poj2112 二分+最大流

Optimal MilkingTime Limit : 4000/2000ms (Java/Other)   Memory Limit : 60000/30000K (Java/Other)Total Submission(s) : 1   Accepted Submission(s) : 1Problem DescriptionFJ has moved his K (1 <= K <= 30) milking machines out into the cow pastures among the C (1 <= C <= 200) cows. A set of paths of various lengths runs among the cows and the milking machines. The milking machine locations are named by ID numbers 1..K; the cow
locations are named by ID numbers K+1..K+C. 

Each milking point can "process" at most M (1 <= M <= 15) cows each day. 

Write a program to find an assignment for each cow to some milking machine so that the distance the furthest-walking cow travels is minimized (and, of course, the milking machines are not overutilized). At least one legal assignment is possible for all input
data sets. Cows can traverse several paths on the way to their milking machine.  

Input* Line 1: A single line with three space-separated integers: K, C, and M. 

* Lines 2.. ...: Each of these K+C lines of K+C space-separated integers describes the distances between pairs of various entities. The input forms a symmetric matrix. Line 2 tells the distances from milking machine 1 to each of the other entities; line 3 tells
the distances from machine 2 to each of the other entities, and so on. Distances of entities directly connected by a path are positive integers no larger than 200. Entities not directly connected by a path have a distance of 0. The distance from an entity
to itself (i.e., all numbers on the diagonal) is also given as 0. To keep the input lines of reasonable length, when K+C > 15, a row is broken into successive lines of 15 numbers and a potentially shorter line to finish up a row. Each new row begins on its
own line.  

OutputA single line with a single integer that is the minimum possible total distance for the furthest walking cow.  

Sample Input

2 3 20 3 2 1 13 0 3 2 02 3 0 1 01 2 1 0 21 0 0 2 0

 

Sample Output

2

 

SourcePKU 和上一題基本一個思路額~~本來想著能不能直接建邊不用floyd 後來覺得不行啊~~

#include <cstring>#include <cstdio>#include <queue>#define MAXN 5000#define MAXM 500000#define inf 0x3f3f3f3fusing namespace std;struct node{    int u,v,f,c;};node edge[MAXM*3];int first[MAXN],next[MAXM*3];int gap[MAXN],d[MAXN],curedge[MAXN],pre[MAXN];int cc;inline void add_edge(int u,int v,int f,int c){    edge[cc].u=u;    edge[cc].v=v;    edge[cc].f=f;    edge[cc].c=c;    next[cc]=first[u];    first[u]=cc;    cc++;    edge[cc].u=v;    edge[cc].v=u;    edge[cc].f=0;    edge[cc].c=0;    next[cc]=first[v];    first[v]=cc;    cc++;}int ISAP(int s,int t,int n){    int cur_flow,flow_ans=0,u,tmp,neck,i,v;    memset(d,0,sizeof(d));    memset(gap,0,sizeof(gap));    memset(pre,-1,sizeof(pre));    for(i=0;i<=n;i++)        curedge[i]=first[i];    gap[0]=n+1;    u=s;    while(d[s]<=n)    {        if(u==t)        {            cur_flow=inf;            for(i=s;i!=t;i=edge[curedge[i]].v)            {                if(cur_flow>edge[curedge[i]].f)                {                    neck=i;                    cur_flow=edge[curedge[i]].f;                }            }            for(i=s;i!=t;i=edge[curedge[i]].v)            {                tmp=curedge[i];                edge[tmp].f-=cur_flow;                edge[tmp^1].f+=cur_flow;            }            flow_ans+=cur_flow;            u=neck;        }        for(i=curedge[u];i!=-1;i=next[i])        {            v=edge[i].v;            if(edge[i].f&&d[u]==d[v]+1)                break;        }        if(i!=-1)        {            curedge[u]=i;            pre[v]=u;            u=v;        }        else        {            if(0==--gap[d[u]])                break;            curedge[u]=first[u];            for(tmp=n+5,i=first[u];i!=-1;i=next[i])                if(edge[i].f)                    tmp=min(tmp,d[edge[i].v]);            d[u]=tmp+1;            ++gap[d[u]];            if(u!=s)                u=pre[u];        }    }    return flow_ans;}void build(int limit,int cnt,int m){    int i;    for(i=0;i<cnt;i=i+2)    {        edge[i].f=m;        edge[i^1].f=0;    }    for(i=cnt;i<cc;i=i+2)    {        edge[i].f=(edge[i].c<=limit);        edge[i^1].f=0;    }}int dist[300][300];void Floyd(int n){    int i,j,k;    for(k=1;k<=n;k++)    {        for(i=1;i<=n;i++)        {            for(j=1;j<=n;j++)            {                if(dist[i][k]!=inf&&dist[k][j]!=inf&&                   dist[i][j]>dist[i][k]+dist[k][j])                {                    dist[i][j]=dist[i][k]+dist[k][j];                }            }        }    }}int main(){    int k,c,m;    while(scanf("%d%d%d",&k,&c,&m)!=EOF)    {        memset(first,-1,sizeof(first));        memset(next,-1,sizeof(next));        cc=0;        int i,j;        int s=0,t=k+c+1;        for(i=1;i<=k;i++)            add_edge(i,t,m,0);        int cnt1=cc;        for(i=1;i<=k+c;i++)        {            for(j=1;j<=k+c;j++)            {                scanf("%d",&dist[i][j]);                if(dist[i][j]==0)                    dist[i][j]=inf;            }        }        Floyd(k+c);        for(i=k+1;i<=k+c;i++)        {            for(j=1;j<=k;j++)                add_edge(i,j,inf,dist[i][j]);        }        //int cnt2=cc;        for(i=k+1;i<=k+c;i++)            add_edge(s,i,1,0);        int l=0,r=400000;        int mid,ans=inf;        while(l<=r)        {            mid=(l+r)>>1;            build(mid,cnt1,m);            int res=ISAP(s,t,t);            if(res<c)                l=mid+1;            else if(res>=c)            {                ans=min(ans,mid);                r=mid-1;            }        }        printf("%d\n",ans);    }    return 0;}