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Luogu P3358 最长k可重区间集问题

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题目描述

https://www.luogu.com.cn/problem/P3358

简要题意：给定 $n$ 个开区间和 $k$，每个区间的价值为区间长度，要求选择尽量多的开区间，使得不存在任何一个点不覆盖了 $k$ 次以上，且选择的区间的价值最大

$n\le 500,k\le 3$

Solution

首先为了方便，将开区间变成闭区间

我们参照 志愿者招募 这题的建图方法，可以得到如下建图

建图：

$s$ 连 $1$，容量为 $k$，费用为 $0$

$i$ 连 $i+1$，$i\in[1,n-1]$，容量为 $k$，费用为 $0$

$n$ 连 $t$，容量为 $k$，费用为 $0$

对于一个区间 $l$ 连 $r+1$，容量为 $1$，费用为 $-(r-l+1)$

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#include <iostream>
#include <cstdio>
#include <algorithm>
#include <queue>
#define maxn 1010
#define INF 1000000000 
using namespace std;

int n, k;

struct Edge {
    int to, next, w, fi;
} e[1000000]; int c1, head[maxn];
inline void add_edge(int u, int v, int w, int fi) {
    e[c1].to = v; e[c1].w = w; e[c1].fi = fi;
    e[c1].next = head[u]; head[u] = c1++;
}

inline void Add_edge(int u, int v, int w, int fi) {
    add_edge(u, v, w, fi); add_edge(v, u, 0, -fi);
}

struct node {
    int x, y;
} a[maxn];

int b[maxn], cnt;
void init_hash() {
    for (int i = 1; i <= n; ++i) b[i] = a[i].x, b[i + n] = a[i].y;
    sort(b + 1, b + 2 * n + 1); cnt = unique(b + 1, b + 2 * n + 1) - b - 1; 
    for (int i = 1; i <= n; ++i) {
        a[i].x = lower_bound(b + 1, b + cnt + 1, a[i].x) - b;
        a[i].y = lower_bound(b + 1, b + cnt + 1, a[i].y) - b;
    }
}

int pre[maxn], la[maxn], dis[maxn], fl[maxn];
bool vis[maxn]; int s, t; 
bool spfa() {
    fill(dis, dis + maxn, INF); dis[s] = 0;
    fill(vis, vis + maxn, 0); vis[s] = 1;
    deque<int> Q; Q.push_front(s); pre[t] = -1; fl[s] = INF;
    while (!Q.empty()) {
        int u = Q.front(); Q.pop_front(); vis[u] = 0; 
        for (int i = head[u]; ~i; i = e[i].next) {
            int v = e[i].to, w = e[i].w, fi = e[i].fi;
            if (w > 0 && dis[v] > dis[u] + fi) {
                dis[v] = dis[u] + fi; pre[v] = u;
                fl[v] = min(fl[u], w); la[v] = i;
                if (vis[v]) continue; vis[v] = 1;
                if (Q.empty() || dis[v] <= dis[Q.front()]) Q.push_front(v);
                else Q.push_back(v); 
            }
        }
    }
    return ~pre[t]; 
}

int mc, mf; 
void mcmf() {
    while (spfa()) {
        int now = t; 
        mc += fl[t] * dis[t]; mf += fl[t];
        while (now != s) {
            e[la[now]].w -= fl[t];
            e[la[now] ^ 1].w += fl[t];
            now = pre[now];
        }
    }
} 

int main() { fill(head, head + maxn, -1); 
    ios::sync_with_stdio(false);
    cin.tie(nullptr); cout.tie(nullptr);

    cin >> n >> k; 
    for (int i = 1; i <= n; ++i) cin >> a[i].x >> a[i].y, --a[i].y;
    init_hash(); s = 0; t = cnt + 1;
    for (int i = 0; i <= cnt; ++i) Add_edge(i, i + 1, k, 0);
    for (int i = 1; i <= n; ++i) Add_edge(a[i].x, a[i].y + 1, 1, b[a[i].x] - b[a[i].y] - 1);
    mcmf(); cout << -mc << "\n"; 
    return 0;
}

这个题目还有另外一种建图方法

我们考虑将一系列不相交的区间称为一条路径

那么我们最多选择 $k$ 条路径

所以我们考虑这样建图：

$s$ 连 $s’$，容量为 $k$，费用为 $0$

$s’$ 连 $i$，容量为 $1$，费用为 $0$

$i$ 连 $i’$，容量为 $1$，费用为 $-(r[i]-l[i]+1)$

$i’$ 连 $j$，容量为 $1$，费用为 $0$，其中 $j$ 的左端点大于 $i$ 的右端点

$i’$ 连 $t$，容量为 $1$，费用为 $0$

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#include <iostream>
#include <cstdio>
#include <algorithm>
#include <queue>
#define cn const node
#define maxn 1010
#define INF 1000000000
using namespace std;

int n, k;

struct Edge {
    int to, next, w, fi;
} e[1000000]; int c1, head[maxn];
inline void add_edge(int u, int v, int w, int fi) {
    e[c1].to = v; e[c1].w = w; e[c1].fi = fi;
    e[c1].next = head[u]; head[u] = c1++;
}

inline void Add_edge(int u, int v, int w, int fi) {
    add_edge(u, v, w, fi); add_edge(v, u, 0, -fi);
}

struct node {
    int x, y;
} a[maxn];

int pre[maxn], la[maxn], dis[maxn], fl[maxn];
bool vis[maxn]; int s, t, _s; 
bool spfa() {
    fill(dis, dis + maxn, INF); dis[s] = 0;
    fill(vis, vis + maxn, 0); vis[s] = 1;
    deque<int> Q; Q.push_front(s); pre[t] = -1; fl[s] = INF;
    while (!Q.empty()) {
        int u = Q.front(); Q.pop_front(); vis[u] = 0; 
        for (int i = head[u]; ~i; i = e[i].next) {
            int v = e[i].to, w = e[i].w, fi = e[i].fi;
            if (w > 0 && dis[v] > dis[u] + fi) {
                dis[v] = dis[u] + fi; pre[v] = u;
                fl[v] = min(fl[u], w); la[v] = i;
                if (vis[v]) continue; vis[v] = 1;
                if (Q.empty() || dis[v] <= dis[Q.front()]) Q.push_front(v);
                else Q.push_back(v); 
            }
        }
    }
    return ~pre[t]; 
}

int mc, mf; 
void mcmf() {
    while (spfa()) {
        int now = t; 
        mc += fl[t] * dis[t]; mf += fl[t];
        while (now != s) {
            e[la[now]].w -= fl[t];
            e[la[now] ^ 1].w += fl[t];
            now = pre[now];
        }
    }
} 

int main() { fill(head, head + maxn, -1); 
    ios::sync_with_stdio(false);
    cin.tie(nullptr); cout.tie(nullptr);

    cin >> n >> k; s = 0; _s = 2 * n + 1; t = 2 * n + 2;
    for (int i = 1; i <= n; ++i) cin >> a[i].x >> a[i].y, --a[i].y;
    sort(a + 1, a + n + 1, [](cn &u, cn &v) { return u.x < v.x; });
    Add_edge(s, _s, k, 0); 
    for (int i = 1; i <= n; ++i) {
        Add_edge(_s, i, 1, 0);
        Add_edge(i, i + n, 1, a[i].x - a[i].y - 1);
        Add_edge(i + n, t, 1, 0);
        for (int j = i + 1; j <= n; ++j)
            if (a[j].x > a[i].y) Add_edge(i + n, j, 1, 0); 
    }
    mcmf(); cout << -mc << "\n"; 
    return 0;
}