Just a Hook
Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 28728    Accepted Submission(s): 14242
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.

這道涉及的就是線段樹的成段替換還要區間求和。
成段替換就是加了一個Lazy_tag延遲標記。
然後由於是求整段的和，所以直接輸出sum[1]就可以了。

#include<iostream>
#include<cstring>
#include<cstdlib>
#include<algorithm>
#include<cctype>
#include<cmath>
#include<ctime>
#include<string>
#include<stack>
#include<deque>
#include<queue>
#include<list>
#include<set>
#include<map>
#include<cstdio>
#include<limits.h>
#define fir first
#define sec second
#define fin freopen("/home/ostreambaba/文件/input.txt", "r", stdin)
#define fout freopen("/home/ostreambaba/文件/output.txt", "w", stdout)
#define mes(x, m) memset(x, m, sizeof(x))
#define pii pair<int, int>
#define Pll pair<ll, ll>
#define INF 1e9+7
#define Pi 4.0*atan(1.0)
#define MOD 1000000007

#define lowbit(x) (x&(-x))
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
#define ls rt<<1
#define rs rt<<1|1

typedef long long ll;
typedef unsigned long long ull;
const double eps = 1e-12;
const int maxn = 100000+5;
using namespace std;

inline int read(){
    int x(0),f(1);
    char ch=getchar();
    while(ch<'0'||ch>'9'){if(ch=='-') f=-1;ch=getchar();}
    while (ch>='0'&&ch<='9') x=x*10+ch-'0',ch=getchar();
    return x*f;
}
int sum[maxn<<2];
int Lazy_tag[maxn<<2];
inline void PushUp(int rt)
{
    sum[rt] = sum[rt<<1] + sum[rt<<1|1];
}
inline void PushDown(int rt, int m)
{
    if(Lazy_tag[rt]){
        Lazy_tag[rt<<1] = Lazy_tag[rt<<1|1] = Lazy_tag[rt];
        sum[rt<<1] = (m-(m>>1))*Lazy_tag[rt];
        sum[rt<<1|1] = (m>>1)*Lazy_tag[rt];
        Lazy_tag[rt] = 0;
    }
}
inline void buildTree(int l, int r, int rt)
{
    Lazy_tag[rt] = 0;
    sum[rt] = 1;
    if(l == r){
        return;
    }
    int m = (l+r)>>1;
    buildTree(lson);
    buildTree(rson);
    PushUp(rt);
}
inline void query(int L, int R, int p, int l, int r, int rt)
{
    if(L <= l && r <= R){
        Lazy_tag[rt] = p;
        sum[rt] = (r-l+1)*p;
        return;
    }
    int m = (l+r)>>1;
    PushDown(rt, r-l+1);
    if(L <= m){
        query(L, R, p, lson);
    }
    if(R > m){
        query(L, R, p, rson);
    }
    PushUp(rt);
}
int main()
{
   // fin;
    int Case;
    Case = read();
    for(int k = 1; k <= Case; ++k){
        int N, Q, a, b, c;
        N = read();
        Q = read();
        buildTree(1, N, 1);
        for(int i = 1; i <= Q; ++i){
            a = read();
            b = read();
            c = read();
            query(a, b, c, 1, N, 1);
        }
        printf("Case %d: The total value of the hook is %d.\n", k, sum[1]);
    }
    return 0;
}