HDU 1698 Just a Hook

题目链接：https://vjudge.net/problem/HDU-1698

题目大意：

　　给定一个 N 个数的序列，初始全为1，现在进行Q次操作，每次操作把 [L, R] 区间内的所有数变为 x，求操作完成后序列的总和。

分析：

　　线段树成段更新模板题。

代码如下：

#pragma GCC optimize("Ofast")
#include <bits/stdc++.h>
using namespace std;
 
#define INIT() std::ios::sync_with_stdio(false);std::cin.tie(0);
#define Rep(i,n) for (int i = 0; i < (n); ++i)
#define For(i,s,t) for (int i = (s); i <= (t); ++i)
#define rFor(i,t,s) for (int i = (t); i >= (s); --i)
#define ForLL(i, s, t) for (LL i = LL(s); i <= LL(t); ++i)
#define rForLL(i, t, s) for (LL i = LL(t); i >= LL(s); --i)
#define foreach(i,c) for (__typeof(c.begin()) i = c.begin(); i != c.end(); ++i)
#define rforeach(i,c) for (__typeof(c.rbegin()) i = c.rbegin(); i != c.rend(); ++i)
 
#define pr(x) cout << #x << " = " << x << "  "
#define prln(x) cout << #x << " = " << x << endl
 
#define LOWBIT(x) ((x)&(-x))
 
#define ALL(x) x.begin(),x.end()
#define INS(x) inserter(x,x.begin())
 
#define ms0(a) memset(a,0,sizeof(a))
#define msI(a) memset(a,inf,sizeof(a))
#define msM(a) memset(a,-1,sizeof(a))

#define MP make_pair
#define PB push_back
#define ft first
#define sd second
 
template<typename T1, typename T2>
istream &operator>>(istream &in, pair<T1, T2> &p) {
    in >> p.first >> p.second;
    return in;
}
 
template<typename T>
istream &operator>>(istream &in, vector<T> &v) {
    for (auto &x: v)
        in >> x;
    return in;
}
 
template<typename T1, typename T2>
ostream &operator<<(ostream &out, const std::pair<T1, T2> &p) {
    out << "[" << p.first << ", " << p.second << "]" << "
";
    return out;
}
 
typedef long long LL;
typedef unsigned long long uLL;
typedef pair< double, double > PDD;
typedef pair< int, int > PII;
typedef pair< LL, LL > PLL;
typedef set< int > SI;
typedef vector< int > VI;
typedef map< int, int > MII;
typedef vector< LL > VL;
typedef vector< VL > VVL;
const double EPS = 1e-10;
const int inf = 1e9 + 9;
const LL mod = 1e9 + 7;
const int maxN = 1e5 + 7;
const LL ONE = 1;
const LL evenBits = 0xaaaaaaaaaaaaaaaa;
const LL oddBits = 0x5555555555555555;

int T, N, Q;

#define lson l , mid , rt << 1 
#define rson mid + 1 , r , rt << 1 | 1
struct SegmentTree{
    int st[maxN << 2];
    int lazy[maxN << 2]; // 懒惰标记，lazy[i] = la 表示节点 i 尚未向下更新 la  
    
    inline void pushUp(int rt) {
        st[rt] = st[rt << 1] + st[rt << 1 | 1];
    }
    
    inline void pushDown(int rt, int l, int mid, int r) {
        if(lazy[rt]) {
            st[rt << 1] = lazy[rt] * (mid - l + 1);
            st[rt << 1 | 1] = lazy[rt] * (r - mid);
            lazy[rt << 1] = lazy[rt << 1 | 1] = lazy[rt];
            lazy[rt] = 0;
        } 
    }
    
    inline void build(int l, int r, int rt) {
        if(l >= r) {
            st[rt] = 1;
            return;
        }
        int mid = (l + r) >> 1;
        build(lson);
        build(rson);
        pushUp(rt);
    }
    
    // 成段更新 
    inline void update(int L, int R, int x, int l, int r, int rt) {
        if(L <= l && r <= R) {
            // 不更新到底，标记一下表明下面的还没更新 
            lazy[rt] = x;
            st[rt] = x * (r - l + 1);
            return;
        }
        int mid = (l + r) >> 1;
        pushDown(rt, l, mid, r); // 如果孩子节点有上一次还没更新的，就更新它，同时转移懒惰标记 
        
        if(L <= mid) update(L, R, x, lson);
        if(R > mid) update(L, R, x, rson);
        pushUp(rt);
    }
};
SegmentTree segTr;

int main(){
    INIT();
    cin >> T;
    For(i, 1, T) {
        cin >> N >> Q;
        segTr.build(1, N, 1);
        ms0(segTr.lazy);
        int L, R, x;
        Rep(j, Q) {
            cin >> L >> R >> x;
            segTr.update(L, R, x, 1, N, 1);
        }
        cout << "Case " << i <<": The total value of the hook is " << segTr.st[1] << "." << endl;
    }
    return 0;
}