P2048题解+总结

主要思路

 

 将[l , r]拆分成[l , pos - 1],[pos + 1 , r]分别弹入堆，弹出点pos

 每次都选最优的子段，这样选 k 次。

另：struct 多元组函数写法如下

点击查看代码

#include <bits/stdc++.h>
using namespace std;
const int MAXN = 500005;
long long sum[MAXN],inrmq[MAXN][21],ans = 0;
void rmq_build(int n){
    for(int i = 1;i <= n;i++){
        inrmq[i][0] = i;
    }
    for(int j = 1;(1 << j) <= n;j++){
        for(int i = 1;i + (1 << j) - 1 <= n;i++){
            int xx = inrmq[i][j - 1];
            int yy = inrmq[i + (1 << (j - 1))][j - 1];
            inrmq[i][j] = sum[xx] > sum[yy] ? xx : yy;
        }
    }
}
int rmq_query(int l,int r){
    int temp = log2(r - l + 1);
    int x = inrmq[l][temp];
    int y = inrmq[r - (1 << temp) + 1][temp];
    return sum[x] > sum[y] ? x : y;
}
struct node{
    int start,l,r,pos;
    node(int x,int y,int z){
        start = x;
        l = y;
        r = z;
        pos = rmq_query(y,z);
    }
    friend bool operator < (const node& a,const node& b){
        return sum[a.pos] - sum[a.start - 1] < sum[b.pos] - sum[b.start - 1];
    }
};
priority_queue<node>q;
int main(){
    freopen("1.out","r",stdin);
    int n,k,leng,r;
    cin >> n >> k >> leng >> r;
    for(int i = 1;i <= n;i++){
        cin >> sum[i];
        sum[i] += sum[i - 1];
    }
    rmq_build(n);
    for(int i = 1;i <= n;i++){
        if(i + leng - 1 <= n){
            node x=node(i,i + leng - 1,min(i + r - 1,n));
            q.push(x);

            //printf("%d %d %d %d\n",x.start,x.l,x.r,x.pos);
        }
    }
    //printf("\n");
    while(k--){
        int tempst = q.top().start;
        int templ = q.top().l;
        int tempr = q.top().r;
        int temppos = q.top().pos;
        //printf("%d %d %d %d\n",tempst,templ,tempr,temppos);
        q.pop();
        ans += sum[temppos] - sum[tempst - 1];
        if(templ != temppos) {
            q.push(node(tempst,templ,temppos - 1));
        }
        if(tempr != temppos){
            q.push(node(tempst,temppos + 1,tempr));
        }
    }
    cout << ans;
    return 0;
}