hdu4902 Nice boat 线段树

题意：给你一个数列，给你两个操作，

1）数列中L-R每个值都赋值为 X

2）数列中L-R每个大与 X 的数都变为 gcd（a[i],X） (L <= i <= R)

解题思路：这个题目按理来说时间复杂度是不严谨的，本来想把gcd 从大到小存到 vector 里面，没想到直接更新就行了。

不过从这场比赛发现自己细节方面太糟糕了。还是需要改进

解题代码：

  1 // File Name: 1006.cpp
  2 // Author: darkdream
  3 // Created Time: 2014年07月31日 星期四 12时46分19秒
  4 
  5 #include<vector>
  6 #include<list>
  7 #include<map>
  8 #include<set>
  9 #include<deque>
 10 #include<stack>
 11 #include<bitset>
 12 #include<algorithm>
 13 #include<functional>
 14 #include<numeric>
 15 #include<utility>
 16 #include<sstream>
 17 #include<iostream>
 18 #include<iomanip>
 19 #include<cstdio>
 20 #include<cmath>
 21 #include<cstdlib>
 22 #include<cstring>
 23 #include<ctime>
 24 
 25 using namespace std;
 26 #define maxn 100005 
 27 struct node{
 28     int l , r ; 
 29     int is1;
 30     int limits;
 31 }tree[maxn << 2];
 32 int a[maxn];
 33 int L(int x)
 34 {
 35     return 2*x; 
 36 }
 37 int gcd(int a, int b)
 38 {
 39     return b == 0 ?a:gcd(b,a%b);
 40 }
 41 int R(int x)
 42 {
 43     return 2 * x + 1; 
 44 }
 45 void pushup(int c)
 46 {
 47     tree[c].limits = max(tree[L(c)].limits,tree[R(c)].limits);
 48 }
 49 void pushdown(int c)
 50 {
 51     if(tree[c].is1)
 52     {
 53         tree[L(c)].is1 = 1; 
 54         tree[R(c)].is1 = 1;
 55         tree[c].is1 = 0;
 56         tree[L(c)].limits = tree[R(c)].limits = tree[c].limits;
 57     }
 58 }
 59 void build(int c , int l ,int r)
 60 {
 61     tree[c].l = l ; 
 62     tree[c].r = r;
 63     tree[c].is1 = 0 ; 
 64     if(l == r)
 65     {
 66         tree[c].limits = a[l];
 67         return ;
 68     }
 69     int m = (l+r)/2;
 70     build(L(c),l,m);
 71     build(R(c),m+1,r);
 72     pushup(c);
 73 }
 74 void update1(int c, int l , int r,int x)
 75 {
 76     if(l <= tree[c].l && r >= tree[c].r)
 77     {
 78         tree[c].is1 = 1;
 79         tree[c].limits = x; 
 80         return ;
 81     }
 82     int m = (tree[c].l+ tree[c].r)/2;
 83     pushdown(c);
 84     if(l <= m)
 85         update1(L(c),l,r,x);
 86     if( r > m )
 87         update1(R(c),l,r,x);
 88     pushup(c);
 89 }
 90 void update2(int c, int l , int r,int x)
 91 {
 92     if(tree[c].limits < x)
 93     {
 94         return ; 
 95     }
 96     if(tree[c].is1 && l <= tree[c].l && r >= tree[c].r)
 97     {
 98         tree[c].limits = gcd(tree[c].limits,x);
 99         return ;
100     }
101     if(tree[c].l == tree[c].r)
102     {
103         tree[c].is1 = 0 ; 
104         tree[c].limits = gcd(x,tree[c].limits);
105         return; 
106     }
107     int m = (tree[c].l+ tree[c].r)/2;
108     pushdown(c);
109     if(l <= m)
110         update2(L(c),l,r,x);
111     if( r > m )
112         update2(R(c),l,r,x);
113     pushup(c);
114 }
115 void put(int l , int r , int x)
116 {
117 //    printf("****%d %d\n",l,r);
118     for(int i =l ;i <= r;i ++ )
119         a[i] = x;
120     return; 
121 }
122 void get(int c )
123 {
124     if(tree[c].is1) 
125     {
126         put(tree[c].l,tree[c].r,tree[c].limits);
127         return; 
128     }
129     if(tree[c].l == tree[c].r)
130     {
131         a[tree[c].l] = tree[c].limits;
132         return; 
133     }
134     pushdown(c);
135     get(L(c));
136     get(R(c));
137 }
138 int main(){
139     int t;
140     //freopen("input.txt","r",stdin);
141     //freopen("output.txt","w",stdout);
142     scanf("%d",&t);
143     while(t--)
144     {
145         //memset(tree,0,sizeof(tree));
146         int n ,m;
147         scanf("%d",&n);
148         for(int i = 1;i <= n;i ++)
149         {        
150             scanf("%d",&a[i]);
151          //   printf("%d ",a[i]);
152         }
153         //printf("\n");
154         build(1,1,n);
155         scanf("%d",&m);
156         int ta,tb,tc,td;
157         for(int i = 1;i <= m;i ++)
158         {
159             scanf("%d %d %d %d",&ta,&tb,&tc,&td);
160             if(ta == 1)
161             {
162                 update1(1,tb,tc,td);
163             }else{
164                 update2(1,tb,tc,td);
165             }
166         }
167         get(1);
168         for(int i =1;i <= n;i ++)
169             printf("%d ",a[i]);
170         printf("\n");
171         //printf("%d\n",t);
172     }
173     return 0;
174 }

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posted on 2014-08-01 15:21 dark_dream 阅读(160) 评论(0) 编辑收藏举报