奔小康赚大钱
Time Limit: 1000ms
Memory Limit: 32768KB
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传说在遥远的地方有一个非常富裕的村落,有一天,村长决定进行制度改革:重新分配房子。
这可是一件大事,关系到人民的住房问题啊。村里共有n间房间,刚好有n家老百姓,考虑到每家都要有房住(如果有老百姓没房子住的话,容易引起不安定因素),每家必须分配到一间房子且只能得到一间房子。
另一方面,村长和另外的村领导希望得到最大的效益,这样村里的机构才会有钱.由于老百姓都比较富裕,他们都能对每一间房子在他们的经济范围内出一定的价格,比如有3间房子,一家老百姓可以对第一间出10万,对第2间出2万,对第3间出20万.(当然是在他们的经济范围内).现在这个问题就是村领导怎样分配房子才能使收入最大.(村民即使有钱购买一间房子但不一定能买到,要看村领导分配的).
Input
输入数据包含多组测试用例,每组数据的第一行输入n,表示房子的数量(也是老百姓家的数量),接下来有n行,每行n个数表示第i个村名对第j间房出的价格(n<=300)。
Output
Sample Input
Sample Output
Source
HDOJ 2008 Summer Exercise(4)- Buffet Dinner
#include <stdio.h>
#include <string.h>
#define p 305
#define m 999999999
int n, tu[p][p];
int link[p], slack[p];
int lx[p], ly[p];
int visitx[p], visity[p];
int MAX(int a, int b)
{
if(a>b)
return a;
return b;
}
int MIN(int a, int b)
{
if(a<b)
return a;
return b;
}
int dfs(int k)
{
int i, d;
visitx[k] = 1;
for(i=0; i<n; i++)
{
if(visity[i])
continue;
d = lx[k]+ly[i]-tu[k][i];
if(d==0)
{
visity[i] = 1;
if(link[i]==-1 || dfs(link[i]))
{
link[i] = k;
return 1;
}
}
else
slack[i] = MIN(slack[i], d);
}
return 0;
}
int KM()
{
int i, j, ans;
for(i=0; i<n; i++)
link[i] = -1;
memset(lx, 0, sizeof(lx));
memset(ly, 0, sizeof(ly));
for(i=0; i<n; i++)
{
for(j=0; j<n; j++)
lx[i] = MAX(lx[i], tu[i][j]);
}
for(i=0; i<n; i++)
{
for(j=0; j<n; j++)
{
slack[j] = m;
}
while(1)
{
memset(visitx, 0, sizeof(visitx));
memset(visity, 0, sizeof(visity));
if(dfs(i))
break;
else
{
int d;
d = m;
for(j=0; j<n; j++)
{
if(!visity[j])
d = MIN(d, slack[j]);
}
for(j=0; j<n; j++)
{
if(visitx[j])
lx[j] -= d;
if(visity[j])
ly[j] += d;
}
}
}
}
ans = 0;
for(i=0; i<n; i++)
ans += tu[link[i]][i];
return ans;
}
int main()
{
int i, j;
while(scanf("%d", &n)!=-1)
{
for(i=0; i<n; i++)
{
for(j=0; j<n; j++)
{
scanf("%d", &tu[i][j]);
}
}
printf("%d\n", KM());
}
return 0;
}