poj 3264 Balanced Lineup

来源：互联网发布：三年数据对比分析编辑：程序博客网时间：2024/05/21 21:34

一个简单的线段树，憋了我好久。问题出在了存数据的数组开得不够，因为我建立线段树时要从D-1开始，但是D那一行的元素个数，可能会超过50000所以，存数据的数组要开的大一些。

 1 #include<stdio.h> 2 #define INF 1000000000 3  4 int n, m,D,a[(50000+1000)<<1],treemin[51000<<2], treemax[51000<<2]; 5  6 void build() 7 { 8     for(int i = D-1; i > 0; i --) 9     {10         if(treemin[i<<1] > n || treemin[i<<1] == 0) a[treemin[i<<1]] = INF;11         if(treemin[i<<1|1] > n || treemin[i<<1|1] == 0) a[treemin[i<<1|1]] = INF;12         treemin[i] = (a[treemin[i<<1]]<a[treemin[i<<1|1]] ? treemin[i<<1] : treemin[i<<1|1]);13         14         if(treemax[i<<1] > n || treemax[i<<1] == 0) a[treemax[i<<1]] = -1;15         if(treemax[i<<1|1] > n || treemax[i<<1|1] == 0) a[treemax[i<<1|1]] = -1;16         treemax[i] = (a[treemax[i<<1]]>a[treemax[i<<1|1]] ? treemax[i<<1] : treemax[i<<1|1]);17     }    18 }19 20 void querymax(int cur, int x, int y, int s, int t, int& ans1)21 {22     int mid = (x + y) >> 1, ls = cur << 1, rs = cur << 1 | 1;23     if(x >= s && y <= t)24     {25         if(a[treemax[cur]] > ans1) ans1 = a[treemax[cur]];26         return ;27     }28     if(mid >= s)29         querymax(ls, x, mid, s, t, ans1);30     if(mid + 1 <= t)31         querymax(rs, mid + 1, y, s, t, ans1);32 }33 34 void querymin(int cur, int x, int y, int s, int t, int& ans2)35 {36     int mid = (x + y) >> 1, ls = cur << 1, rs = cur << 1 | 1;37     if(x >= s && y <= t)38     {39         if(ans2 > a[treemin[cur]]) ans2 = a[treemin[cur]];40         return;41     }42     if(mid >= s)43         querymin(ls, x, mid, s, t, ans2);44     if(mid+1<= t)45         querymin(rs, mid+1, y, s, t, ans2);46 }47 48 int solve(int x, int y)49 {50     int ans1 = -INF;51     querymax(1,0,D-1,x,y,ans1);52     int ans2 = INF;53     querymin(1,0,D-1,x,y,ans2);54     //printf("treemax[1]=%d\n",treemax[1]);55     //printf("ans1 = %d  ans2 = %d\n",ans1, ans2);56     return ans1 - ans2;57 }58 void init()59 {60     while(~scanf("%d%d",&n,&m))61     {62         for(int i = 1; i <= n; i ++)63             scanf("%d",&a[i]);64         for(D = 1; D < n + 2; D <<= 1);65         for(int i = 0; i < D; i ++)66         {67             treemax[D+i] = i;68             treemin[D+i] = i;69         }70         build();71         int A, B;72         while(m --)73         {74             scanf("%d%d",&A,&B);75             printf("%d\n",solve(A,B));76         }77     }78 }79 int main()80 {81     init();82     return 0;83 }

0 0