牛多校1I Points Division（DP）题解

Posted 2022-05-03 kirinsb

题意：n个点，分成两组A，B，如果点i在A中，那么贡献值\(a_i\)，反之为\(b_i\)。
现要求任意\(i \in A，j \in B\)不存在 \(x_i >= x_j\) 且 \(y_i <= y_j\)，也就是说A中点不在B中点的右下方。
思路：https://blog.nowcoder.net/n/7205418146f3446eb0b1ecec8d2ab1da
代码：

#include<cmath>
#include<set>
#include<map>
#include<queue>
#include<cstdio>
#include<vector>
#include<cstring>
#include <iostream>
#include<algorithm>
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
const int maxn = 1e5 + 5;
const int M = 50 + 5;
const ull seed = 131;
const int INF = 0x3f3f3f3f;
const int MOD = 1000000007;
struct Node
    int x, y, a, b;
p[maxn];
bool cmp(Node x, Node y)
    if(x.x == y.x) return x.y > y.y;
    return x.x < y.x;

vector<int> vv;
ll Max[maxn << 2], lazy[maxn << 2];
void pushup(int rt)
    Max[rt] = max(Max[rt << 1], Max[rt << 1 | 1]);

void pushdown(int rt)
    if(lazy[rt])
        lazy[rt << 1] += lazy[rt];
        lazy[rt << 1 | 1] += lazy[rt];
        Max[rt << 1] += lazy[rt];
        Max[rt << 1 | 1] += lazy[rt];
        lazy[rt] = 0;
    

void build(int l, int r, int rt)
    Max[rt] = lazy[rt] = 0;
    if(l == r) return;
    int m = (l + r) >> 1;
    build(l, m, rt << 1);
    build(m + 1, r, rt << 1 | 1);

void update(int L, int R, int l, int r, int v, int rt)
    if(L > R) return;
    if(L <= l && R >= r)
        Max[rt] += v;
        lazy[rt] += v;
        return;
    
    pushdown(rt);
    int m = (l + r) >> 1;
    if(L <= m)
        update(L, R, l, m, v, rt << 1);
    if(R > m)
        update(L, R, m + 1, r, v, rt << 1 | 1);
    pushup(rt);

void change(int pos , int l, int r, ll v, int rt)
    if(l == r)
        Max[rt] = max(Max[rt], v);
        return;
    
    pushdown(rt);
    int m = (l + r) >> 1;
    if(pos <= m)
        change(pos, l, m, v, rt << 1);
    else
        change(pos, m + 1, r, v, rt << 1 | 1);
    pushup(rt);

ll query(int L, int R, int l, int r, int rt)
    if(L <= l && R >= r)
        return Max[rt];
    
    pushdown(rt);
    int m = (l + r) >> 1;
    ll MAX = -1;
    if(L <= m)
        MAX = max(MAX, query(L, R, l, m, rt << 1));
    if(R > m)
        MAX = max(MAX, query(L, R, m + 1, r, rt << 1 | 1));
    pushup(rt);
    return MAX;

int main()
    int n;
    while(~scanf("%d", &n))
        vv.clear();
        for(int i = 1; i <= n; i++)
            scanf("%d%d%d%d", &p[i].x, &p[i].y, &p[i].a, &p[i].b);
            vv.push_back(p[i].y);
        
        vv.push_back(-1);
        sort(vv.begin(), vv.end());
        vv.erase(unique(vv.begin(), vv.end()), vv.end());
        for(int i = 1; i <= n; i++) p[i].y = lower_bound(vv.begin(), vv.end(), p[i].y) - vv.begin() + 1;

        sort(p + 1, p + n + 1, cmp);
        build(1, vv.size(), 1);
        for(int i = 1; i <= n; i++)
            ll tmp = query(1, p[i].y, 1, vv.size(), 1);
            update(p[i].y, vv.size(), 1, vv.size(), p[i].b, 1);
            update(1, p[i].y - 1, 1, vv.size(), p[i].a, 1);
            change(p[i].y, 1, vv.size(), tmp + p[i].b, 1);
        
        printf("%lld\n", Max[1]);
    
    return 0;