SPOJ KGSS Maximum Sum （线段树）

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SPOJ KGSS Maximum Sum

题意：求区间最大值与第二大值之和

思路：线段树维护两个最大值

/*************************************************************************     File Name: B.cpp     ID: obsoles1     PROG:      LANG: C++      Mail: 384099319@qq.com      Created Time: 2016年07月13日 星期三 18时04分37秒 ************************************************************************///#include<bits/stdc++.h>#include<cstdio>#include<cstring>#define Max(x,y) ((x)>(y)?(x):(y))#define Min(x,y) ((x)<(y)?(x):(y))#define each(it,v) for(__typeof((v).begin()) it=(v).begin();it!=(v).end();++it)#define Abs(x,y) ((x)>(y)?((x)-(y)):((y)-(x)))#define ll long long#define Mem0(x) memset(x,0,sizeof(x))#define Mem1(x) memset(x,-1,sizeof(x))#define MemX(x) memset(x,0x3f,sizeof(x))#define pb push_back#define left t<<1#define right t<<1|1using namespace std;const int N=1e5+10;struct node{  int max1,max2;}tree[N<<2];int max1,max2;void push_up(int t){  if(tree[left].max1>tree[right].max1){    tree[t].max1=tree[left].max1;    tree[t].max2=Max(tree[left].max2,tree[right].max1);  }else{    tree[t].max1=tree[right].max1;    tree[t].max2=Max(tree[right].max2,tree[left].max1);  }  //cout<<"tree["<<t<<"]="<<tree[t].max1<<" , "<<tree[t].max2<<endl;}void build(int t,int L,int R){  if(L==R){    scanf("%d",&tree[t].max1);    tree[t].max2=0;    return;  }  int mid=(L+R)>>1;  build(left,L,mid);  build(right,mid+1,R);  push_up(t);}void update(int t,int L,int R,int x,int v){  if(L==R){    tree[t].max1=v;    return;  }  int mid=(L+R)>>1;  if(x<=mid)update(left,L,mid,x,v);  else update(right,mid+1,R,x,v);  push_up(t);}void query(int t,int L,int R,int l,int r){  if(l<=L && R<=r){    if(tree[t].max1>max1){      max2=Max(max2,max1);      max1=tree[t].max1;      max2=Max(max2,tree[t].max2);    }else{      max2=Max(max2,tree[t].max1);    }    return;  }  int mid=(L+R)>>1;  if(l<=mid)query(left,L,mid,l,r);  if(r>mid)query(right,mid+1,R,l,r);}int main(){  int n,m,l,r;  char op[2];  while(~scanf("%d",&n)){    Mem0(tree);    build(1,1,n);    scanf("%d",&m);    while(m--){      scanf("%s%d%d",op,&l,&r);      if(op[0]=='U')update(1,1,n,l,r);      else{        max1=max2=0;        query(1,1,n,l,r);        //cout<<"max1="<<max1<<" max2="<<max2<<endl;        printf("%d\n",max1+max2);      }    }  }}

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