网络流--最大流--POJ 2139（超级源汇+拆点建图+二分+Floyd）

Description

FJ's cows really hate getting wet so much that the mere thought of getting caught in the rain makes them shake in their hooves. They have decided to put a rain siren on the farm to let them know when rain is approaching. They intend to create a rain evacuation plan so that all the cows can get to shelter before the rain begins. Weather forecasting is not always correct, though. In order to minimize false alarms, they want to sound the siren as late as possible while still giving enough time for all the cows to get to some shelter.

The farm has F (1 <= F <= 200) fields on which the cows graze. A set of P (1 <= P <= 1500) paths connects them. The paths are wide, so that any number of cows can traverse a path in either direction.

Some of the farm's fields have rain shelters under which the cows can shield themselves. These shelters are of limited size, so a single shelter might not be able to hold all the cows. Fields are small compared to the paths and require no time for cows to traverse.

Compute the minimum amount of time before rain starts that the siren must be sounded so that every cow can get to some shelter.

Input

* Line 1: Two space-separated integers: F and P

* Lines 2..F+1: Two space-separated integers that describe a field. The first integer (range: 0..1000) is the number of cows in that field. The second integer (range: 0..1000) is the number of cows the shelter in that field can hold. Line i+1 describes field i.

* Lines F+2..F+P+1: Three space-separated integers that describe a path. The first and second integers (both range 1..F) tell the fields connected by the path. The third integer (range: 1..1,000,000,000) is how long any cow takes to traverse it.

Output

* Line 1: The minimum amount of time required for all cows to get under a shelter, presuming they plan their routes optimally. If it not possible for the all the cows to get under a shelter, output "-1".

Sample Input

3 4
7 2
0 4
2 6
1 2 40
3 2 70
2 3 90
1 3 120

Sample Output

110

这个沙雕题，我建图建立了一天。

题意：

每个点有一个羊蓬容量，有一个羊的数量。每个点之间的连线还有花费。问你是否能将所有的羊都赶到羊圈里，能，就输出最小花费。

思路：

每个点拆成i和N+i两个点，建立超级源点，源点到每一个点的距离都是他们现在样的数量，控制满流时的流量。N+i到汇点的距离设成点的容量。点与点之间的距离就变成了点与拆出的N+I的关系了，二分枚举时间花费，条件是能使原图满流。完事撒花。

#include<cstdio>
#include<cstring>
#include<algorithm>
#include<vector>
#include<queue>
#define INF 1e9
#define INFLL 1LL<<60
using namespace std;
const int maxn=500+10;

struct Edge
{
    int from,to,cap,flow;
    Edge(){}
    Edge(int f,int t,int c,int fl):from(f),to(t),cap(c),flow(fl){}
};

struct Dinic
{
    int n,m,s,t;
    vector<Edge> edges;
    vector<int> G[maxn];
    int d[maxn];
    int cur[maxn];
    bool vis[maxn];

    void init(int n,int s,int t)
    {
        this->n=n, this->s=s, this->t=t;
        edges.clear();
        for(int i=0;i<n;i++) G[i].clear();
    }

    void AddEdge(int from,int to,int cap)
    {
        edges.push_back( Edge(from,to,cap,0) );
        edges.push_back( Edge(to,from,0,0) );
        m=edges.size();
        G[from].push_back(m-2);
        G[to].push_back(m-1);
    }

    bool BFS()
    {
        queue<int> Q;
        memset(vis,0,sizeof(vis));
        vis[s]=true;
        d[s]=0;
        Q.push(s);
        while(!Q.empty())
        {
            int x=Q.front(); Q.pop();
            for(int i=0;i<G[x].size();i++)
            {
                Edge e=edges[G[x][i]];
                if(!vis[e.to] && e.cap>e.flow)
                {
                    vis[e.to]=true;
                    d[e.to] = d[x]+1;
                    Q.push(e.to);
                }
            }
        }
        return vis[t];
    }

    int DFS(int x,int a)
    {
        if(x==t || a==0) return a;
        int flow=0,f;

        for(int& i=cur[x];i<G[x].size();++i)
        {
            Edge& e=edges[G[x][i]];
            if(d[e.to]==d[x]+1 && (f=DFS(e.to, min(a,e.cap-e.flow) ) )>0 )
            {
                e.flow+=f;
                edges[G[x][i]^1].flow-=f;
                flow+=f;
                a-=f;
                if(a==0) break;
            }
        }
        return flow;
    }

    int Max_Flow()
    {
        int flow=0;
        while(BFS())
        {
            memset(cur,0,sizeof(cur));
            flow += DFS(s,INF);
        }
        return flow;
    }
}DC;

int n,m;
int now[maxn],can[maxn];//存放每个牛栏还能放下的牛数. 为0则不能放了,>0则还有空位,<0则需要转移
long long dist[maxn][maxn];
void floyd(int n)
{
    for(int k=1;k<=n;k++)
    for(int i=1;i<=n;i++)
    for(int j=1;j<=n;j++)
        dist[i][j]=min(dist[i][j], dist[i][k]+dist[k][j]);
}

bool solve(long long limit,int MF)//判断只走长度<=limit的路看是否有解
{
    DC.init(2*n+2,0,2*n+1);
    for(int i=1;i<=n;i++)
    {
        DC.AddEdge(0,i,now[i]);
        DC.AddEdge(i+n,2*n+1,can[i]);
    }

    for(int i=1;i<=n;i++)
    for(int j=1;j<=n;j++)
    if(dist[i][j]<=limit)
        DC.AddEdge(i,j+n,INF);
    return DC.Max_Flow() == MF;//判断是否满流
}

int main()
{
    while(scanf("%d%d",&n,&m)==2)
    {
        long long L=0,R=0;//二分的上下界
        int MF = 0;
        memset(dist,0x3f,sizeof(dist));
        for(int i=1;i<=n;i++)
            dist[i][i]=0;
        for(int i=1;i<=n;i++)
        {
            int v1,v2;
            scanf("%d%d",&now[i],&can[i]);
            MF +=now[i];//记录满流量
        }
        for(int i=1;i<=m;i++)
        {
            int u,v;
            long long w;
            scanf("%d%d%I64d",&u,&v,&w);
            dist[u][v]=dist[v][u]=min(dist[u][v],w);
        }
        floyd(n);//计算最短路径
        for(int i=1;i<=n;i++)
        for(int j=1;j<=n;j++)
            if(dist[i][j]<INFLL)
                R=max(R,dist[i][j]);
        if(!solve(R, MF)) printf("-1
");
        else
        {
            while(R>L)
            {
                long long mid = L+(R-L)/2;
                if(solve(mid,MF)) R=mid;
                else L=mid+1;
                //cout<<mid<<endl;
            }
            printf("%I64d
",L);
        }
    }
    return 0;
}