CF1321E - World of Darkraft: Battle for Azathoth

攻击武器和怪兽是单调的，只用关心防御武器够打的怪兽，是一个前缀和。

我用线段树上的叶子值表示前缀和，这样每次插入一个怪兽，需要进行区间修改操作。

 

/*
* @Author: chenkexing
* @Date:   2020-03-06 11:53:04
* @Last Modified by:   chenkexing
* @Last Modified time: 2020-03-06 14:12:58
*/
#include <algorithm>
#include <iterator>
#include <iostream>
#include <cstring>
#include <cstdlib>
#include <iomanip>
#include <bitset>
#include <cctype>
#include <cstdio>
#include <string>
#include <vector>
#include <stack>
#include <cmath>
#include <queue>
#include <list>
#include <map>
#include <set>
#include <cassert>
//#include <unordered_set>
//#include <unordered_map>
// #include<bits/extc++.h>
// using namespace __gnu_pbds;
using namespace std;
#define pb push_back
#define fi first
#define se second
#define debug(x) cerr << #x << " := " << x << endl;
#define bug cerr << "-----------------------" << endl;
#define FOR(a, b, c) for (int a = b; a <= c; ++a)

typedef long long ll;
typedef long double ld;
typedef pair<int, int> pii;
typedef pair<ll, ll> pll;

const int inf = 0x3f3f3f3f;
const ll inff = 0x3f3f3f3f3f3f3f3f;
const int mod = 1e9 + 7;

template <typename T>
inline T read(T &x) {
    x = 0;
    int f = 0;
    char ch = getchar();
    while (ch < '0' || ch > '9') f |= (ch == '-'), ch = getchar();
    while (ch >= '0' && ch <= '9') x = x * 10 + ch - '0', ch = getchar();
    return x = f ? -x : x;
}

/**********showtime************/
const int maxn = 2e5 + 9;
const int maxm = 1e6 + 9;
pii a[maxn], b[maxn];
bool cmp1(pii a, pii b) {
    if (a.fi == b.fi) {
        return a.se > b.se;
    }
    return a.fi < b.fi;
}
struct Mon {
    int x, y;
    int val;
} mon[maxn];

bool cmp2(Mon a, Mon b) {
    return a.x < b.x;
}

ll tree[maxm];
ll mx[maxm << 2], lazy[maxm << 2];
void build(int le, int ri, int rt) {
    lazy[rt] = 0;

    if (le == ri) {
        mx[rt] = -tree[le];
        return;
    }
    int mid = (le + ri) >> 1;
    build(le, mid, rt << 1);
    build(mid + 1, ri, rt << 1 | 1);
    mx[rt] = max(mx[rt << 1], mx[rt << 1 | 1]);
}
void pushdown(int rt) {
    lazy[rt << 1] += lazy[rt];
    lazy[rt << 1 | 1] += lazy[rt];
    mx[rt << 1] += lazy[rt];
    mx[rt << 1 | 1] += lazy[rt];
    lazy[rt] = 0;
}
void update(int L, int R, int val, int le, int ri, int rt) {
    if (le >= L && ri <= R) {
        lazy[rt] += val;
        mx[rt] += val;
        return;
    }
    if (lazy[rt]) pushdown(rt);
    int mid = (le + ri) >> 1;
    if (mid >= L) update(L, R, val, le, mid, rt << 1);
    if (mid < R) update(L, R, val, mid + 1, ri, rt << 1 | 1);
    mx[rt] = max(mx[rt << 1], mx[rt << 1 | 1]);
}

int main() {
    memset(tree, inff, sizeof(tree));
    int n, m, k;
    scanf("%d%d%d", &n, &m, &k);
    for (int i = 1; i <= n; i++) {
        scanf("%d%d", &a[i].fi, &a[i].se);
    }

    for (int i = 1; i <= m; i++) {
        scanf("%d%d", &b[i].fi, &b[i].se);
        tree[b[i].fi] = min(tree[b[i].fi], 1ll * b[i].se);
    }

    for (int i = 1; i <= k; i++) {
        scanf("%d%d%d", &mon[i].x, &mon[i].y, &mon[i].val);
    }
    int N = 1e6;
    build(1, N, 1);

    sort(a + 1, a + 1 + n, cmp1);

    sort(mon + 1, mon + 1 + k, cmp2);

    ll ans = -inff;
    int j = 1;
    for (int i = 1; i <= n; i++) {
        while (j <= k && mon[j].x < a[i].fi) {
            if (mon[j].y + 1 <= N)
                update(mon[j].y + 1, N, mon[j].val, 1, N, 1);
            j++;
        }
        ans = max(ans, mx[1] - a[i].se);
        // cout<<i << " :: " << mx[1]<<endl;
    }

    printf("%lld\n", ans);
    return 0;
}