P3097 [USACO13DEC]最优挤奶Optimal Milking

P3097 [USACO13DEC]最优挤奶Optimal Milking

题意简述：给定n个点排成一排，每个点有一个点权，多次改变某个点的点权并将最大点独立集计入答案，输出最终的答案

感谢@zht467 提供翻译

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错误日志： 又双叒叕没开long long

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Solution

考虑线段树维护
只有四种情况， 选择左端点与否 \(*\) 选择右端点与否
共四种情况
维护这四个便可以上推了

void pushup(LL id){
    tree[id].ans[0][0] = max(tree[lid].ans[0][0] + tree[rid].ans[1][0], tree[lid].ans[0][1] + tree[rid].ans[0][0]);
    tree[id].ans[0][1] = max(tree[lid].ans[0][0] + tree[rid].ans[1][1], tree[lid].ans[0][1] + tree[rid].ans[0][1]);
    tree[id].ans[1][0] = max(tree[lid].ans[1][0] + tree[rid].ans[1][0], tree[lid].ans[1][1] + tree[rid].ans[0][0]);
    tree[id].ans[1][1] = max(tree[lid].ans[1][0] + tree[rid].ans[1][1], tree[lid].ans[1][1] + tree[rid].ans[0][1]);
    }

Code

#include<iostream>
#include<cstdio>
#include<queue>
#include<cstring>
#include<algorithm>
#include<climits>
#define LL long long
using namespace std;
LL RD(){
    LL out = 0,flag = 1;char c = getchar();
    while(c < '0' || c >'9'){if(c == '-')flag = -1;c = getchar();}
    while(c >= '0' && c <= '9'){out = out * 10 + c - '0';c = getchar();}
    return flag * out;
    }
const LL maxn = 80019;
LL num, na, a[maxn];
#define lid (id << 1)
#define rid (id << 1) | 1
struct seg_tree{
    LL l, r;
    LL ans[2][2];//0为选，1为不选
    }tree[maxn << 2];
void pushup(LL id){
    tree[id].ans[0][0] = max(tree[lid].ans[0][0] + tree[rid].ans[1][0], tree[lid].ans[0][1] + tree[rid].ans[0][0]);
    tree[id].ans[0][1] = max(tree[lid].ans[0][0] + tree[rid].ans[1][1], tree[lid].ans[0][1] + tree[rid].ans[0][1]);
    tree[id].ans[1][0] = max(tree[lid].ans[1][0] + tree[rid].ans[1][0], tree[lid].ans[1][1] + tree[rid].ans[0][0]);
    tree[id].ans[1][1] = max(tree[lid].ans[1][0] + tree[rid].ans[1][1], tree[lid].ans[1][1] + tree[rid].ans[0][1]);
    }
void build(LL id, LL l, LL r){
    tree[id].l = l, tree[id].r = r;
    if(l == r){
        tree[id].ans[1][1] = a[l];
        return ;
        }
    LL mid = (l + r) >> 1;
    build(lid, l, mid), build(rid, mid + 1, r);
    pushup(id);
    }
void update(LL id, LL val, LL l, LL r){
    if(tree[id].l == l && tree[id].r == r){
        tree[id].ans[1][1] = val;
        return ;
        }
    LL mid = (tree[id].l + tree[id].r) >> 1;
    if(mid < l)update(rid, val, l, r);
    else if(mid >= r)update(lid, val, l, r);
    pushup(id);
    }
LL sum;
int main(){
    num = RD(), na = RD();
    for(LL i = 1;i <= num;i++)a[i] = RD();
    build(1, 1, num);
    for(LL i = 1;i <= na;i++){
        LL p = RD(), val = RD();
        update(1, val, p, p);
        LL ans = max(tree[1].ans[0][0], tree[1].ans[0][1]);
        ans = max(ans, tree[1].ans[1][0]);
        ans = max(ans, tree[1].ans[1][1]);
        sum += ans;
        }
    
    printf("%lld\n", sum);
    return 0;
    }