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：

  1 #pragma GCC optimize("Ofast")
  2 #include <bits/stdc++.h>
  3 using namespace std;
  4  
  5 #define INIT() std::ios::sync_with_stdio(false);std::cin.tie(0);
  6 #define Rep(i,n) for (int i = 0; i < (n); ++i)
  7 #define For(i,s,t) for (int i = (s); i <= (t); ++i)
  8 #define rFor(i,t,s) for (int i = (t); i >= (s); --i)
  9 #define ForLL(i, s, t) for (LL i = LL(s); i <= LL(t); ++i)
 10 #define rForLL(i, t, s) for (LL i = LL(t); i >= LL(s); --i)
 11 #define foreach(i,c) for (__typeof(c.begin()) i = c.begin(); i != c.end(); ++i)
 12 #define rforeach(i,c) for (__typeof(c.rbegin()) i = c.rbegin(); i != c.rend(); ++i)
 13  
 14 #define pr(x) cout << #x << " = " << x << "  "
 15 #define prln(x) cout << #x << " = " << x << endl
 16  
 17 #define LOWBIT(x) ((x)&(-x))
 18  
 19 #define ALL(x) x.begin(),x.end()
 20 #define INS(x) inserter(x,x.begin())
 21  
 22 #define ms0(a) memset(a,0,sizeof(a))
 23 #define msI(a) memset(a,inf,sizeof(a))
 24 #define msM(a) memset(a,-1,sizeof(a))
 25 
 26 #define MP make_pair
 27 #define PB push_back
 28 #define ft first
 29 #define sd second
 30  
 31 template<typename T1, typename T2>
 32 istream &operator>>(istream &in, pair<T1, T2> &p) {
 33     in >> p.first >> p.second;
 34     return in;
 35 }
 36  
 37 template<typename T>
 38 istream &operator>>(istream &in, vector<T> &v) {
 39     for (auto &x: v)
 40         in >> x;
 41     return in;
 42 }
 43  
 44 template<typename T1, typename T2>
 45 ostream &operator<<(ostream &out, const std::pair<T1, T2> &p) {
 46     out << "[" << p.first << ", " << p.second << "]" << "
";
 47     return out;
 48 }
 49  
 50 typedef long long LL;
 51 typedef unsigned long long uLL;
 52 typedef pair< double, double > PDD;
 53 typedef pair< int, int > PII;
 54 typedef pair< LL, LL > PLL;
 55 typedef set< int > SI;
 56 typedef vector< int > VI;
 57 typedef map< int, int > MII;
 58 typedef vector< LL > VL;
 59 typedef vector< VL > VVL;
 60 const double EPS = 1e-10;
 61 const int inf = 1e9 + 9;
 62 const LL mod = 1e9 + 7;
 63 const int maxN = 1e5 + 7;
 64 const LL ONE = 1;
 65 const LL evenBits = 0xaaaaaaaaaaaaaaaa;
 66 const LL oddBits = 0x5555555555555555;
 67 
 68 int T, N, Q;
 69 
 70 #define lson l , mid , rt << 1 
 71 #define rson mid + 1 , r , rt << 1 | 1
 72 struct SegmentTree{
 73     int st[maxN << 2];
 74     int lazy[maxN << 2]; // 懒惰标记，lazy[i] = la 表示节点 i 尚未向下更新 la  
 75     
 76     inline void pushUp(int rt) {
 77         st[rt] = st[rt << 1] + st[rt << 1 | 1];
 78     }
 79     
 80     inline void pushDown(int rt, int l, int mid, int r) {
 81         if(lazy[rt]) {
 82             st[rt << 1] = lazy[rt] * (mid - l + 1);
 83             st[rt << 1 | 1] = lazy[rt] * (r - mid);
 84             lazy[rt << 1] = lazy[rt << 1 | 1] = lazy[rt];
 85             lazy[rt] = 0;
 86         } 
 87     }
 88     
 89     inline void build(int l, int r, int rt) {
 90         if(l >= r) {
 91             st[rt] = 1;
 92             return;
 93         }
 94         int mid = (l + r) >> 1;
 95         build(lson);
 96         build(rson);
 97         pushUp(rt);
 98     }
 99     
100     // 成段更新 
101     inline void update(int L, int R, int x, int l, int r, int rt) {
102         if(L <= l && r <= R) {
103             // 不更新到底，标记一下表明下面的还没更新 
104             lazy[rt] = x;
105             st[rt] = x * (r - l + 1);
106             return;
107         }
108         int mid = (l + r) >> 1;
109         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 
110         
111         if(L <= mid) update(L, R, x, lson);
112         if(R > mid) update(L, R, x, rson);
113         pushUp(rt);
114     }
115 };
116 SegmentTree segTr;
117 
118 int main(){
119     INIT();
120     cin >> T;
121     For(i, 1, T) {
122         cin >> N >> Q;
123         segTr.build(1, N, 1);
124         ms0(segTr.lazy);
125         int L, R, x;
126         Rep(j, Q) {
127             cin >> L >> R >> x;
128             segTr.update(L, R, x, 1, N, 1);
129         }
130         cout << "Case " << i <<": The total value of the hook is " << segTr.st[1] << "." << endl;
131     }
132     return 0;
133 }
View Code