【LOJ】#3097. 「SNOI2019」通信

LOJ#3097. 「SNOI2019」通信

费用流，有点玄妙

显然按照最小路径覆盖那题的建图思路，把一个点拆成两种点，一种是从这个点出去，标成(x_{i})，一种是输入到这个点，使得两条路径合成一条（或者是新建一条），标成(y_i)

源点向每个(x_i)流一条容量为1，费用为0的边

然后向每个(y_{i})流一条容量为1，费用为W的边

每个(y_i)向汇点连一条容量为1，费用为0的边

这个时候，如果你充满梦想，你可以把所有的(x_{i})到(y_j)((i < j))连一条(|a_{i} - a_{j}|)的边

然而你没有梦想，你觉得似乎不太优秀

然后你可以分治，把每一层的点分成左右两部分，然后把左边点权值离散化后建出一个前缀最大值节点，串成一条链，并在对应的位置连上左边的(a_{i})

另一种情况把右边的权值离散化（或者按相同的方式给左边连负数也行），建出前缀最大值节点，串成一条链，在对应位置连上右边的(a_{i})

这样边数就被优化成(nlog n)了

#include <bits/stdc++.h>
#define fi first
#define se second
#define pii pair<int, int>
#define mp make_pair
#define pb push_back
#define space putchar(' ')
#define enter putchar('
')
#define eps 1e-10
#define ba 47
#define MAXN 1005
//#define ivorysi
using namespace std;
typedef long long int64;
typedef unsigned int u32;
typedef double db;
template <class T>
void read(T &res) {
    res = 0;
    T f = 1;
    char c = getchar();
    while (c < '0' || c > '9') {
        if (c == '-')
            f = -1;
        c = getchar();
    }
    while (c >= '0' && c <= '9') {
        res = res * 10 + c - '0';
        c = getchar();
    }
    res *= f;
}
template <class T>
void out(T x) {
    if (x < 0) {
        x = -x;
        putchar('-');
    }
    if (x >= 10) {
        out(x / 10);
    }
    putchar('0' + x % 10);
}
struct node {
    int to, next, cap;
    int64 val;
} E[3000005];
int head[2000005], sumE = 1;
int x[1005], y[1005], N, a[1005], W, S, T, Ncnt = 0;
void add(int u, int v, int c, int64 a) {
    E[++sumE].to = v;
    E[sumE].next = head[u];
    E[sumE].cap = c;
    E[sumE].val = a;
    head[u] = sumE;
}
void addtwo(int u, int v, int c, int64 a) {
    add(u, v, c, a);
    add(v, u, 0, -a);
}
int val[1005], tot, pos[1005];
void build(int l, int r) {
    if (l == r)
        return;
    if (r - l + 1 == 2) {
        addtwo(x[l], y[r], 1, abs(a[l] - a[r]));
        return;
    }
    int mid = (l + r) >> 1;
    build(l, mid);
    build(mid + 1, r);
    tot = 0;
    for (int i = l; i <= mid; ++i) {
        val[++tot] = a[i];
    }
    sort(val + 1, val + tot + 1);
    tot = unique(val + 1, val + tot + 1) - val - 1;
    for (int i = tot; i >= 1; --i) {
        pos[i] = ++Ncnt;
        if (i != tot)
            addtwo(pos[i + 1], pos[i], 1e9, 0);
    }
    for (int i = l; i <= mid; ++i) {
        int t = lower_bound(val + 1, val + tot + 1, a[i]) - val;
        addtwo(x[i], pos[t], 1, a[i]);
    }
    for (int i = mid + 1; i <= r; ++i) {
        int t = lower_bound(val + 1, val + tot + 1, a[i]) - val;
        if (t <= tot)
            addtwo(pos[t], y[i], 1, -a[i]);
    }
    tot = 0;
    for (int i = mid + 1; i <= r; ++i) val[++tot] = a[i];
    sort(val + 1, val + tot + 1);
    tot = unique(val + 1, val + tot + 1) - val - 1;
    for (int i = 1; i <= tot; ++i) {
        pos[i] = ++Ncnt;
        if (i != 1)
            addtwo(pos[i - 1], pos[i], 1e9, 0);
    }
    for (int i = mid + 1; i <= r; ++i) {
        int t = lower_bound(val + 1, val + tot + 1, a[i]) - val;
        addtwo(pos[t], y[i], 1, a[i]);
    }
    for (int i = l; i <= mid; ++i) {
        int t = lower_bound(val + 1, val + tot + 1, a[i]) - val;
        if (t <= tot)
            addtwo(x[i], pos[t], 1, -a[i]);
    }
}
int64 dis[100005];
int preE[100005];
bool inq[100005];
queue<int> Q;
bool SPFA() {
    for (int i = 1; i <= Ncnt; ++i) dis[i] = 1e18;
    memset(inq, 0, sizeof(inq));
    inq[S] = 1;
    dis[S] = 0;
    Q.push(S);
    while (!Q.empty()) {
        int u = Q.front();
        Q.pop();
        inq[u] = 0;
        for (int i = head[u]; i; i = E[i].next) {
            if (E[i].cap > 0) {
                int v = E[i].to;
                if (dis[v] > dis[u] + E[i].val) {
                    dis[v] = dis[u] + E[i].val;
                    preE[v] = i;
                    if (!inq[v]) {
                        inq[v] = 1;
                        Q.push(v);
                    }
                }
            }
        }
    }
    return dis[T] != 1e18;
}
void Init() {
    read(N);
    read(W);
    S = 2 * N + 1, T = 2 * N + 2;
    for (int i = 1; i <= N; ++i) {
        read(a[i]);
        x[i] = ++Ncnt;
        y[i] = ++Ncnt;
        addtwo(S, x[i], 1, 0);
        addtwo(S, y[i], 1, W);
        addtwo(y[i], T, 1, 0);
    }
    Ncnt += 2;
    build(1, N);
}
void Solve() {
    Init();
    int64 ans = 0;
    while (SPFA()) {
        ans += dis[T];
        int p = T;
        while (p != S) {
            int t = preE[p];
            E[t].cap -= 1;
            E[t ^ 1].cap += 1;
            p = E[t ^ 1].to;
        }
    }
    out(ans);
    enter;
}
int main() {
#ifdef ivorysi
    freopen("f1.in", "r", stdin);
#endif
    Solve();
}