"Poj2122" Optimal milking

http://poj.org/problem?id=2112 (Topic link)

Test instructions

There are K can squeeze M-head cows milking machine and C-head cows, tell some milking machine and the distance between cows, to find the optimal allocation scheme to minimize the maximum distance.

Solution

First Floyd run out of 22 points between the shortest distance, two answers, maximum flow.

Details

Note that distances not exceeding 200 are Floyd before the distance between two points is less than 200.

Code

poj2112#include<algorithm> #include <iostream> #include <cstdlib> #include <cstring># include<cstdio> #include <cmath> #include <queue> #define LL long long#define inf 10000000#define Pi ACOs ( -1.0) #define FREE (a) freopen (a ".", "R", stdin), Freopen (a ". Out", "w", stdout); using namespace Std;const int maxn= 500010;struct Edge {int to,next,w;} E[maxn<<1];int head[maxn],d[maxn],f[300][300];int n,m,cnt=1,es,et,c,k,m,ans;void Link (int u,int v,int W) {e[++ Cnt]= (Edge) {v,head[u],w};head[u]=cnt;e[++cnt]= (edge) {u,head[v],w};head[v]=cnt;} 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++) f[i][j]=min (f[i][j],f[i ][K]+F[K][J]);} BOOL BFs () {memset (d,-1,sizeof (d));queue<int> Q;q.push (es);d [Es]=0;while (!q.empty ()) {int X=q.front (); Q.pop () ; for (int i=head[x];i;i=e[i].next) if (e[i].w && d[e[i].to]<0) {D[e[i].to]=d[x]+1;q.push (e[i].to);}} return d[et]>0;} int dfs (int x,int f) {if (X==et | | f==0) return f;int W,used=0;for (int i=head[x];i;i=e[i].next) if (e[i].w && d[e[i].to]==d[x]+1) {W=dfs (E[i].to,min (e[i].w,f-used)) ; Used+=w;e[i].w-=w;e[i^1].w+=w;if (used==f) return used;} if (!used) D[x]=-1;return used;} BOOL Dinic (int mid) {int Ans=0;memset (head,0,sizeof (head)); cnt=1;for (int i=k+1;i<=k+c;i++) {for (int j=1;j<=k;j+ +) if (f[i][j]<=mid) link (i,j,1); link (es,i,1);} for (int i=1;i<=k;i++) Link (i,et,m), while (BFS ()) Ans+=dfs (Es,inf); return ans==c;} int main () {scanf ("%d%d%d", &k,&c,&m), es=k+c+1;et=es+1;for (int i=1;i<=k+c;i++) for (int j=1;j<=k+c; J + +) {scanf ("%d", &f[i][j]); if (f[i][j]==0) F[i][j]=inf;} Floyd (); int L=0,r=60000,res;while (l<=r) {int mid= (l+r) >>1;if (Dinic (mid)) Res=mid,r=mid-1;else l=mid+1;} printf ("%d", res); return 0;}

　　

"Poj2122" Optimal milking