思路
费用流水题
对每月拆点，入点向出点连cap=ui的边，s向入点连cost=di的边，i的入点向i+1的入点连cap=S的边即可
代码
#include <cstdio>
#include <algorithm>
#include <cstring>
#include <queue>
#include <vector>
using namespace std;
struct Edge{
int u,v,cap,flow,cost;
};
const int MAXN = 5001;
const int INF = 0x3f3f3f3f;
vector <Edge> edges;
vector <int> G[MAXN];
int n,m,s,t,S;
int a[MAXN],p[MAXN],d[MAXN];
bool vis[MAXN];
void addedge(int u,int v,int cap,int cost){
edges.push_back((Edge){u,v,cap,0,cost});
edges.push_back((Edge){v,u,0,0,-cost});
int cnt=edges.size();
G[u].push_back(cnt-2);
G[v].push_back(cnt-1);
}
bool spfa(int s,int t,int &cost,int &flow){
memset(d,0x3f,sizeof(d));
memset(vis,0,sizeof(vis));
d[s]=0;p[s]=0;a[s]=0x3f3f3f3f;
queue<int> q;
q.push(s);
vis[s]=true;
while(!q.empty()){
int x=q.front();q.pop();
vis[x]=false;
for(int i=0;i<G[x].size();i++){
Edge &e=edges[G[x][i]];
if(e.cap>e.flow&&d[e.v]>d[x]+e.cost){
d[e.v]=d[x]+e.cost;
p[e.v]=G[x][i];
a[e.v]=min(a[x],e.cap-e.flow);
if(!vis[e.v]){
q.push(e.v);
vis[e.v]=true;
}
}
}
}
if(d[t]==0x3f3f3f3f)
return false;
flow+=a[t];
cost+=d[t]*a[t];
for(int u=t;u!=s;u=edges[p[u]].u){
edges[p[u]].flow+=a[t];
edges[p[u]^1].flow-=a[t];
}
return true;
}
void MFMC(int s,int t,int &flow,int &cost){
flow=0;
cost = 0;
while(spfa(s,t,cost,flow));
return;
}
int main(){
s=MAXN-2;
t=MAXN-3;
scanf("%d %d %d",&n,&m,&S);
for(int i=1;i<=n;i++){
int mid;
scanf("%d",&mid);
addedge(i,i+n,mid,0);
addedge(i+n,t,INF,0);
}
for(int i=1;i<=n;i++){
int mid;
scanf("%d",&mid);
addedge(s,i,INF,mid);
if(i!=n)
addedge(i,i+1,S,m);
}
int flow,cost;
MFMC(s,t,flow,cost);
printf("%d\n",cost);
return 0;
}