[POI2006] SZK-Schools - 费用流

差不多就是个二分图带权匹配？（我还是敲费用流吧）

每个点向着自己能到的学校连边，费用按题意设定

跑最小费用最大流即可

#include <bits/stdc++.h>
using namespace std;

// Init: init()
// Input: make(u,v,cap,cost)
// Solver: solve(s,t)
// Output: ans, cost
namespace flow {
const int N = 100005;
const int M = 1000005;
const int inf = 1e+9;
struct Edge {
    int p, c, w, nxt = -1;
} e[N];
int s, t, tans, ans, cost, ind, bus[N], qhead = 0, qtail = -1, qu[M],vis[N], dist[N];

void graph_link(int p, int q, int c, int w) {
    e[ind].p = q;
    e[ind].c = c;
    e[ind].w = w;
    e[ind].nxt = bus[p];
    bus[p] = ind;
    ++ind;
}
void make(int p, int q, int c, int w) {
    graph_link(p, q, c, w);
    graph_link(q, p, 0, -w);
}
int dinic_spfa() {
    qhead = 0;
    qtail = -1;
    memset(vis, 0x00, sizeof vis);
    memset(dist, 0x3f, sizeof dist);
    vis[s] = 1;
    dist[s] = 0;
    qu[++qtail] = s;
    while (qtail >= qhead) {
        int p = qu[qhead++];
        vis[p] = 0;
        for (int i = bus[p]; i != -1; i = e[i].nxt)
            if (dist[e[i].p] > dist[p] + e[i].w && e[i].c > 0) {
                dist[e[i].p] = dist[p] + e[i].w;
                if (vis[e[i].p] == 0)
                    vis[e[i].p] = 1, qu[++qtail] = e[i].p;
            }
    }
    return dist[t] < inf;
}
int dinic_dfs(int p, int lim) {
    if (p == t)
        return lim;
    vis[p] = 1;
    int ret = 0;
    for (int i = bus[p]; i != -1; i = e[i].nxt) {
        int q = e[i].p;
        if (e[i].c > 0 && dist[q] == dist[p] + e[i].w && vis[q] == 0) {
            int res = dinic_dfs(q, min(lim, e[i].c));
            cost += res * e[i].w;
            e[i].c -= res;
            e[i ^ 1].c += res;
            ret += res;
            lim -= res;
            if (lim == 0)
                break;
        }
    }
    return ret;
}
void solve(int _s,int _t) {
    s=_s; t=_t;
    while (dinic_spfa()) {
        memset(vis, 0x00, sizeof vis);
        ans += dinic_dfs(s, inf);
    }
}
void init() {
    memset(bus, 0xff, sizeof bus);
}
}

const int N = 505;
int n,m[N],a[N],b[N],k[N];

int id_school(int i) {
    return i+2;
}

int id_num(int i) {
    return i+n+2;
}

void make(int u,int v,int w,int c) {
    flow::make(u,v,w,c);
}

signed main() {
    cin>>n;
    flow::init();
    for(int i=1;i<=n;i++) cin>>m[i]>>a[i]>>b[i]>>k[i];
    for(int i=1;i<=n;i++) {
        make(1,id_school(i),1,0);
        make(id_num(i),2,1,0);
        for(int j=a[i];j<=b[i];j++) {
            make(id_school(i),id_num(j),1,k[i]*abs(m[i]-j));
        }
    }
    flow::solve(1,2);
    if(flow::ans!=n) puts("NIE");
    else cout<<flow::cost;
}

来源：https://www.cnblogs.com/mollnn/p/12293829.html