POJ 3680 最小费用最大流

建图如下

若限制为k

首先将坐标离散化。

对每个区间，左端点对应的编号到右端点对应的编号建一条边，容量为k，费用为负权

然后对每个编号i,建立i到i+1的边，容量为k，费用为0

起点和编号1连边，容量为k，费用0

最后一个编号和终点连边,容量为k，费用0

然后跑费用流即可

#include <iostream>
#include <algorithm>
#include <cstring>
#include <string>
#include <cstdio>
#include <cmath>
#include <queue>
#include <map>
#include <set>
#define eps 1e-5
#define MAXN 555
#define MAXM 55555
#define INF 100000007
using namespace std;
struct EDGE
{
    int v, cap, cost, next, re;    //  re记录逆边的下标。
} edge[MAXM];
int n, m, ans, flow, src, des;
int e, head[MAXN];
int que[MAXN], pre[MAXN], dis[MAXN];
bool vis[MAXN];
void init()
{
    e = ans = flow = 0;
    memset(head, -1, sizeof(head));
}
void add(int u, int v, int cap, int cost)
{
    edge[e].v = v;
    edge[e].cap = cap;
    edge[e].cost = cost;
    edge[e].next = head[u];
    edge[e].re = e + 1;
    head[u] = e++;
    edge[e].v = u;
    edge[e].cap = 0;
    edge[e].cost = -cost;
    edge[e].next = head[v];
    edge[e].re = e - 1;
    head[v] = e++;
}
bool spfa()
{
    int i, h = 0, t = 1;
    for(i = 0; i <= n; i ++)
    {
        dis[i] = INF;
        vis[i] = false;
    }
    dis[src] = 0;
    que[0] = src;
    vis[src] = true;
    while(t != h)
    {
        int u = que[h++];
        h %= n;
        vis[u] = false;
        for(i = head[u]; i != -1; i = edge[i].next)
        {
            int v = edge[i].v;
            if(edge[i].cap && dis[v] > dis[u] + edge[i].cost)
            {
                dis[v] = dis[u] + edge[i].cost;
                pre[v] = i;
                if(!vis[v])
                {
                    vis[v] = true;
                    que[t++] = v;
                    t %= n;
                }
            }
        }
    }
    if(dis[des] == INF) return false;
    return true;
}
void end()
{
    int u, p, mi = INF;
    for(u = des; u != src; u = edge[edge[p].re].v)
    {
        p = pre[u];
        mi = min(mi, edge[p].cap);
    }
    for(u = des; u != src; u = edge[edge[p].re].v)
    {
        p = pre[u];
        edge[p].cap -= mi;
        edge[edge[p].re].cap += mi;
        ans += mi * edge[p].cost;     //  cost记录的为单位流量费用，必须得乘以流量。
    }
    flow += mi;
}
int nt, k;
int a[MAXN], b[MAXN], w[MAXN];
int x[MAXN];
int bin(int low, int high, int v)
{
    while(low <= high)
    {
        int mid = (low + high) >> 1;
        if(x[mid] == v) return mid;
        else if(x[mid] > v) high = mid - 1;
        else low = mid + 1;
    }
    return -1;
}
void build()
{
    scanf("%d%d", &nt, &k);
    int cnt = 0;
    for(int i = 1; i <= nt; i++)
    {
        scanf("%d%d%d", &a[i], &b[i], &w[i]);
        x[cnt++] = a[i];
        x[cnt++] = b[i];
    }
    sort(x, x + cnt);
    cnt = unique(x, x + cnt) - x;
    for(int i = 1; i <= nt; i++)
    {
        int lx = bin(0, cnt - 1, a[i]) + 1;
        int rx = bin(0, cnt - 1, b[i]) + 1;
        add(lx, rx, 1, -w[i]);
    }
    for(int i = 1; i < cnt; i++)
        add(i, i + 1, k, 0);
    src = cnt + 1;
    des = cnt + 2;
    n = des;
    add(src, 1, k, 0);
    add(cnt, des, k, 0);
}
void MCMF()
{
    init();
    build();
    while(spfa()) end();
}
int main()
{
    int T;
    scanf("%d", &T);
    while(T--)
    {
        MCMF();
        printf("%d\n", -ans);
    }
    return 0;
}

zkw版   貌似比原作者写的指针版本的慢了不少，不过这种图zkw费用流本来就不擅长

#include <iostream>
#include <algorithm>
#include <cstring>
#include <string>
#include <cstdio>
#include <cmath>
#include <queue>
#include <deque>
#include <map>
#include <set>
#define eps 1e-5
#define MAXN 5555
#define MAXM 55555
#define INF 100000007
using namespace std;
struct EDGE
{
    int cost, cap, v;
    int next, re;
}edge[MAXM];
int head[MAXN], e;
int vis[MAXN], d[MAXN];
int ans, cost, src, des, n;
void init()
{
    memset(head, -1, sizeof(head));
    e = 0;
    ans = cost = 0;
}
void add(int u, int v, int cap, int cost)
{
    edge[e].v = v;
    edge[e].cap = cap;
    edge[e].cost = cost;
    edge[e].re = e + 1;
    edge[e].next = head[u];
    head[u] = e++;
    edge[e].v = u;
    edge[e].cap = 0;
    edge[e].cost = -cost;
    edge[e].re = e - 1;
    edge[e].next = head[v];
    head[v] = e++;
}
int aug(int u, int f)
{
    if(u == des)
    {
        ans += cost * f;
        return f;
    }
    vis[u] = 1;
    int tmp = f;
    for(int i = head[u]; i != -1; i = edge[i].next)
        if(edge[i].cap && !edge[i].cost && !vis[edge[i].v])
        {
            int delta = aug(edge[i].v, tmp < edge[i].cap ? tmp : edge[i].cap);
            edge[i].cap -= delta;
            edge[edge[i].re].cap += delta;
            tmp -= delta;
            if(!tmp) return f;
        }
    return f - tmp;
}
bool modlabel()
{
    for(int i = 0; i <= n; i++) d[i] = INF;
    d[des] = 0;
    deque<int>Q;
    Q.push_back(des);
    while(!Q.empty())
    {
        int u = Q.front(), tmp;
        Q.pop_front();
        for(int i = head[u]; i != -1; i = edge[i].next)
            if(edge[edge[i].re].cap && (tmp = d[u] - edge[i].cost) < d[edge[i].v])
                (d[edge[i].v] = tmp) <= d[Q.empty() ? src : Q.front()] ? Q.push_front(edge[i].v) : Q.push_back(edge[i].v);
    }
    for(int u = 1; u <= n; u++)
        for(int i = head[u]; i != -1; i = edge[i].next)
            edge[i].cost += d[edge[i].v] - d[u];
    cost += d[src];
    return d[src] < INF;
}
void costflow()
{
    while(modlabel())
    {
        do
        {
            memset(vis, 0, sizeof(vis));
        }while(aug(src, INF));
    }
}
int nt, k;
int a[MAXN], b[MAXN], w[MAXN];
int x[MAXN];
int bin(int low, int high, int v)
{
    while(low <= high)
    {
        int mid = (low + high) >> 1;
        if(x[mid] == v) return mid;
        else if(x[mid] > v) high = mid - 1;
        else low = mid + 1;
    }
    return -1;
}
void build()
{
    scanf("%d%d", &nt, &k);
    int cnt = 0;
    for(int i = 1; i <= nt; i++)
    {
        scanf("%d%d%d", &a[i], &b[i], &w[i]);
        x[cnt++] = a[i];
        x[cnt++] = b[i];
    }
    sort(x, x + cnt);
    cnt = unique(x, x + cnt) - x;
    for(int i = 1; i <= nt; i++)
    {
        int lx = bin(0, cnt - 1, a[i]) + 1;
        int rx = bin(0, cnt - 1, b[i]) + 1;
        add(lx, rx, 1, -w[i]);
    }
    for(int i = 1; i < cnt; i++)
        add(i, i + 1, k, 0);
    src = cnt + 1;
    des = cnt + 2;
    n = des;
    add(src, 1, k, 0);
    add(cnt, des, k, 0);
}
int main()
{
    int T;
    scanf("%d", &T);
    while(T--)
    {
        init();
        build();
        costflow();
        printf("%d\n", -ans);
    }
    return 0;
}