【HDOJ】4902 Nice boat

区间线段树。题目还不错。
  1 /*  */
  2 #include <iostream>
  3 #include <string>
  4 #include <map>
  5 #include <queue>
  6 #include <set>
  7 #include <vector>
  8 #include <algorithm>
  9 #include <cstdio>
 10 #include <cmath>
 11 #include <cstring>
 12 #include <climits>
 13 #include <cctype>
 14 using namespace std;
 15 
 16 #define lson l, mid, rt<<1
 17 #define rson mid+1, r, rt<<1|1
 18 #define MAXN 100005
 19 
 20 typedef struct {
 21     int mx, val;
 22 } node_t;
 23 
 24 node_t T[MAXN<<2];
 25 int n, m, x;
 26 int l;
 27 
 28 int gcd(int n, int m) {
 29     if (m == 0)
 30         return n;
 31     if (m > n)
 32         return gcd(m, n);
 33     return gcd(m, n%m);
 34 }
 35 
 36 void PushUp(int rt) {
 37     T[rt].val = T[rt<<1].val==T[rt<<1|1].val ? T[rt<<1].val:-1;
 38     T[rt].mx = max(T[rt<<1].mx, T[rt<<1|1].mx);
 39 }
 40 
 41 void PushDown(int rt) {
 42     if (T[rt].val >= 0) {
 43         T[rt<<1].val = T[rt<<1|1].val = T[rt].val;
 44         T[rt<<1].mx = T[rt<<1|1].mx = T[rt].mx;
 45     }
 46 }
 47 
 48 void build(int l, int r, int rt) {
 49     if (l == r) {
 50         scanf("%d", &T[rt].val);
 51         T[rt].mx = T[rt].val;
 52         return ;
 53     }
 54     int mid = (l+r)>>1;
 55     build(lson);
 56     build(rson);
 57     PushUp(rt);
 58 }
 59 
 60 void update1(int ll, int rr, int l, int r, int rt) {
 61     if (ll<=l && rr>=r) {
 62         T[rt].val = x;
 63         T[rt].mx = x;
 64         return ;
 65     }
 66     int mid = (l+r)>>1;
 67     PushDown(rt);
 68     if (rr <= mid) {
 69         update1(ll, rr, lson);
 70     } else if (ll > mid) {
 71         update1(ll, rr, rson);
 72     } else {
 73         update1(ll, mid, lson);
 74         update1(mid+1, rr, rson);
 75     }
 76     PushUp(rt);
 77 }
 78 
 79 void update2(int ll, int rr, int l, int r, int rt) {
 80     if (T[rt].mx <= x)
 81         return ;
 82     int mid = (l+r)>>1;
 83     if (ll<=l && rr>=r) {
 84         if (T[rt].val >= 0) {
 85             T[rt].val = gcd(T[rt].val, x);
 86             T[rt].mx = T[rt].val;
 87         } else {
 88             update2(ll, rr, lson);
 89             update2(ll, rr, rson);
 90             PushUp(rt);
 91         }
 92         return ;
 93     }
 94     PushDown(rt);
 95     if (rr <= mid) {
 96         update2(ll, rr, lson);
 97     } else if (ll > mid) {
 98         update2(ll, rr, rson);
 99     } else {
100         update2(ll, mid, lson);
101         update2(mid+1, rr, rson);
102     }
103     PushUp(rt);
104 }
105 
106 void printAll(int l, int r, int rt) {
107     if (T[rt].val >= 0) {
108         for (int i=l; i<=r; ++i)
109             printf("%d ", T[rt].val);
110         return ;
111     }
112     int mid = (l+r)>>1;
113     printAll(lson);
114     printAll(rson);
115 }
116 
117 int main() {
118     int i, j, k;
119     int l, r;
120     int t;
121 
122     #ifndef ONLINE_JUDGE
123         freopen("data.in", "r", stdin);
124         freopen("data.out", "w", stdout);
125     #endif
126 
127     scanf("%d", &t);
128     while (t--) {
129         scanf("%d", &n);
130         build(1, n, 1);
131         scanf("%d", &m);
132         while (m--) {
133             scanf("%d %d %d %d", &k, &l, &r, &x);
134             if (k == 1) {
135                 update1(l, r, 1, n, 1);
136             } else {
137                 update2(l, r, 1, n, 1);
138             }
139         }
140         printAll(1, n, 1);
141         putchar('
');
142     }
143 
144     return 0;
145 }
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原文地址：https://www.cnblogs.com/bombe1013/p/4392469.html