poj 3264 Balanced Lineup

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。

#include<stdio.h>
#define INF 1000000000

int n, m,D,a[(50000+1000)<<1],treemin[51000<<2], treemax[51000<<2];

void build()
{
    for(int i = D-1; i > 0; i --)
    {
        if(treemin[i<<1] > n || treemin[i<<1] == 0) a[treemin[i<<1]] = INF;
        if(treemin[i<<1|1] > n || treemin[i<<1|1] == 0) a[treemin[i<<1|1]] = INF;
        treemin[i] = (a[treemin[i<<1]]<a[treemin[i<<1|1]] ? treemin[i<<1] : treemin[i<<1|1]);
        
        if(treemax[i<<1] > n || treemax[i<<1] == 0) a[treemax[i<<1]] = -1;
        if(treemax[i<<1|1] > n || treemax[i<<1|1] == 0) a[treemax[i<<1|1]] = -1;
        treemax[i] = (a[treemax[i<<1]]>a[treemax[i<<1|1]] ? treemax[i<<1] : treemax[i<<1|1]);
    }    
}

void querymax(int cur, int x, int y, int s, int t, int& ans1)
{
    int mid = (x + y) >> 1, ls = cur << 1, rs = cur << 1 | 1;
    if(x >= s && y <= t)
    {
        if(a[treemax[cur]] > ans1) ans1 = a[treemax[cur]];
        return ;
    }
    if(mid >= s)
        querymax(ls, x, mid, s, t, ans1);
    if(mid + 1 <= t)
        querymax(rs, mid + 1, y, s, t, ans1);
}

void querymin(int cur, int x, int y, int s, int t, int& ans2)
{
    int mid = (x + y) >> 1, ls = cur << 1, rs = cur << 1 | 1;
    if(x >= s && y <= t)
    {
        if(ans2 > a[treemin[cur]]) ans2 = a[treemin[cur]];
        return;
    }
    if(mid >= s)
        querymin(ls, x, mid, s, t, ans2);
    if(mid+1<= t)
        querymin(rs, mid+1, y, s, t, ans2);
}

int solve(int x, int y)
{
    int ans1 = -INF;
    querymax(1,0,D-1,x,y,ans1);
    int ans2 = INF;
    querymin(1,0,D-1,x,y,ans2);
    //printf("treemax[1]=%d\n",treemax[1]);
    //printf("ans1 = %d  ans2 = %d\n",ans1, ans2);
    return ans1 - ans2;
}
void init()
{
    while(~scanf("%d%d",&n,&m))
    {
        for(int i = 1; i <= n; i ++)
            scanf("%d",&a[i]);
        for(D = 1; D < n + 2; D <<= 1);
        for(int i = 0; i < D; i ++)
        {
            treemax[D+i] = i;
            treemin[D+i] = i;
        }
        build();
        int A, B;
        while(m --)
        {
            scanf("%d%d",&A,&B);
            printf("%d\n",solve(A,B));
        }
    }
}
int main()
{
    init();
    return 0;
}