[线性规划 费用流]Codeforces Gym101190D.Delight for a cat

先写出线性规划标准型，然后差分建网络流

具体可以看这个题解

#include <cstdio>#include <iostream>#include <algorithm>#include <queue>using namespace std;typedef long long ll;const int N=1010;const ll inf=1LL<<60;int n,k,ms,me,cnt=1,a[N],b[N],G[N];ll ans;struct edge{    int s,t,nx,w,f;}E[N*10],*p[N];inline char nc(){    static char buf[100000],*p1=buf,*p2=buf;    return p1==p2&&(p2=(p1=buf)+fread(buf,1,100000,stdin),p1==p2)?EOF:*p1++;}inline void rea(int &x){    char c=nc(); x=0;    for(;c>'9'||c<'0';c=nc());for(;c>='0'&&c<='9';x=x*10+c-'0',c=nc());}inline void add(int x,int y,int f,int w){    E[++cnt].t=y; E[cnt].nx=G[x]; E[cnt].f=f; E[cnt].s=x; E[cnt].w=w; G[x]=cnt;    E[++cnt].t=x; E[cnt].nx=G[y]; E[cnt].f=0; E[cnt].s=y; E[cnt].w=-w; G[y]=cnt;}queue<int> Q;ll dis[N];int vis[N],pre[N];int S,T;inline bool spfa(){    for(int i=0;i<=T;i++) dis[i]=inf;    dis[S]=0; vis[S]=1; Q.push(S);    while(!Q.empty()){        int x=Q.front(); Q.pop(); vis[x]=0;        for(int i=G[x];i;i=E[i].nx)            if(E[i].f && dis[E[i].t]>dis[x]+E[i].w){                dis[E[i].t]=dis[x]+E[i].w; pre[E[i].t]=i;                if(!vis[E[i].t]) vis[E[i].t]=1,Q.push(E[i].t);            }    }    if(dis[T]==inf) return false;    int mc=1<<30;    for(int i=T;i;i=E[pre[i]].s) mc=min(mc,E[pre[i]].f);    ans-=1LL*mc*dis[T];    for(int i=T;i;i=E[pre[i]].s) E[pre[i]].f-=mc,E[pre[i]^1].f+=mc;    return true;}int main(){    freopen("delight.in","r",stdin);    freopen("delight.out","w",stdout);    rea(n); rea(k); rea(ms); rea(me);    for(int i=1;i<=n;i++) rea(a[i]);    for(int i=1;i<=n;i++) rea(b[i]),ans+=b[i];    S=0; T=n-k+3;    add(S,n-k+1,k-me,0); add(n-k+2,T,k-me,0);     for(int i=1;i<=n;i++){        int x=i-k,y=i;        if(x<1) x=n-k+1; if(y>n-k) y=n-k+2;        add(x,y,1,b[i]-a[i]); p[i]=E+cnt-1;    }    add(n-k+1,1,k-ms-me,0); add(n-k,n-k+2,k-ms-me,0);    for(int i=1;i<=n-k-1;i++) add(i,i+1,k-ms-me,0);    while(spfa());    printf("%lld\n",ans);    ll cur=0;    for(int i=1;i<=n;i++){        //if(p[i]->f) cur+=b[i]; else cur+=a[i];        putchar(p[i]->f?'E':'S');    }    //printf("\n%lld\n",cur);    return 0;}