[BZOJ1061][NOI2008]志愿者招募 费用流

列出几个线性规划的式子
添加变量把不等式变成流量恒等式
按照式子建图，跑最小费用最大流
答案即为费用
https://www.byvoid.com/zhs/blog/noi-2008-employee

/**************************************************************    Problem: 1061    User: di4CoveRy    Language: C++    Result: Accepted    Time:2152 ms    Memory:5256 kb****************************************************************/#include <bits/stdc++.h>#define INF (1<<29)#define N 2050 #define M 200050using namespace std;typedef long long LL;int head[N],cnt=1,tot,S,T;int p[N],rp[N],L[N],a,b,c[N],d;LL dis[N];int n,m,k;LL ans;inline int rd() {    int x=0,f=1;char ch=getchar();    while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}    while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}    return x*f;}struct Edge{ int a,b,v,cost,next; }e[M],id;inline void add(int a,int b,int v,int cost) {    if (!a || !b) return ;     e[++cnt] = (Edge){ a,b,v,cost,head[a] }, head[a] = cnt;    e[++cnt] = (Edge){ b,a,0,-cost,head[b] },head[b] = cnt;}#define cp e[i].v  #define B e[i].b  bool SPFA() {      bool flag = false;    for (int i=1;i<=tot;i++) p[i] = 0, dis[i] = (1LL<<62);    dis[S] = 0;    queue<int> q; q.push(S);      while (!q.empty()) {          int u = q.front(); q.pop();          if (u == T) flag = true;        for (int i=head[u];i;i=e[i].next)              if (cp > 0 && dis[u] + e[i].cost < dis[B]) {                  dis[B] = dis[u] + e[i].cost;                  p[B] = i;                  q.push(B);              }       }      return flag;  }  void mcf() {      int g = p[T] , flow = INF;      while (g) {        flow = min(flow , e[g].v);          g = p[ e[g].a ];      }    g = p[T];      while (g) {          e[g  ].v -= flow;          e[g^1].v += flow;          ans += 1LL * e[g].cost * flow;          g = p[ e[g].a ];      }}int main() {    n = rd(), m = rd();    for (int _=1;_<=n;_++) p[_] = rd();    for (int _=1;_<=n;_++) rp[_] = p[_]-p[_-1];    S = ++tot, T = ++tot;    for (int _=0;_<=n+1;_++) L[_] = ++tot;    for (int _=1;_<=n;_++)        if (rp[_] > 0) add(S, L[_], rp[_], 0); else add(L[_], T, -rp[_], 0);    for (int _=1;_<=n;_++) add(L[_+1], L[_], INF, 0);    add(L[n+1], T, p[n], 0);    for (int _=1;_<=m;_++) {        int a = rd(), b = rd(), v = rd();        add(L[a], L[b+1], INF, v);    }    while (SPFA()) mcf();    cout << ans << endl;    return 0;}