最长k可重线段集问题

时间：2023-03-08 21:25:47

和那道可重区间集一样
不过这道题可能有垂直于x轴的线段，这就很烦了，直接连会有负环，判掉又会WA
可以想办法把r端点和l端点分开，又要保证答案不变
那么直接把区间l，r都乘以2，l=r时r++，否则l++，这样r就与l分开，并且对其它没有影响（相当于在x轴上多加了点）

这道题在LOJ上可以切

如果看到了的并且有数据可以卡掉我的代码的请在下面评论

# include <bits/stdc++.h>

# define RG register

# define IL inline

# define Fill(a, b) memset(a, b, sizeof(a))

# define Copy(a, b) memcpy(a, b, sizeof(a))

# define Sqr(a) (1LL * (a) * (a))

using namespace std;

typedef long long ll;

# define int ll

const int _(1010), __(1e6 + 10), INF(2e9);

IL ll Read(){

    char c = '%'; ll x = 0, z = 1;

    for(; c > '9' || c < '0'; c = getchar()) if(c == '-') z = -1;

    for(; c >= '0' && c <= '9'; c = getchar()) x = x * 10 + c - '0';

    return x * z;

}

int n, k, l[_], y_1[_], r[_], y_2[_], _w[_], o[_], len;

int cnt, fst[_], w[__], to[__], nxt[__], dis[_], vis[_], S, T, cost[__], pe[_], pv[_], max_flow, max_cost;

queue <int> Q;

IL void Add(RG int u, RG int v, RG int f, RG int co){

    cost[cnt] = co; w[cnt] = f; to[cnt] = v; nxt[cnt] = fst[u]; fst[u] = cnt++;

    cost[cnt] = -co; w[cnt] = 0; to[cnt] = u; nxt[cnt] = fst[v]; fst[v] = cnt++;

}

IL bool Bfs(){

    Q.push(S); Fill(dis, 127); dis[S] = 0; vis[S] = 1;

    while(!Q.empty()){

        RG int u = Q.front(); Q.pop();

        for(RG int e = fst[u]; e != -1; e = nxt[e]){

            if(!w[e] || dis[to[e]] <= dis[u] + cost[e]) continue;

            dis[to[e]] = dis[u] + cost[e];

            pe[to[e]] = e; pv[to[e]] = u;

            if(!vis[to[e]]) vis[to[e]] = 1, Q.push(to[e]);

        }

        vis[u] = 0;

    }

    if(dis[T] >= dis[T + 1]) return 0;

    RG int ret = INF;

    for(RG int u = T; u != S; u = pv[u]) ret = min(ret, w[pe[u]]);

    for(RG int u = T; u != S; u = pv[u]) w[pe[u]] -= ret, w[pe[u] ^ 1] += ret;

    max_cost -= ret * dis[T]; max_flow += ret;

    return 1;

}

# undef int

int main(RG int argc, RG char *argv[]){

# define int ll

    Fill(fst, -1); n = Read(); k = Read();

    for(RG int i = 1; i <= n; ++i){

        l[i] = Read(); y_1[i] = Read(); r[i] = Read(); y_2[i] = Read();

        _w[i] = sqrt(Sqr(l[i] - r[i]) + Sqr(y_1[i] - y_2[i]));

        if(l[i] > r[i]) swap(l[i], r[i]), swap(y_1[i], y_2[i]);

        l[i] <<= 1; r[i] <<= 1;

        if(l[i] == r[i]) ++r[i];

        else ++l[i];

    }

    for(RG int i = 1; i <= n; ++i) o[++len] = l[i], o[++len] = r[i];

    sort(o + 1, o + len + 1); len = unique(o + 1, o + len + 1) - o - 1;

    T = len + 1;

    for(RG int i = 0; i <= len; ++i) Add(i, i + 1, k, 0);

    for(RG int i = 1; i <= n; ++i){

        l[i] = lower_bound(o + 1, o + len + 1, l[i]) - o;

        r[i] = lower_bound(o + 1, o + len + 1, r[i]) - o;

        Add(l[i], r[i], 1, -_w[i]);

    }

    while(Bfs()); printf("%lld\n", max_cost);

    return 0;

}

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