Hdu 5316 Magician

求区间奇偶交错最大子序列

线段树模板题

单点更新，区间查询

就是区间合并的时候麻烦一点点，但是也不难

代码如下

#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
#define LL long long
const long long INFF = 0x3f3f3f3f3f3f3f3fll;

struct Info{
    LL v[2][2];
    void maintain(LL x){
        v[0][0]=v[0][1]=v[1][0]=v[1][1]=x;
    }
    Info(){
        maintain(-INFF);
    }
    LL value(){
        LL a=max(v[0][0],v[0][1]);
        LL b=max(v[1][0],v[1][1]);
        return max(a,b);
    }
};

Info operator + (const Info & a,const Info & b){
    Info ret;
    for(int i=0;i<=1;i++)
        for(int j=0;j<=1;j++){
            LL & V=ret.v[i][j];
            V=max(a.v[i][j],b.v[i][j]);
            V=max(V,a.v[i][0]+b.v[1][j]);
            V=max(V,a.v[i][1]+b.v[0][j]);
        }
    return ret;
}

const int maxn = 112345;

Info val[maxn*4];

#define Now int o,int l,int r
#define Mid int m = l + (r - l)/2
#define lson o<<1,l,m
#define rson o<<1|1,m+1,r
#define root 1,1,n

void update(Now,int p,LL v){
    if(l==r){
        val[o].maintain(-INFF);
        val[o].v[l%2][l%2]=v;
        return;
    }
    Mid;
    if(p<=m)
        update(lson,p,v);
    else
        update(rson,p,v);
    val[o]=val[o<<1]+val[o<<1|1];
}

Info query(Now,int ql,int qr){
    if(ql<=l && r<=qr){
        return val[o];
    }
    Mid;
    Info ret;
    if(ql<=m)
        ret = ret + query(lson,ql,qr);
    if(m+1<=qr)
        ret = ret + query(rson,ql,qr);
    return ret;
}

int main(){
    int T;
    int n,m;
    scanf("%d",&T);
    while(T-- && ~scanf("%d %d",&n,&m)){
        LL x;
        for(int i=1;i<=n;i++){
            scanf("%lld",&x);
            update(root,i,x);
        }
        int l,r;
        int p,ord;
        while(m--){
            scanf("%d",&ord);
            if(ord==0){
                scanf("%d %d",&l,&r);
                Info ans = query(root,l,r);
                printf("%lld\n",ans.value());
            }
            else{
                scanf("%d %lld",&p,&x);
                update(root,p,x);
            }
        }
    }
    return 0;
}