【最大流】【HDU3338】【Kakuro Extension】
题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=3338
题目大意：填数字，使白色区域的值得和等于有值得黑色区域的相对应的值，用网络流来做
题目思路：增加一个源点和汇点，然后左进上出，用源点连左面，容量为相对应的值，汇点连上面，容量是相对应的值，然后左面连空白区域的，容量为8，因为数字为1-9，有下限，但因每个空白必须有值，所以至少为一，所以容量空间为0-8；然后空白区域来连上面，容量同样为8；
#include<iostream>
#include<cstring>
#include<cstdio>
using namespace std;
#define INF 999999999
#define maxn 14000
#define RE(x) (x)^1
int head[maxn];
int st,ed;//开始节点，结束节点
struct Edge
{
int v,next;
int val;
Edge() {}
Edge( int V , int NEXT , int W = 0 ):v(V),next(NEXT),val(W) {}
} edge[500000];
int lvl[maxn], gap[maxn];
int cnt_edge;
int map[maxn][maxn];
struct gg
{
int x,y;
int val;
} row[maxn],col[maxn];
int emp,row_num,col_num;
int n,m,T;
void Insert_Edge( int u , int v , int flow = 0 )
{
edge[cnt_edge] = Edge(v,head[u],flow);
head[u] = cnt_edge++;
edge[cnt_edge] = Edge(u,head[v]);
head[v] = cnt_edge++;
}
void Init()
{
cnt_edge = 0;
memset(head,-1,sizeof(head));
memset(lvl, 0, sizeof (lvl));
memset(gap, 0, sizeof (gap));
}
int dfs(int u, int flow)
{
if (u==ed)
{
return flow;
}
int tf = 0, sf, mlvl = ed-1;
for (int i= head[u]; i != -1; i = edge[i].next)
{
if (edge[i].val > 0)
{
if (lvl[u] ==lvl[edge[i].v]+1)
{
sf = dfs(edge[i].v, min(flow-tf, edge[i].val));
edge[i].val -= sf;
edge[RE(i)].val += sf;
tf += sf;
if (lvl[st] >=ed)
{
return tf;
}
if (tf == flow)
{
break;
}
}
mlvl = min(mlvl, lvl[edge[i].v]);
}
}
if (tf == 0)
{
--gap[lvl[u]];
if (!gap[lvl[u]])
{
lvl[st] =ed;
}
else
{
lvl[u] = mlvl+1;
++gap[lvl[u]];
}
}
return tf;
}
int sap()
{
int ans = 0;
gap[0]=ed;
while (lvl[st] <ed)
{
ans += dfs(st, INF);
}
return ans;
}
int print( int tp )
{
int ans = 0;
int id = tp + row_num+1;
for( int i = head[id] ; i != -1 ; i = edge[i].next )
{
int v = edge[i].v;
if( v <=row_num+1 )
{
ans+= edge[i].val;
break;
}
}
return ans+1;
}
int main()
{
int i,j;
char s[15];
//freopen("F://ACMInput/input.txt","r",stdin);
while(scanf("%d%d",&n,&m)!=-1)
{
emp=row_num=col_num=0;
for(i=0; i<n; i++)
for(j=0; j<m; j++)
{
scanf("%s",s);
if(s[0]=='.')
{
map[i][j]=++emp;
}
else
{
map[i][j]=-1;
if(s[4]!='X')
{
int tp=(s[4]-'0')*100+(s[5]-'0')*10+s[6]-'0';
row[++row_num].x=i;
row[row_num].y=j;
row[row_num].val=tp;
}
if(s[0]!= 'X' )
{
int tp = (s[0]-'0')*100+(s[1]-'0')*10+s[2]-'0';
col[++col_num].x = i;
col[col_num].y = j;
col[col_num].val = tp;
}
}
}
T=emp+col_num+row_num+2;
st=1;
ed=T;
Init();
for(i=1; i<=row_num; i++)
{
int pos = i;
int x = row[i].x;
int y = row[i].y;
int cnt_len = 0;
for( y=y+1; y <m ; y++ )
{
if( map[x][y] != -1 )
{
cnt_len++;
Insert_Edge(i+1, row_num+ map[x][y]+1,8);
}
else break;
}
Insert_Edge(st,pos+1,row[i].val-cnt_len);
}
for( i = 1 ; i <=col_num ; i++ )
{
int pos =i+1+row_num+emp;
int x = col[i].x;
int y = col[i].y;
int cnt_len = 0;
for( x=x+1 ; x < n ; x++ )
{
if( map[x][y] != -1 )
{
cnt_len++;
Insert_Edge(row_num+ map[x][y]+1,pos,8);
}
else break;
}
Insert_Edge(pos,ed,col[i].val-cnt_len);
}
sap();
for(i=0; i<n; i++)
{
for(j=0; j<m; j++)
{
if(map[i][j]==-1)
printf("_ ");
else
printf("%d ",print(map[i][j]));
}
printf("\n");
}
}
return 0;
}