bzoj [Usaco2005 mar]Ombrophobic Bovines 发抖的牛 最大流

很好的题目= =并没有想出正解= =
首先二分最短的路径x
然后每次二分的时候跑最大流。
每个牛拆成i和i’,起点向i连容量为数量的边，i’向终点连容量为容量的边。
如果i,j最短路小于二分的答案x就连i和j’容量为inf。

#include<cstdio>#include<cstring>#include<algorithm>#define fo(i,a,b) for(int i=a;i<=b;i++)#define fd(i,a,b) for(int i=a;i>=b;i--)#define inf 1000000000using namespace std;const int N=1e5+5;int n,m;typedef long long ll;int a[N],b[N];int head[N],next[N],go[N];int val[N];int q[N];ll f[1005][1005];int dis[N];int s,T,ans,tot,sum;inline void add(int x,int y,int z){    go[++tot]=y;    next[tot]=head[x];    val[tot]=z;    head[x]=tot;}inline void ins(int x,int y,int z){    add(x,y,z);    add(y,x,0);}inline bool bfs(){    int t=0,w=1;    memset(dis,-1,sizeof(dis));    q[1]=s,dis[s]=0;    while (t<w)    {        int x=q[++t];        for(int i=head[x];i;i=next[i])        {            int v=go[i];            if (val[i]&&dis[v]==-1)            {                dis[v]=dis[x]+1;                q[++w]=v;            }        }    }    return dis[T]!=-1;}inline int dfs(int x,int f){    if (x==T)return f;    int w,used=0;    for(int i=head[x];i;i=next[i])    {        int v=go[i];        if (dis[v]==dis[x]+1)        {            w=f-used;            w=dfs(v,min(w,val[i]));            val[i]-=w;            val[i^1]+=w;            used+=w;            if (used==f)return f;         }    }    if (!used)dis[x]=-1;    return used;}inline void dinic(){    while (bfs())ans+=dfs(s,inf);}inline bool check(ll x){    s=0,T=2*n+1;    tot=1;ans=0;    memset(head,0,sizeof(head));    fo(i,1,n)ins(s,i,a[i]),ins(n+i,T,b[i]);    fo(i,1,n)    fo(j,1,n)    if (f[i][j]<=x)ins(i,j+n,inf);    dinic();    if (ans==sum)return 1;    else return 0;}int main(){    scanf("%d%d",&n,&m);    fo(i,1,n)    {        scanf("%d%d",&a[i],&b[i]);        sum+=a[i];    }    fo(i,1,n)    fo(j,1,n)f[i][j]=inf*200ll;    fo(i,1,m)    {        int x,y;        ll z;        scanf("%d%d%lld",&x,&y,&z);        f[x][y]=f[y][x]=min(f[x][y],z);    }    fo(i,1,n)f[i][i]=0;    fo(k,1,n)    fo(i,1,n)    fo(j,1,n)    f[i][j]=min(f[i][j],f[i][k]+f[k][j]);    ll l=0,r=inf*200ll-1;    ll ans=-1;    while (l<=r)    {        ll mid=(l+r)>>1;        if (check(mid))        {            ans=mid;            r=mid-1;        }        else l=mid+1;    }    printf("%lld\n",ans);}