Codeforces Round #222 (Div. 1) D. Developing Game

D - Developing Game

思路：我们先枚举左边界，把合法的都扣出来，那么对于这些合法的来说值有v 和 r两维了，把v, r看成线段的两端，

问题就变成了，最多能选多少线段 使得不存在这样两条(l1 r1)  (l2 r2) l2 > r1，我们把线段l1 r1 看成二维平面内的点(l1, r1)

那么答案就变成了分别在(1, 1)，  (2, 2)，  (3, 3)，  (4, 4)， .....， (n, n)的左上的点的个数的最大值，我们用这个建

线段树然后就变成了区间加减问题。

#include<bits/stdc++.h>
#define LL long long
#define fi first
#define se second
#define mk make_pair
#define PII pair<int, int>
#define y1 skldjfskldjg
#define y2 skldfjsklejg

using namespace std;

const int N = 3e5 + ;
const int M = 1e7 + ;
const int inf = 0x3f3f3f3f;
const LL INF = 0x3f3f3f3f3f3f3f3f;
const int mod = ;

int n;
struct segmentTree {
    int mx[N << ], lazy[N << ], id[N << ];

    void pushdown(int rt) {
        if(lazy[rt]) {
            lazy[rt << ] += lazy[rt];
            lazy[rt <<  | ] += lazy[rt];
            mx[rt << ] += lazy[rt];
            mx[rt <<  | ] += lazy[rt];
            lazy[rt] = ;
        }
    }

    void pushup(int rt) {
        if(mx[rt << ] >= mx[rt <<  | ]) {
            mx[rt] = mx[rt << ];
            id[rt] = id[rt << ];
        } else {
            mx[rt] = mx[rt <<  | ];
            id[rt] = id[rt <<  | ];
        }
    }

    void build(int l, int r, int rt) {
        mx[rt] = lazy[rt] = ;
        if(l == r) {
            id[rt] = l;
            return;
        }
        int mid = l + r >> ;
        build(l, mid, rt << );
        build(mid + , r, rt <<  | );
        pushup(rt);
    }

    void update(int L, int R, int val, int l, int r, int rt) {
        if(l >= L && r <= R) {
            mx[rt] += val;
            lazy[rt] += val;
            return;
        }
        int mid = l + r >> ;
        pushdown(rt);
        if(L <= mid) update(L, R, val, l, mid, rt << );
        if(R > mid) update(L, R, val, mid + , r, rt <<  | );
        pushup(rt);
    }
} seg;

struct node {
    int l, v, r, id;
    bool operator < (const node &rhs) const {
        return l < rhs.l;
    }
} a[N];

int main() {
    seg.build(, , );
    scanf("%d", &n);
    for(int i = ; i <= n; i++)
        scanf("%d%d%d", &a[i].l, &a[i].v, &a[i].r), a[i].id = i;
    sort(a + , a +  + n);

    int ans = , ret1, ret2;
    priority_queue<PII, vector<PII>, greater<PII> > que;
    for(int i = , j = ; i <= ; i++) {

        while(j <= n && a[j].l <= i) {
            int x = a[j].v, y = a[j].r;
            que.push(mk(x, y));
            seg.update(x, y, , , , );
            j++;
        }

        while(!que.empty() && que.top() < mk(i, )) {
            PII now = que.top(); que.pop();
            seg.update(now.fi, now.se, -, , , );
        }

        if(ans < seg.mx[]) {
            ans = seg.mx[];
            ret1 = i;
            ret2 = seg.id[];
        }
    }

    vector<int> vec;
    for(int i = ; i <= n; i++) {
        if(a[i].l <= ret1 && a[i].v >= ret1 && a[i].v <= ret2 && a[i].r >= ret2)
            vec.push_back(a[i].id);
    }

    sort(vec.begin(), vec.end());
    printf("%d\n", vec.size());
    for(int i = ; i < vec.size(); i++)
        printf("%d ", vec[i]);
    puts("");
    return ;
}

/*
*/