Codeforces Round #608 (Div. 2) D. Portals

链接：

https://codeforces.com/contest/1271/problem/D

题意：

You play a strategic video game (yeah, we ran out of good problem legends). In this game you control a large army, and your goal is to conquer n castles of your opponent.

Let's describe the game process in detail. Initially you control an army of k warriors. Your enemy controls n castles; to conquer the i-th castle, you need at least ai warriors (you are so good at this game that you don't lose any warriors while taking over a castle, so your army stays the same after the fight). After you take control over a castle, you recruit new warriors into your army — formally, after you capture the i-th castle, bi warriors join your army. Furthermore, after capturing a castle (or later) you can defend it: if you leave at least one warrior in a castle, this castle is considered defended. Each castle has an importance parameter ci, and your total score is the sum of importance values over all defended castles. There are two ways to defend a castle:

if you are currently in the castle i, you may leave one warrior to defend castle i;

there are m one-way portals connecting the castles. Each portal is characterised by two numbers of castles u and v (for each portal holds u>v). A portal can be used as follows: if you are currently in the castle u, you may send one warrior to defend castle v.

Obviously, when you order your warrior to defend some castle, he leaves your army.

You capture the castles in fixed order: you have to capture the first one, then the second one, and so on. After you capture the castle i (but only before capturing castle i+1) you may recruit new warriors from castle i, leave a warrior to defend castle i, and use any number of portals leading from castle i to other castles having smaller numbers. As soon as you capture the next castle, these actions for castle i won't be available to you.

If, during some moment in the game, you don't have enough warriors to capture the next castle, you lose. Your goal is to maximize the sum of importance values over all defended castles (note that you may hire new warriors in the last castle, defend it and use portals leading from it even after you capture it — your score will be calculated afterwards).

Can you determine an optimal strategy of capturing and defending the castles?

思路：

计算到每个城堡最少多少人，计算出每个位置可以派出去多少人。

记录每个点最右边的坐标，因为左边的人可以带到右边去用，所以往右边放是较优的。

这样就可以用一个堆来维护。

代码：

#include<bits/stdc++.h>
using namespace std;

const int MAXN = 5e3+10;

struct Node
{
    int v, i;
    bool operator < (const Node & rhs) const
    {
        return this->v < rhs.v;
    }
};

priority_queue<Node> que;
int a[MAXN], b[MAXN], c[MAXN];
int need[MAXN], las[MAXN], use[MAXN];
int n, m, k;

int main()
{
    cin >> n >> m >> k;
    for (int i = 1;i <= n;i++)
        cin >> a[i] >> b[i] >> c[i], las[i] = i;
    int u, v;
    for (int i = 1;i <= m;i++)
    {
        cin >> u >> v;
        las[v] = max(las[v], u);
    }
    need[n] = a[n];
    for (int i = n-1;i >= 1;i--)
        need[i] = max(a[i], need[i+1]-b[i]);
    int sum = k;
    for (int i = 1;i <= n;i++)
    {
        if (sum < a[i])
        {
            puts("-1");
            return 0;
        }
        sum += b[i];
    }
    sum = k;
    for (int i = 1;i <= n;i++)
    {
        use[i] = sum+b[i]-need[i+1];
        sum = need[i+1];
        que.push(Node{c[i], i});
    }
    int ans = 0;
    while(!que.empty())
    {
        int val = que.top().v;
        int x = que.top().i;
        que.pop();
        int y = las[x];
        while(use[y] == 0 && y > 0)
            y--;
        if (y == 0)
            continue;
        use[y]--;
        ans += val;
    }
    cout << ans << endl;

    return 0;
}