P2034 选择数字

P2034 选择数字

题目描述
给定一行n个非负整数a[1]..a[n]。如今你能够选择其中若干个数，但不能有超过k个连续的数字被选择。你的任务是使得选出的数字的和最大。

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错误日志： longlong 的 \(inf\) 没有设为 0xfffffffffffffff

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Solution

正难则反
正难则反
复习看到了双倍经验就回来看看
求最大值有点难， 转化为求最小去除值
设 \(dp[i]\) 为 \(i\) 为断点， 考虑到 \(i\) 处的最小去除值
发现每隔 \(K + 1\) 必有一个断点
因此 \(dp[i] = \min_{j = i - k - 1}^{i - 1}dp[j] + a[i]\)
而后单调队列优化， 答案在区间 \([n - K, n]\) 内
总值减一下便可

Code

#include<iostream>
#include<cstdio>
#include<queue>
#include<cstring>
#include<algorithm>
#include<climits>
#define LL long long
#define REP(i, x, y) for(LL i = (x);i <= (y);i++)
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 = 200019, inf = 0xfffffffffffffff;
LL num, K;
LL a[maxn], sum;
LL dp[maxn];
struct Que{
    LL Index, val;
    }Q[maxn];
LL head = 1, tail;
void push_back(LL Index, LL val){
    while(head <= tail && Q[tail].val >= val)tail--;
    Q[++tail] = (Que){Index, val};
    }
LL get_min(LL p){
    while(head <= tail && p - Q[head].Index > K + 1)head++;
    return Q[head].val;
    }
void init(){
    num = RD(), K = RD();
    REP(i, 1, num)a[i] = RD(), sum += a[i];
    memset(dp, 127, sizeof(dp));
    push_back(0, 0);
    }
void solve(){
    REP(i, 1, num){
        dp[i] = get_min(i) + a[i];
        push_back(i, dp[i]);
        }
    LL ans = inf;
    REP(i, num - K, num)ans = min(ans, dp[i]);
    printf("%lld\n", sum - ans);
    return ;
    REP(i, 1, num)printf("%lld ", dp[i]);
    }
int main(){
    init();
    solve();
    return 0;
    }