[BZOJ2245][SDOI2011]工作安排（费用流）

题目描述

题解

代码

#include<iostream>
#include<cstring>
#include<cstdio>
#include<queue>
using namespace std;
#define LL long long

const int max_n=300;
const int max_N=max_n*10+2;
const int max_m=max_N*100;
const int max_e=max_m*2;
const LL INF=1e18;

LL m,n,sum,N,mincost;
LL A[max_n][max_n],s[max_n],ss[max_n],C[max_n],T[max_n][max_n],W[max_n][max_n];
LL tot,point[max_N],nxt[max_e],v[max_e],remain[max_e],c[max_e];
LL dis[max_N],last[max_N]; bool vis[max_N];
queue <LL> q;

inline void addedge(LL x,LL y,LL cap,LL z)
{
++tot; nxt[tot]=point[x]; point[x]=tot; v[tot]=y; remain[tot]=cap; c[tot]=z;
++tot; nxt[tot]=point[y]; point[y]=tot; v[tot]=x; remain[tot]=0; c[tot]=-z;
}
inline LL addflow(LL s,LL t)
{
LL now=t,ans=INF;
while (now!=s)
{
ans=min(ans,remain[last[now]]);
now=v[last[now]^1];
}
now=t;
while (now!=s)
{
remain[last[now]]-=ans;
remain[last[now]^1]+=ans;
now=v[last[now]^1];
}
return ans;
}
inline bool bfs(LL s,LL t)
{
for (LL i=1;i<=N;++i) dis[i]=INF; dis[s]=0;
memset(vis,0,sizeof(vis)); vis[s]=true;
while (!q.empty()) q.pop(); q.push(s);

while (!q.empty())
{
LL now=q.front(); q.pop();
vis[now]=false;
for (LL i=point[now];i!=-1;i=nxt[i])
if (dis[v[i]]>dis[now]+c[i]&&remain[i])
{
dis[v[i]]=dis[now]+c[i];
last[v[i]]=i;
if (!vis[v[i]])
{
vis[v[i]]=true;
q.push(v[i]);
}
}
}
if (dis[t]==INF) return false;
LL flow=addflow(s,t);
mincost+=flow*dis[t];
return true;
}
inline void mincost_flow(LL s,LL t)
{
while (bfs(s,t));
}
int main()
{
tot=-1;
memset(point,-1,sizeof(point));
memset(nxt,-1,sizeof(nxt));

scanf("%lld%lld",&m,&n);
for (LL i=1;i<=n;++i) scanf("%lld",&C[i]);
for (LL i=1;i<=m;++i)
for (LL j=1;j<=n;++j)
scanf("%lld",&A[i][j]);
for (LL i=1;i<=m;++i)
{
scanf("%lld",&s[i]);
for (LL j=1;j<=s[i];++j) scanf("%lld",&T[i][j]); T[i][s[i]+1]=INF;
for (LL j=1;j<=s[i]+1;++j) scanf("%lld",&W[i][j]);
}

N=n+m+2;
for (LL i=1;i<=n;++i)
addedge(1,1+i,C[i],0);
for (LL i=1;i<=n;++i)
for (LL j=1;j<=m;++j)
if (A[j][i])
addedge(1+i,1+n+j,INF,0);
for (LL i=1;i<=m;++i)
for (LL j=1;j<=s[i]+1;++j)
addedge(1+n+i,N,T[i][j]-T[i][j-1],W[i][j]);

mincost_flow(1,N);
printf("%lld\n",mincost);
}

总结