POJ2391：Ombrophobic Bovines floyd+二分答案+最大流

最新推荐文章于 2017-03-30 18:43:16 发布

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本文详细介绍了如何运用Floyd算法与网络流技术解决任意两点间的最短路径问题，并通过二分查找优化解决方案。通过构建拆点网络流模型，实现高效的路径搜索。案例分析与代码实现提供了实际操作的指导。

Floyd都写错了我还是退役吧

floyd求出任意两点间的最短路，然后二分答案拆点网络流验证。

#include<iostream>
#include<cstdio>
#include<cstring>
#define N 405
#define ll long long 
#define inf 1000000007
using namespace std;
int n,m,S,T,total,cnt;
int head[N],q[N],dis[N],cur[N],a[N>>1],b[N>>1];
ll d[N][N];
int next[100005],list[100005],key[100005];
inline ll read()
{
    ll a=0,f=1; char c=getchar();
    while (c<'0'||c>'9') {if (c=='-') f=-1; c=getchar();}
    while (c>='0'&&c<='9') {a=a*10+c-'0'; c=getchar();}
    return a*f;
}
inline void insert(int x,int y,int z)
{
    next[++cnt]=head[x];
    head[x]=cnt;
    list[cnt]=y;
    key[cnt]=z;
}
inline bool BFS()
{
    int t=0,w=1,x;
    memset(dis,-1,sizeof(dis));
    q[1]=0; dis[0]=1;
    while (t<w)
    {
        x=q[++t];
        for (int i=head[x];i;i=next[i])
            if (key[i]&&dis[list[i]]==-1)
                dis[q[++w]=list[i]]=dis[x]+1;
    }
    return dis[T]!=-1;
}
int find(int x,int flow)
{
    if (x==T) return flow;
    int w,used=0;
    for (int i=cur[x];i;i=next[i])
        if (key[i]&&dis[list[i]]==dis[x]+1)
        {
            w=find(list[i],min(key[i],flow-used));
            key[i]-=w; key[i^1]+=w; used+=w;
            if (key[i]) cur[x]=i;
            if (used==flow) return used;
        }
    if (!used) dis[x]=-1;
    return used;
}
inline int dinic()
{
    int ans=0;
    while (BFS())
    {
        for (int i=0;i<=T;i++) cur[i]=head[i];
        ans+=find(S,inf);
    }
    return ans;
}
inline void build(ll mid)
{
    memset(head,0,sizeof(head)); cnt=1;
    for (int i=1;i<=n;i++)
        insert(S,i,a[i]),insert(i,S,0),insert(i+n,T,b[i]),insert(T,i+n,0),insert(i,i+n,inf),insert(i+n,i,0);
    for (int i=1;i<=n;i++)
        for (int j=1;j<=n;j++)
            if (d[i][j]<=mid&&d[i][j]!=-1&&i!=j)
                insert(i,j+n,inf),insert(j+n,i,0);
}
int main()
{
    n=read(); m=read(); T=n+n+1;
    for (int i=1;i<=n;i++) a[i]=read(),b[i]=read(),total+=a[i];
    for (int i=1;i<=n;i++)
        for (int j=1;j<=n;j++)
            d[i][j]=(i==j?0:-1);
    ll l=0,r=0,ans=-1;
    for (int i=1;i<=m;i++)
    {
        int u=read(),v=read(),w=read();
        if (d[u][v]==-1||w<d[u][v])
            d[u][v]=d[v][u]=w,r=max(r,d[u][v]);
    }
    for (int k=1;k<=n;k++)
        for (int i=1;i<=n;i++)
            for (int j=1;j<=n;j++)
            {
                if (d[i][k]==-1||d[k][j]==-1) continue;
                if (d[i][j]==-1||d[i][j]>d[i][k]+d[k][j])
                {
                    d[i][j]=d[i][k]+d[k][j];
                    r=max(r,d[i][j]);
                }
            }
    while (l<=r)
    {
        ll mid=l+r>>1;
        build(mid);
        if (dinic()==total) ans=mid,r=mid-1; else l=mid+1;
    }
    cout << ans << endl;
    return 0;
}