POJ 2112
//题目类型：最大流+二分搜索
//本题与09年东北四省赛Power Line题目类似
//注意:最大值不要使用0x7fffffff,否则会一直wa
#include <iostream>
//#include <conio.h>
#include <queue>
using namespace std;
#define narray 240
#define karray 31
#define carray 201
const int INF = 10000000; //不要使用0x7fffffff,否则会一直wa
int graph[narray][narray];
int dist[carray*karray];
int k,c,m;
int distnum;
int length;
int vn;
int cnt;
int capacity[narray][narray];
int src,des;
int d[narray]; //标号
int num[narray]; //num[i]表示标号为i的顶点数有多少
int pre[narray]; //记录前驱
void Floyd()
{
int t,i,j;
for(t=1; t<=k+c;t++)
for(i=1;i<=k+c;i++)
for(j=1; j<=k+c; j++)
{
if(graph[i][t]<INF && graph[t][j]<INF && graph[i][j]>graph[i][t]+graph[t][j])
graph[i][j] = graph[i][t]+graph[t][j];
}
//graph[i][j]=min(graph[i][t]+graph[t][j],graph[i][j]);
}
void init(int src,int des)
{
int i,j;
queue<int> myqueue;
myqueue.push(des);
memset(num,0,sizeof(num));
for(i=0;i<vn;++i)
d[i] = INF;
d[des] = 0;
num[0] = 1;
int frontint;
while(!myqueue.empty())
{
frontint = myqueue.front();myqueue.pop();
for(i=0;i<vn;++i)
{
if(d[i]>=vn && capacity[i][frontint]>0)
{
d[i] = d[frontint]+1;
myqueue.push(i);
num[d[i]]++;
}
}
}
}
int findarc(int t)
{
int i,j;
for(i=0;i<vn;++i)
{
if(capacity[t][i]>0 && d[t]==d[i]+1) return i;
}
return -1;
}
int relabel(int t)
{
int i,j;
int mm = INF;
for(i=0;i<vn;++i)
{
if(capacity[t][i]>0) mm = min(mm,d[i]+1);
}
return mm==INF?vn:mm;
}
int maxFlow(int src,int des)
{
int sumflow = 0;
int delta;
int i=src;
int j;
memset(pre,-1,sizeof(pre));
while(d[src]<vn)
{
j = findarc(i);
if(j>=0)
{
pre[j] = i;
i = j;
if(i==des)
{
delta = INF;
for (i=des;i!=src;i=pre[i]) delta=min(delta,capacity[pre[i]][i]);
for (i=des;i!=src;i=pre[i]) capacity[pre[i]][i] -= delta, capacity[i][pre[i]] += delta;
sumflow += delta;
}
}
else
{
int x = relabel(i);
num[x]++;
num[d[i]]--;
if(num[d[i]]==0) return sumflow;
d[i] = x;
if(i!=src) i =pre[i];
}
}
return sumflow;
}
bool build()
{
int i,j;
memset(capacity,0,sizeof(capacity));
for(i=1;i<=k;++i)
{
capacity[src][i]= m;
}
for(i=k+1;i<=k+c;++i)
{
capacity[i][des] = 1;
}
for(i=1;i<=k;++i)
{
for(j=k+1;j<=k+c;++j)
{
if(graph[i][j]<=length)
capacity[i][j] =1;
else
capacity[i][j] =0;
}
}
init(src,des);
if(maxFlow(src,des)>=c) return true;
else return false;
}
int main()
{
//freopen("1.txt","r",stdin);
int i,j;
while(scanf("%d%d%d",&k,&c,&m)!=-1)
{
vn = k+c+2;
distnum = 0;
src = 0;
des = k+c+1;
for(i=1;i<=k+c;++i)
{
for(j=1;j<=k+c;++j)
{
scanf("%d",&graph[i][j]);
if(graph[i][j]==0) graph[i][j]=INF; //输入的数据中0代表无穷大(此处注意变换)
}
}
Floyd(); //使用Floyd算法求出最短距离
for(i=k+1;i<=k+c;++i)
{
for(j=1;j<=k;++j)
dist[distnum++] = graph[i][j];
}
sort(dist,dist+distnum); //对奶牛和奶牛机之间的距离排序
int l=0,r=distnum-1,mid;
while(l+1<r) //枚举
{
mid = (l+r)/2;
length = dist[mid];
if(build()) r = mid;
else l = mid;
}
if(distnum==1) //注意考虑只有一个元素的情况
{
printf("%d\n",dist[0]);
}
else if(l+1==r)
{
length = dist[l];
if(build())
{
printf("%d\n",length);
}
else
{
length = dist[r];
if(build())
printf("%d\n",length);
}
}
/* 也可采用这种方法进行枚举 （但是时间效率比较低,极易超时）
int l=0,r=10000,mid;
while(l<r)
{
mid = (l+r)/2;
length = mid;
if(build()) r = mid; //此处不要写成mid+1，因为流量正好等于c
else l = mid+1;
}
printf("%d\n",r);
*/
}
//getch();
return 0;
}
本题亦可采取二分匹配+二分搜索的方法
//二分匹配（匈牙利算法）+二分搜索
#include <iostream>
//#include <conio.h>
using namespace std;
#define narray 240
const int INF = 10000000;
int map[narray][narray];
int graph[narray][narray];
bool final[narray];
int cap[narray];
int match[narray][narray];
int k,c,m;
void Floyd()
{
int i,j,t;
for(t=1;t<=k+c;++t)
{
for(i=1;i<=k+c;++i)
{
for(j=1;j<=k+c;++j)
{
if(map[i][t]<INF && map[t][j]<INF && map[i][j]>map[i][t]+map[t][j])
map[i][j] = map[i][t]+map[t][j];
}
}
}
}
void build(int length)
{
int i,j;
memset(graph,0,sizeof(graph));
for(i=k+1;i<=k+c;++i)
{
for(j=1;j<=k;++j)
{
if(map[i][j]<=length)
graph[i-k][j] = 1;
}
}
}
bool DFS(int p)
{
int i,j;
for(i=1;i<=k;++i)
{
if(graph[p][i] && !final[i])
{
final[i] = true;
if(cap[i]) //如果奶牛机的容量不为0
{
cap[i]--;
match[i][++match[i][0]]=p;
return true;
}
else
{
for(j=1;j<=match[i][0];++j) //match[i][0]记录第i个奶牛机匹配的奶牛数
{
if(DFS(match[i][j]))
{
match[i][j] = p;
return true;
}
}
}
}
}
return false;
}
bool mat()
{
int i,j;
memset(match,0,sizeof(match)); //注意初始化match
for(i=1;i<=k;++i) cap[i]=m; //记录奶牛机的容量
for(i=1;i<=c;++i) //从奶牛进行匹配，不是从奶牛机
{
memset(final,0,sizeof(final));
if(!DFS(i)) return false;
}
return true;
}
int main()
{
//freopen("1.txt","r",stdin);
int i,j;
while(scanf("%d%d%d",&k,&c,&m)!=-1)
{
for(i=1;i<=k+c;++i)
{
for(j=1;j<=k+c;++j)
{
scanf("%d",&map[i][j]);
if(map[i][j]==0) map[i][j]=INF;
}
}
Floyd();
int l= 0,r=10000,mid;
while(l<r)
{
mid = (l+r)/2;
build(mid);
if(mat()) r= mid;
else l = mid+1;
}
printf("%d\n",r);
}
//getch();
return 0;
}
