【LOJ】#2306. 「NOI2017」蔬菜

题解

从后往前递推
如果我们知道了第i天的最优方案和第i天选择的蔬菜，加入第i天选择的蔬菜数量为S，我们只需要减去最小的S - (i - 1) * M 个蔬菜即可
所以我们只要求出最后一天的蔬菜选择
我们把每个蔬菜拆成c - 1个价值为a和1个价值为a + s，从大到小排序，然后用并查集维护可以选择的位置

代码

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <ctime>
#include <vector>
//#define ivorysi
#define MAXN 100005
#define eps 1e-7
#define mo 974711
#define pb push_back
#define mp make_pair
using namespace std;
typedef long long int64;
typedef unsigned int u32;
typedef double db;
const int P = 100000;
int N,M,K;
int num[MAXN];
int fa[MAXN];
struct node {
    int x,val,c,t;
    friend bool operator < (const node &a,const node &b) {
    return a.val < b.val;
    }
}veg[MAXN * 2];
int MK[MAXN * 10],tot = 0;
int64 ans[MAXN];
bool cmp(node a,node b) {
    return b < a;
}
void Init() {
    scanf("%d%d%d",&N,&M,&K);
    int a,s,c,x;
    for(int i = 1 ; i <= N ; ++i) {
    scanf("%d%d%d%d",&a,&s,&c,&x);
    if(x == 0) {
        veg[i * 2 - 1] = (node){x,a,c - 1,P};
        veg[i * 2] = (node){0,a + s,1,P};
    }
    else {
        int t = (c - 1) / x + 1;
        veg[i * 2 - 1] = (node){x,a,c - 1 - (t - 1)* x,t};
        veg[i * 2] = (node){0,a + s,1,t};
    }
    }
    sort(veg + 1,veg + 2 * N + 1,cmp);
}
int find_pos(int x) {
    return fa[x] == x ? x : fa[x] = find_pos(fa[x]);
}
void Solve() {
    for(int i = 1 ; i <= P + 1; ++i) fa[i] = i;
    for(int i = 1 ; i <= 2 * N ; ++i) {
    int s = find_pos(veg[i].t);
    int used = 0;
    while(s && used < veg[i].c + veg[i].x * (veg[i].t - 1)) {
        int now = veg[i].c + veg[i].x * (veg[i].t - s);
        now -= used;
        while(now && num[s] < M) {
        --now;++used;num[s]++;
        MK[++tot] = veg[i].val;
        }
        if(num[s] == M) fa[s] = find_pos(s - 1);
        s = find_pos(s - 1);
    }
    }
    sort(MK + 1,MK + tot + 1);
    for(int i = 1 ; i <= tot ; ++i) ans[P] += MK[i];
    int poi = 1,S = 0;
    for(int i = 1 ; i <= P ; ++i) S += num[i];
    
    for(int i = P - 1 ; i >= 1 ; --i) {
    ans[i] = ans[i + 1];
    int del = max(S - i * M,0);
    S -= del;
    for(int j = poi ; j <= poi + del - 1; ++j) {
        ans[i] -= MK[j];
    }
    poi += del;
    }
    int p;
    for(int i = 1 ; i <= K ; ++i) {
    scanf("%d",&p);
    printf("%lld\n",ans[p]);
    }
}
int main() {
#ifdef ivorysi
    freopen("f1.in","r",stdin);
#endif
    Init();
    Solve();
}