POJ 2112 Optimal Milking

最大流+二分答案+Floyd

#include<cstdio>
#include<cstring>
#include<string>
#include<cmath>
#include<vector>
#include<queue>
#include<algorithm>
using namespace std;

const int maxn = 300 + 10;
const int INF = 0x7FFFFFFF;
struct Edge
{
    int from, to, cap, flow;
    Edge(int u, int v, int c, int f) :from(u), to(v), cap(c), flow(f) {}
};
vector<Edge>edges;
vector<int>G[maxn];
bool vis[maxn];
int d[maxn];
int cur[maxn];
int n, m, s, t;
int K,C,M;
int R[maxn][maxn];
int Se[maxn*maxn];


void init()
{
    for (int i = 0; i < maxn; 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));
    int w = edges.size();
    G[from].push_back(w - 2);
    G[to].push_back(w - 1);
}
bool BFS()
{
    memset(vis, 0, sizeof(vis));
    queue<int>Q;
    Q.push(s);
    d[s] = 0;
    vis[s] = 1;
    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] = 1;
                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[x]+1 == d[e.to]&&(f=DFS(e.to,min(a,e.cap-e.flow)))>0)
        {
            edges[G[x][i]].flow+=f;
            edges[G[x][i] ^ 1].flow-=f;
            flow+=f;
            a-=f;
            if(a==0) break;
        }
    }
    if(!flow) d[x] = -1;
    return flow;
}
int dinic(int s, int t)
{
    int flow = 0;
    while (BFS())
    {
        memset(cur, 0, sizeof(cur));
        flow += DFS(s, INF);
    }
    return flow;
}

void floyd()
{
    for(int k=1; k<=K+C; k++)
        for(int i=1; i<=K+C; i++)
            for(int j=1; j<=K+C; j++)
                if(R[k][j]!=999999999&&R[i][k]!=999999999)
                    if(R[i][j]>R[i][k]+R[k][j])
                        R[i][j]=R[i][k]+R[k][j];
}

void solve()
{
    s=0;
    t=K+C+1;
    int To=0;
    for(int i=K+1; i<=K+C; i++)
        for(int j=1; j<=K; j++)
            if(R[i][j]!=999999999)
            {
                Se[To]=R[i][j];
                To++;
            }
    sort(Se,Se+To);
    int Min=0,Max=To-1;
    int Mid=(Min+Max)/2;
    while(1)
    {
        int Lim=Se[Mid];
        init();
        for(int i=K+1; i<=K+C; i++)
            for(int j=1; j<=K; j++)
                if(R[i][j]!=999999999)
                    if(R[i][j]<=Lim)
                        AddEdge(j,i,INF);
        for(int i=1; i<=K; i++) AddEdge(s,i,M);
        for(int i=K+1; i<=K+C; i++) AddEdge(i,t,1);
        if(dinic(s,t)<C)
        {
            Min=Mid+1;
            Mid=(Min+Max)/2;
        }
        else
        {
            Max=Mid;
            Mid=(Min+Max)/2;
        }
        if(Min==Max) break;
    }
    printf("%d\n",Se[Mid]);
}

int main()
{
    while(~scanf("%d%d%d",&K,&C,&M))
    {
        for(int i=1; i<=K+C; i++)
            for(int j=1; j<=K+C; j++)
            {
                scanf("%d",&R[i][j]);
                if(R[i][j]==0) R[i][j]=999999999;
            }
        floyd();
        solve();
    }
    return 0;
}