poj 3264 Balanced Lineup

来源：互联网发布：安卓app源码编辑：程序博客网时间：2024/06/04 14:53
就是简单的线段树最大值减去最小值，这里没有单点更新
#include <iostream>#include <cstdio>#include <cstring>using namespace std;const int maxn=50005;int num[maxn];struct{    int l,r,maxx,minn;}tree[4*maxn];int max(int m, int n){    if(m>n)      return m;    return n;}int min(int m,int n){    if(m<n)      return m;    return n;}void build (int root, int l, int r){    tree[root].l=l;    tree[root].r=r;    if(tree[root].l == tree[root].r)      {          tree[root].maxx=num[l];          tree[root].minn=num[l];          return;      }    int mid=(l+r)/2;    build(2*root, l, mid);    build(2*root+1, mid+1, r);    tree[root].maxx=max(tree[2*root].maxx,tree[2*root+1].maxx);    tree[root].minn=min(tree[2*root].minn,tree[2*root+1].minn);}/*void update(int root, int pos, int val){    if(tree[root].l == tree[root].r&&tree[root].l==pos)    {        tree[root].sum=val;        return;    }    int mid=(tree[root].l + tree[root].r)/2;    if(pos <= mid)        update(2*root, pos, val);    else        update(2*root+1, pos, val);    tree[root].sum=max(tree[2*root].sum,tree[2*root+1].sum);}*/int query1(int root, int L, int R){    int s;    if(L == tree[root].l && R == tree[root].r)      return tree[root].maxx;    if(R <= tree[2*root].r)       s=query1(2*root, L, R);    else if(L >= tree[2*root+1].l)       s=query1(2*root+1, L, R);    else    {        s=max(query1(2*root, L, tree[2*root].r),query1(2*root+1, tree[2*root+1].l, R));    }    return s;}int query2(int root, int L, int R){    int s;    if(L == tree[root].l && R == tree[root].r)      return tree[root].minn;    if(R <= tree[2*root].r)       s=query2(2*root, L, R);    else if(L >= tree[2*root+1].l)       s=query2(2*root+1, L, R);    else    {        s=min(query2(2*root, L, tree[2*root].r),query2(2*root+1, tree[2*root+1].l, R));    }    return s;}char str[20];int main(){    int t, m, n, cas=1, a, b,ans1,ans2;    while(~scanf("%d%d",&m,&n))    {        for(int i=1; i<=m; i++)        scanf("%d",&num[i]);        build(1, 1, maxn);        for(int i=1; i<=n; i++)        {            scanf("%d%d",&a,&b);            ans1=query1(1,a,b);            ans2=query2(1,a,b);            cout<<ans1-ans2<<endl;        }    }    return 0;}
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