POJ 2391 Ombrophobic Bovines ( 经典最大流 && Floyd && 二分 && 拆点建图)

#include<bits/stdc++.h>
#define LL long long
using namespace std;
const int maxn = 450;
const int INF  = 1e9;
const LL  INF_LL = 1LL<<60;
LL Dist[maxn][maxn];
int Have[maxn], Can[maxn];
int N, M;
int Full_Flow;
LL Limit;

struct Edge
{
    Edge(){}
    Edge(int from,int to,int cap,int flow):from(from),to(to),cap(cap),flow(flow){}
    int from,to,cap,flow;
};

struct Dinic
{
    int n,m,s,t;            //结点数,边数(包括反向弧),源点与汇点编号
    vector<Edge> edges;     //边表 edges[e]和edges[e^1]互为反向弧
    vector<int> G[maxn];    //邻接表,G[i][j]表示结点i的第j条边在e数组中的序号
    bool vis[maxn];         //BFS使用,标记一个节点是否被遍历过
    int d[maxn];            //d[i]表从起点s到i点的距离(层次)
    int cur[maxn];          //cur[i]表当前正访问i节点的第cur[i]条弧

    void init(int n,int s,int t)
    {
        this->n=n,this->s=s,this->t=t;
        for(int i=1;i<=n;i++) G[i].clear();
        edges.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()
    {
        memset(vis,0,sizeof(vis));
        queue<int> Q;//用来保存节点编号的
        Q.push(s);
        d[s]=0;
        vis[s]=true;
        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];
    }

    //a表示从s到x目前为止所有弧的最小残量
    //flow表示从x到t的最小残量
    int DFS(int x,int a)
    {
        //printf("%d %d\n", x, a);
        if(x==t || a==0)return a;
        int flow=0,f;//flow用来记录从x到t的最小残量
        for(int& i=cur[x]; i<G[x].size(); i++)
        {
            Edge& e=edges[G[x][i]];
            if(d[x]+1==d[e.to] && (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 Maxflow()
    {
        int flow=0;
        while(BFS())
        {
            memset(cur,0,sizeof(cur));
            flow += DFS(s,INF);
        }
        return flow;
    }
}DC;

bool OK(LL Upper)
{
    DC.init(2*N+2, 0, 2*N+1);
    for(int i=1; i<=N; i++){
        DC.AddEdge(0, i, Have[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] <= Upper)
                DC.AddEdge(i, j+N, INF);

    return (Full_Flow == DC.Maxflow());
}

int main(void)
{
    while(~scanf("%d %d", &N, &M)){
        Full_Flow = 0;
        Limit = 0;

        for(int i=1; i<=N; i++){
            scanf("%d %d", &Have[i], &Can[i]);
            Full_Flow += Have[i];///记录所有牛的数量
        }

        for(int i=1; i<=N; i++)
            for(int j=1; j<=N; j++)///跑 Floyd 前的初始化
                if(i==j) Dist[i][j] = 0;
                else Dist[i][j] = INF_LL;

        int From, To;
        LL Cost;
        for(int i=1; i<=M; i++){
            scanf("%d %d %lld", &From, &To, &Cost);
            Dist[From][To] = Dist[To][From] = min(Dist[From][To], Cost);///有重边，只需记录最小的花费那条边
        }

        for(int k=1; k<=N; k++)///Floyd 算法
            for(int i=1; i<=N; i++)
                for(int j=1; j<=N; j++)
                    if(Dist[i][k] < INF_LL && Dist[k][j] < INF_LL)
                        Dist[i][j] = min(Dist[i][j], Dist[i][k]+Dist[k][j]);

        Limit = 0;///再次强调这里的时间是覆盖时间，所以我们找出花费最大的两点互大花费作为二分上界
        for(int i=1; i<=N; i++)
            for(int j=1; j<=N; j++)
                if(Dist[i][j] < INF_LL)
                    Limit = max(Limit, Dist[i][j]);

        if(!OK(Limit)) puts("-1");///给出最充裕的时间都无法满流肯定是无解了
        else{
            LL L = 0, R = Limit, ans;
            while(L <= R){///二分答案
                LL mid = L + ((R-L)>>1);
                if(!OK(mid)) L = mid+1;
                else ans = mid, R = mid-1;
            }
            printf("%lld\n", ans);
        }
    }
    return 0;
}