[luogu3980]志愿者招募

记$x_{i}$为第$i$类志愿者数量$,y_{j}=\sum_{j\in [s_{i},t_{i}]}x_{i}-a_{j}$​，则问题即
$$
\forall i\in [1,m],x_{i}\ge 0\\
\forall j\in [1,n],y_{j}\ge 0\\
y_{1}-\sum_{s_{i}=1}x_{i}=-a_{1}\\\sum_{t_{i}=n}x_{i}-y_{n}=a_{n}\\
\forall j\in [2,n],y_{j}+\sum_{t_{i}=j-1}x_{j}-y_{j-1}-\sum_{s_{i}=j}x_{i}=a_{j-1}-a_{j}\\
\min \sum_{i=1}^{m}c_{i}x_{i}
$$
显然可以转化为费用流，具体方式参考[loj6079]养猫

时间复杂度为$o({\rm MCMF}(n,m))$，可以通过

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
namespace MCMF{
    const int N=1005,M=20005;
    int S,T,E,head[N],vis[N],from[N];
    ll ans1,ans2,d[N];queue<int>q;
    struct edge{
        int nex,to;ll len,cost;
    }e[M<<1];
    void init(){
        E=0;
        memset(head,-1,sizeof(head));
    }
    void add(int x,int y,ll z,ll w){
        e[E]=edge{head[x],y,z,w},head[x]=E++;
        e[E]=edge{head[y],x,0,-w},head[y]=E++;
    }
    bool spfa(){
        memset(d,0x3f,sizeof(d));
        d[S]=0,q.push(S);
        while (!q.empty()){
            int k=q.front();q.pop();
            for(int i=head[k];i!=-1;i=e[i].nex){
                int u=e[i].to;
                if ((e[i].len)&&(d[u]>d[k]+e[i].cost)){
                    d[u]=d[k]+e[i].cost,from[u]=i;
                    if (!vis[u])q.push(u),vis[u]=1;
                }
            }
            vis[k]=0;
        }
        return d[T]<=1e18;
    }
    pair<ll,ll> query(int s,int t){
        S=s,T=t,ans1=ans2=0;
        while (spfa()){
            ll s=1e18;
            for(int i=T;i!=S;i=e[from[i]^1].to)s=min(s,e[from[i]].len);
            ans1+=s,ans2+=s*d[T];
            for(int i=T;i!=S;i=e[from[i]^1].to){
                e[from[i]].len-=s;
                e[from[i]^1].len+=s;
            }
        }
        return make_pair(ans1,ans2);
    }
};
const int N=1005;
int n,m,s,t,c,a[N];
int main(){
    using namespace MCMF;
    init();
    scanf("%d%d",&n,&m);
    for(int i=1;i<=n;i++){
        scanf("%d",&a[i]);
        add(i,i+1,1e18,0);
    }
    for(int i=1;i<=m;i++){
        scanf("%d%d%d",&s,&t,&c);
        add(t+1,s,1e18,c);
    }
    add(1,n+2,a[1],0),add(0,n+1,a[n],0);
    for(int i=2;i<=n;i++){
        if (a[i-1]-a[i]>0)add(0,i,a[i-1]-a[i],0);
        else add(i,n+2,a[i]-a[i-1],0);
    }
    printf("%lld\n",query(0,n+2).second);
    return 0;
}