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## 哪位大牛看一下我这哪错了

Posted by yygy at 2009-02-27 15:59:04 on Problem 2195
```#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;

const int size = 160;
const int INF = 100000000;  //  相对无穷大

bool map[size][size];         // 二分图的相等子图, map[i][j] = true 代表Xi与Yj有边
bool xckd[size], yckd[size];  // 标记在一次DFS中，Xi与Yi是否在交错树上
int match[size];             // 保存匹配信息，其中i为Y中的顶点标号，match[i]为X中顶点标号
int abs(int x)
{
if(x<0)
return -x;
return x;
}
bool DFS(int, const int);
int min(int a,int b)
{
if(a<b)
return a;
return b;
}
int max(int a,int b)
{
if(a>b)
return a;
return b;
}
void KM_Perfect_Match(const int n, const int edge[][size]) {
int i, j;
int lx[size], ly[size];   //  KM算法中Xi与Yi的标号
for(i = 0; i < n; i++)
{
lx[i] = INF;
ly[i] = 0;
for(j = 0; j < n; j++)
{
lx[i] = min(lx[i], edge[i][j]);
}
}
bool perfect = false;
while(!perfect)
{
//  初始化邻接矩阵
for(i = 0; i < n; i++)
{
for(j = 0; j < n; j++)
{
if(lx[i]+ly[j] == edge[i][j])
map[i][j] = true;
else
map[i][j] = false;
}
}
// 匹配过程
int live = 0;
memset(match, -1, sizeof(match));
for(i = 0; i < n; i++)
{
memset(xckd, false, sizeof(xckd));
memset(yckd, false, sizeof(yckd));
if(DFS(i, n))
live++;
else
{
xckd[i] = true;
break;
}
}
if(live == n)
perfect = true;
else
{
// 修改标号过程
int ex = INF;
for(i = 0; i < n; i++)
{
for(j = 0; xckd[i] && j < n; j++)
{
if(!yckd[j])
ex = min(ex, lx[i]+ly[j]-edge[i][j]);
}
}
for(i = 0; i < n; i++)
{
if(xckd[i])
lx[i] -= ex;
if(yckd[i])
ly[i] += ex;
}
}
}
}

// 此函数用来寻找是否有以Xp为起点的增广路径，返回值为是否含有增广路

bool DFS(int p, const int n)
{
int i;
for(i = 0; i < n; i++)
{
if(!yckd[i] && map[p][i])
{
yckd[i] = true;
int t = match[i];
match[i] = p;
if(t == -1 || DFS(t, n))
{
return true;
}
match[i] = t;
if(t != -1)
xckd[t] = true;
}
}
return false;
}

int main()
{
int n, edge[size][size],a,b,c; //  edge[i][j]为连接Xi与Yj的边的权值
int i,man[105][2],house[105][2],j;
char s[105];
while(scanf("%d%d",&a,&b)!=EOF&&a&&b)
{

c=n=0;
for(i=0;i<a;i++)
{
scanf("%s",s);
for(j=0;s[j]!='\0';j++)
{
if(s[j]=='m')
{
man[n][0]=i;
man[n++][1]=j;
}
if(s[j]=='H')
{
house[c][0]=i;
house[c++][1]=j;
}
}
}
for(i=0;i<n;i++)
{
for(j=0;j<n;j++)
{
edge[i][j]=abs(man[i][0]-house[j][0])+abs(man[i][1]-house[j][1]);
}
}
KM_Perfect_Match(n, edge);
int cost = 0;
for(i = 0; i < n; i++)
cost += edge[match[i]][i];
printf("%d\n",cost);
}
// cost 为最大匹配的总和, match[]中保存匹配信息

return 0;
}
```

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