z。求最后钩子总长度。

解题思路：就是线段树，区间的变换变成了区间的覆盖，最后输出总长度也就是sum[1]即可。理解后也就是模板题

#include<cstdio>
#include<algorithm>
using namespace std;
const int N=100005;
const int maxn=4*N;
int sum[maxn],lazy[maxn];
int n,x,y,z;

void pushup(int num)
{
    sum[num] = sum[num*2]+sum[num*2+1];
}

void pushdown(int num,int l)
{
    if(lazy[num])
    {
        //lazy、sum对应的值均是直接覆盖
        lazy[num*2] = lazy[num];
        lazy[num*2+1] = lazy[num];

        sum[num*2] = lazy[num]*(l-l/2);
        sum[num*2+1] = lazy[num]*(l/2);

        lazy[num] = 0;
    }
}

void build(int num,int l,int r)
{
    lazy[num] = 0;
    sum[num] = 1;//默认每段为1
    if(l==r)
    {
        return ;
    }
    int mid=(l+r)/2;
    build(num*2,l,mid);
    build(num*2+1,mid+1,r);
    pushup(num);
}

void update(int num,int l,int r)
{
    if(x<=l&&y>=r)
    {
        //lazy、sum对应的值均是直接覆盖
        lazy[num] = z;
        sum[num] = z*(r-l+1);
        return ;
    }
    pushdown(num,r-l+1);
    int mid=(l+r)/2;
    if(x<=mid)
        update(num*2,l,mid);
    if(y>mid)
        update(num*2+1,mid+1,r);
    pushup(num);
}

int query(int num,int l,int r)
{
    if(x<=l&&y>=r)
    {
        return sum[num];
    }
    pushdown(num,r-l+1);
    int mid=(l+r)/2,ans=0;
    if(mid>=x)
        ans += query(num*2,l,mid);
    if(mid<y)
        ans += query(num*2,mid+1,r);
    return ans;
}

int main()
{
    int T;
    int Case=1;
    scanf("%d",&T);
    while(T--)
    {
        scanf("%d",&n);
        int m;
        build(1,1,n);
        scanf("%d",&m);
        for(int i=1;i<=m;i++)
        {
            scanf("%d %d %d",&x,&y,&z);
            update(1,1,n);
        }
        //x=1;y=n;
        //int ans = query(1,1,n);
        printf("Case %d: The total value of the hook is %d.\n",Case++,sum[1]);
    }
}