【最小割】【bzoj 3511】: 土地划分

来源：互联网发布：js rows函数编辑：程序博客网时间：2024/06/04 08:24
http://www.lydsy.com/JudgeOnline/problem.php?id=3511
为什么推出来就是不一样！！！
//#define _TEST _TEST#include <cstdio>#include <cstring>#include <cstdlib>#include <iostream>#include <cmath>#include <algorithm>using namespace std;/************************************************Code By willinglive    Blog:http://willinglive.cf************************************************/#define rep(i,l,r) for(int i=(l),___t=(r);i<=___t;i++)#define per(i,r,l) for(int i=(r),___t=(l);i>=___t;i--)#define MS(arr,x) memset(arr,x,sizeof(arr))#define LL long long#define INE(i,u,e) for(int i=head[u];~i;i=e[i].next)inline const int read(){int r=0,k=1;char c=getchar();for(;c<'0'||c>'9';c=getchar())if(c=='-')k=-1;for(;c>='0'&&c<='9';c=getchar())r=r*10+c-'0';return k*r;}/////////////////////////////////////////////////const int inf=0x3f3f3f3f;const int N=11111;const int M=233333;int n,m;struct edge{int v,flow,cap,next;}e[M];int head[N],k;int va[N],vb[N];int S,T;int cur[N],q[N],d[N];int sum;/////////////////////////////////////////////////void adde(int u,int v,int f){e[k].v=v;e[k].flow=0;e[k].cap=f;e[k].next=head[u];head[u]=k++;}void ins(int u,int v,int f1,int f2){adde(u,v,f1);adde(v,u,f2);}   inline int F(int i){return e[i].cap-e[i].flow;}    bool bfs()    {        MS(d,-1); d[q[0]=S]=0; memcpy(cur,head,sizeof(cur));        for(int u,v,l=0,r=1;l<r;)        {            u=q[l++];            INE(i,u,e) if(F(i)&&d[v=e[i].v]==-1)            {                q[r++]=v; d[v]=d[u]+1;                if(v==T) return 1;            }        }        return 0;    }    int dfs(int u,int a)    {        if(u==T||a==0) return a;        for(int &i=cur[u];~i;i=e[i].next) if(F(i)&&d[e[i].v]==d[u]+1)        {            if(int t=dfs(e[i].v,min(a,F(i))))            {                e[i].flow+=t; e[i^1].flow-=t;                return t;            }        }        return 0;    }    int dinic()    {        int f=0;        while(bfs()) while(int t=dfs(S,inf)) f+=t;        return f;    }   /////////////////////////////////////////////////void input(){MS(head,-1);    n=read(); m=read();    S=0; T=n+1;    va[1]=inf; vb[1]=0;    vb[n]=0; vb[n]=inf;    rep(i,2,n-1) va[i]=read()<<1;    rep(i,2,n-1) vb[i]=read()<<1;    rep(i,2,n-1) sum+=va[i]+vb[i];    rep(i,1,m)    {    int u,v,ea,eb,ec;    u=read(); v=read(); ea=read(); eb=read(); ec=read();    ins(u,v,ea+eb+2*ec,0); ins(v,u,ea+eb+2*ec,0);    va[u]+=ea+ec; va[v]+=ea+ec;    vb[u]+=eb+ec; vb[v]+=eb+ec;    sum+=ea+eb+ec<<1;    }    rep(i,1,n) ins(S,i,va[i],0), ins(i,T,vb[i],0);}void solve(){cout<<(sum-dinic()>>1)<<endl;}/////////////////////////////////////////////////int main(){    #ifndef _TEST    freopen("std.in","r",stdin); freopen("std.out","w",stdout);    #endif    input(),solve();    return 0;}
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