luogu P3358 最长k可重区间集问题

网络流建图好难，这题居然是网络流(雾，一般分析来说，有限制的情况最大流情况可以拆点通过capacity来限制，比如只使用一次，把一个点拆成入点出点，capacity为1即可，这题是限制最大k重复，可以联想到最大流问题，设源点汇点，限制的k就是其最大的capacity，其最大流一定<=k，跑出来一定满足条件，但如何计算长度呢，就使用费用流吧，最大费用流就把边权取反即可，我们不知道输入的数据范围，只知道个数，就离散化一下，每个区间只能选一次，就对离散化后的l对r连一条capacity为1，权为-len的边，源点对1连一条capacity为k的边，n对汇点连一条capacity为k的边，因为判断是否重复是从头到尾，那么源点汇点就分别连接头尾，但是这样连图不一定保证每个点都考虑进去了，可能会出现图不连通的情况，题目是开区间，那么(i,i+1)和(i+1,i+2)一定是没有交点的，就对i与i+1连一条capacity为无穷，权为0的边，这样就能保证图连通且不会影响到答案，因为如果有断层，其会顺着往下搜索，权为0，capacity为无穷，对答案无影响

#include<bits/stdc++.h>
using namespace std;
#define lowbit(x) ((x)&(-x))
typedef long long LL;

const int maxm = 5e3+5;
const int INF = 0x3f3f3f3f;

struct edge{
    int u, v, cap, flow, cost, nex;
} edges[maxm];

struct Points{
    int l, r, len;
} point[505];

int head[maxm], cur[maxm], cnt, fa[1024], d[1024], n, allx[1024];
bool inq[1024];

void init() {
    memset(head, -1, sizeof(head));
}

void add(int u, int v, int cap, int cost) {
    edges[cnt] = edge{u, v, cap, 0, cost, head[u]};
    head[u] = cnt++;
}

void addedge(int u, int v, int cap, int cost) {
    add(u, v, cap, cost), add(v, u, 0, -cost);
}

bool spfa(int s, int t, int &flow, LL &cost) {
    for(int i = 0; i <= ((n<<1)|1); ++i) d[i] = INF; //init()
    memset(inq, false, sizeof(inq));
    d[s] = 0, inq[s] = true;
    fa[s] = -1, cur[s] = INF;
    queue<int> q;
    q.push(s);
    while(!q.empty()) {
        int u = q.front();
        q.pop();
        inq[u] = false;
        for(int i = head[u]; i != -1; i = edges[i].nex) {
            edge& now = edges[i];
            int v = now.v;
            if(now.cap > now.flow && d[v] > d[u] + now.cost) {
                d[v] = d[u] + now.cost;
                fa[v] = i;
                cur[v] = min(cur[u], now.cap - now.flow);
                if(!inq[v]) {q.push(v); inq[v] = true;}
            }
        }
    }
    if(d[t] == INF) return false;
    flow += cur[t];
    cost += 1LL*d[t]*cur[t];
    for(int u = t; u != s; u = edges[fa[u]].u) {
        edges[fa[u]].flow += cur[t];
        edges[fa[u]^1].flow -= cur[t];
    }
    return true;
}

int MincostMaxflow(int s, int t, LL &cost) {
    cost = 0;
    int flow = 0;
    while(spfa(s, t, flow, cost));
    return flow;
}

void run_case() {
    init();
    int l, r, k, xcnt = 0;
    cin >> n >> k;
    for(int i = 1; i <= n; ++i) {
        cin >> l >> r;
        allx[++xcnt] = l, allx[++xcnt] = r, point[i] = Points{l, r, r-l};
    }
    sort(allx+1,allx+1+xcnt);
    int len = unique(allx+1,allx+1+xcnt)-allx;
    for(int i = 1; i <= n; ++i) {
        point[i].l = lower_bound(allx+1,allx+len,point[i].l)-allx;
        point[i].r = lower_bound(allx+1,allx+len,point[i].r)-allx;
    }
    for(int i = 1; i < len-1; ++i)
        addedge(i, i+1, INF, 0);
    int s = 0, t = len;
    for(int i = 1; i <= n; ++i) {
        addedge(point[i].l, point[i].r, 1, -point[i].len);
    }
    addedge(s, 1, k, 0), addedge(len-1, t, k, 0);
    LL cost = 0;
    MincostMaxflow(s, t, cost);
    cout << -cost;
}

int main() {
    ios::sync_with_stdio(false), cin.tie(0);
    run_case();
    cout.flush();
    return 0;
}