2019CCPC哈尔滨E. Exchanging Gifts

发表于 | 阅读数

给定$n$个格子

  • 对于前M1个约束，区间$[L,R]$内取的数量必须严格不少于K

  • 对于后M2个约束，区间$[L,R]$外取的数量必须严格不少于K

  • 满足所有M1+M2个约束的前提下使得取的个数最少，输出最少取的个数

    转化一下

然后答案满足单调性。
注意！！！

跑一遍差分约束即可，

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cnt[to] = cnt[x] + 1;
if(dis[to]<0)
    return false;

注意这两个剪枝点即可。

代码 
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#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
#define pii pair<int, int>
#define mk make_pair
const int N = 1e6 + 10;
const int mod = 1e9 + 7;
int read()
{
    int x = 0, f = 1;
    char c = getchar();
    while (c < '0' || c > '9')
    {
        if (c == '-')
            f = -1;
        c = getchar();
    }
    while (c >= '0' && c <= '9')
        x = (x << 1) + (x << 3) + c - '0', c = getchar();
    return x * f;
}
struct Node
{
    int l, r, w;
} q[N];
int n, m1, m2;
int vis[N], cnt[N], dis[N];
vector<pii> g[N];
bool check(int mid)
{
    for (int i = 0; i <= n; i++)
        g[i].clear(), vis[i] = 0, dis[i] = 1e9, cnt[i] = 0;
    for (int i = 1; i <= n; i++)
    {
        g[i - 1].push_back(mk(i, 1));
        g[i].push_back(mk(i - 1, 0));
    }
    for (int i = 1; i <= m1; i++)
    {
        g[q[i].r].push_back(mk(q[i].l, -q[i].w));
    }

    for (int i = m1 + 1; i <= m2 + m1; i++)
    {

        if (mid - q[i].w < 0)
            return false;
        g[q[i].l].push_back(mk(q[i].r, mid - q[i].w));
    }
    g[0].push_back({n, mid});
    g[n].push_back({0, -mid});
    queue<int> q;

    q.push(0);
    dis[0] = 0;
    vis[0] = 1;

    while (!q.empty())
    {
        int x = q.front();
        q.pop();
        vis[x] = 0;
        for (pii now : g[x])
        {
            int to = now.first;

            if (dis[to] > dis[x] + now.second)
            {
                dis[to] = dis[x] + now.second;

                if (!vis[to])
                {
                    cnt[to] = cnt[x] + 1;
                    if (cnt[to] > n)
                        return false;
                    q.push(to);
                    vis[to] = 1;
                }
            }
        }
    }

    return true;
}
int main()
{
    int T = read();
    while (T--)
    {
        n = read(), m1 = read(), m2 = read();
        for (int i = 1; i <= m1 + m2; i++)
            q[i].l = read() - 1, q[i].r = read(), q[i].w = read();
        int l = 0, r = n, ans = n;
        while (l <= r)
        {
            int mid = (l + r) >> 1;
            if (check(mid))
            {
                ans = mid;
                r = mid - 1;
            }
            else
            {
                l = mid + 1;
            }
        }

        printf("%d\n", ans);
    }
}