Cardboard Box

貪了個半天貪不對， 我發現我根本就不會貪心。

我們先按b排序， 然後枚舉選兩顆心的b的最大值， 在這個之前的肯定都要選一個， 因為前面的要是一個都沒選的話，

你可以把當前選兩顆心的替換成前面選兩顆心， 然後用平衡樹或者線段樹維護一下前k大和就好啦。

#include<bits/stdc++.h>
#define LL long long
#define fi first
#define se second
#define mk make_pair
#define PLL pair<LL, LL>
#define PLI pair<LL, int>
#define
 
 PII pair<int, int>
#define SZ(x) ((int)x.size())
#define ull unsigned long long

using namespace std;

const int N = 3e5 + 7;
const int inf = 0x3f3f3f3f;
const LL INF = 0x3f3f3f3f3f3f3f3f;
const int mod = 1e9 + 7;
const double eps = 1e-9;
const double PI = acos(-1);

#define lson l, mid, rt->ls
#define
 
 rson mid + 1, r, rt->rs
struct Node {
    Node() {
        ls = rs = NULL;
        sum = cnt = 0;
    }
    Node *ls, *rs;
    LL sum;
    int cnt;
};

inline void pull(Node* rt) {
    rt->cnt = rt->ls->cnt + rt->rs->cnt;
    rt->sum = rt->ls->sum + rt->rs->sum;
    
 
if(!rt->ls->cnt) delete rt->ls, rt->ls = NULL;
    if(!rt->rs->cnt) delete rt->rs, rt->rs = NULL;
}
inline void push(Node* rt) {
    if(!rt->ls) rt->ls = new Node();
    if(!rt->rs) rt->rs = new Node();
}
void update(int p, int c, int l, int r, Node* rt) {
    if(l == r) {
        rt->cnt += c;
        rt->sum += 1LL * p * c;
        return;
    }
    push(rt);
    int mid = l + r >> 1;
    if(p <= mid) update(p, c, lson);
    else update(p, c, rson);
    pull(rt);
}
LL query(int k, int l, int r, Node* rt) {
    if(!k) return 0;
    if(rt->cnt <= k) return rt->sum;
    if(l == r) return 1LL * k * l;
    push(rt);
    int mid = l + r >> 1;
    if(rt->ls->cnt >= k) return query(k, lson);
    else return query(rt->ls->cnt, lson) + query(k - rt->ls->cnt, rson);
}

int n, w, a[N], b[N], id[N], fin[N];
LL ans = INF;

bool cmpa(int x, int y) {
    return a[x] < a[y];
}
bool cmpb(int x, int y) {
    return b[x] < b[y];
}
int main() {
    Node* Rt = new Node();
    scanf("%d%d", &n, &w);
    for(int i = 1; i <= n; i++) scanf("%d%d", &a[i], &b[i]);
    for(int i = 1; i <= n; i++) id[i] = i;
    int where = -1;
    sort(id + 1, id + 1 + n, cmpa);
    if(w <= n) {
        LL ret = 0;
        for(int i = 1; i <= w; i++) ret += a[id[i]];
        ans = min(ans, ret);
        where = 0;
    }
    sort(id + 1, id + 1 + n, cmpb);
    for(int i = 1; i <= n; i++) update(a[id[i]], 1, 1, inf, Rt);
    LL prefix = 0;
    LL ret = 0;
    for(int i = 1; i <= n; i++) {
        int x = id[i];
        update(a[x], -1, 1, inf, Rt);
        ret = prefix + b[x];
        int need = w - (i + 1);
        if(Rt->cnt >= need) {
            ret += query(need, 1, inf, Rt);
            if(ret < ans) {
                ans = ret;
                where = i;
            }
        }
        prefix += a[x];
        update(b[x] - a[x], 1, 1, inf, Rt);
    }
    if(!where) {
        sort(id + 1, id + 1 + n, cmpa);
        for(int i = 1; i <= w; i++) fin[id[i]] = 1;
    } else {
        priority_queue<PII, vector<PII>, greater<PII> > que;
        fin[id[where]] = 2;
        w -= where + 1;
        for(int i = 1; i < where; i++) que.push(mk(b[id[i]] - a[id[i]], i)), fin[id[i]] = 1;
        for(int i = where + 1; i <= n; i++) que.push(mk(a[id[i]], i));
        while(w--) {
            int who = que.top().se;
            que.pop();
            if(who < where) fin[id[who]] = 2;
            else fin[id[who]] = 1;
        }
    }
    printf("%lld\n", ans);
    for(int i = 1; i <= n; i++) printf("%d", fin[i]);
    puts("");
    return 0;
}

/*
*/

Codeforces 436E Cardboard Box (看題解)