# HDU 3338 Kakuro Extension （网络流，最大流）

### Description

If you solved problem like this, forget it.Because you need to use a completely different algorithm to solve the following one.
Kakuro puzzle is played on a grid of "black" and "white" cells. Apart from the top row and leftmost column which are entirely black, the grid has some amount of white cells which form "runs" and some amount of black cells. "Run" is a vertical or horizontal maximal one-lined block of adjacent white cells. Each row and column of the puzzle can contain more than one "run". Every white cell belongs to exactly two runs — one horizontal and one vertical run. Each horizontal "run" always has a number in the black half-cell to its immediate left, and each vertical "run" always has a number in the black half-cell immediately above it. These numbers are located in "black" cells and are called "clues".The rules of the puzzle are simple:

1.place a single digit from 1 to 9 in each "white" cell
2.for all runs, the sum of all digits in a "run" must match the clue associated with the "run"

Given the grid, your task is to find a solution for the puzzle.

Picture of the first sample input

Picture of the first sample output

### Input

The first line of input contains two integers n and m (2 ≤ n,m ≤ 100) — the number of rows and columns correspondingly. Each of the next n lines contains descriptions of m cells. Each cell description is one of the following 7-character strings:

.......— "white" cell;
XXXXXXX— "black" cell with no clues;
AAA\BBB— "black" cell with one or two clues. AAA is either a 3-digit clue for the corresponding vertical run, or XXX if there is no associated vertical run. BBB is either a 3-digit clue for the corresponding horizontal run, or XXX if there is no associated horizontal run.
The first row and the first column of the grid will never have any white cells. The given grid will have at least one "white" cell.It is guaranteed that the given puzzle has at least one solution.

### Output

Print n lines to the output with m cells in each line. For every "black" cell print '_' (underscore), for every "white" cell print the corresponding digit from the solution. Delimit cells with a single space, so that each row consists of 2m-1 characters.If there are many solutions, you may output any of them.

### Sample Input

6 6
XXXXXXX XXXXXXX 028\XXX 017\XXX 028\XXX XXXXXXX
XXXXXXX 022\022 ....... ....... ....... 010\XXX
XXX\034 ....... ....... ....... ....... .......
XXX\014 ....... ....... 016\013 ....... .......
XXX\022 ....... ....... ....... ....... XXXXXXX
XXXXXXX XXX\016 ....... ....... XXXXXXX XXXXXXX
5 8
XXXXXXX 001\XXX 020\XXX 027\XXX 021\XXX 028\XXX 014\XXX 024\XXX
XXX\035 ....... ....... ....... ....... ....... ....... .......
XXXXXXX 007\034 ....... ....... ....... ....... ....... .......
XXX\043 ....... ....... ....... ....... ....... ....... .......
XXX\030 ....... ....... ....... ....... ....... ....... XXXXXXX

_ _ 5 8 9
7 6 9 8 4
_ 6 8 _ 7 6
_ 9 2 7 4
_ 7 9 _
_ _ _ _ _ _
1 9 9 1 1 8 6
_ _ 1 7 7 9 1 9
_ 1 3 9 9 9 3 9
_ 6 7 2 4 9 2 _

## 代码

#include<iostream>
#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<algorithm>
using namespace std;

const int maxN=101*101*2;
const int maxM=maxN*100*2;
const int inf=2147483647;
//这两个define分别是把(x,y)对应成一维数组中的位置，即对于每一个格子分配的编号，第二个是把一维数组里的编号转换成二维坐标，并输出，这是为了方便调试
#define pos(x,y) ((x-1)*m+(y-1)%m+1)
#define inpos(x) '['<<x/m+(int)(x%m!=0)<<','<<(x-1)%m+1<<']'

class Edge
{
public:
int u,v,flow;
};

int n,m;
int cnt;
int Next[maxM];
Edge E[maxM];
int Mat1[101][101];//第一个矩阵，存黑格中的左数
int Mat2[101][101];//第二个矩阵，存黑格中的右数
int Ans[101][101];//答案矩阵
char str[10];
int cur[maxN];
int Q[maxN];
int depth[maxN];

bool bfs();
int dfs(int u,int flow);

int main()
{
while (cin>>n>>m)
{
cnt=-1;//多组数据，首先清空
memset(Mat1,-1,sizeof(Mat1));
memset(Mat2,-1,sizeof(Mat2));
for (int i=1;i<=n;i++)
for (int j=1;j<=m;j++)//读入，同时预处理出Mat1和Mat2
{
scanf("%s",str);
if ((str[0]!='X')&&(str[0]!='.'))//如果左数存在
Mat1[i][j]=(str[0]-'0')*100+(str[1]-'0')*10+(str[2]-'0')*1;
if ((str[4]!='X')&&(str[4]!='.'))//如果右数存在
Mat2[i][j]=(str[4]-'0')*100+(str[5]-'0')*10+(str[6]-'0')*1;
if (str[0]=='.')//如果为白格
Mat1[i][j]=0;
if (str[4]=='.')//如果为白格
Mat2[i][j]=0;
}
/*
for (int i=1;i<=n;i++)//输出检查
{
for (int j=1;j<=m;j++)
cout<<Mat1[i][j]<<'/'<<Mat2[i][j]<<" ";
cout<<endl;
}
//*/
for (int i=1;i<=n;i++)//网络流建图，首先建的是黑格左数的
for (int j=1;j<=m;j++)
if (Mat1[i][j]>0)//当找到一个黑格左数存在
{
int k=i+1;//要向下扫描所有的白格，k就是横坐标
while ((Mat1[k][j]==0)&&(k<=n))//只要还是白格并且没有超出范围
{
k=k+1;
}
}
for (int i=1;i<=n;i++)//然后建的是黑格右数
for (int j=1;j<=m;j++)
if (Mat2[i][j]>0)//当找到一个黑格右数存在
{
int k=j+1;//要向右扫描所有的白格，k为其纵坐标
while ((Mat2[i][k]==0)&&(k<=m))
{
k=k+1;
}
}
/*
for (int i=0;i<=cnt;i++)
if (E[i].flow>0)
cout<<inpos(E[i].u)<<" -> "<<inpos(E[i].v)<<" "<<E[i].flow<<endl;
//*/
int Tot=0;//求解最大流
while (bfs())
{
for (int i=0;i<=n*m*2+1;i++)
while (int di=dfs(0,inf))
Tot+=di;
}
//cout<<Tot<<endl;
{
int u=E[i].v;//找到一个黑格左数点
//cout<<inpos(u)<<" "<<E[i].flow<<endl;
{
int v=E[j].v;
int x=v/m+(int)(v%m!=0);//将白格的真实横纵坐标求出来
int y=(v-1)%m+1;
if ((v!=0)&&(E[j].flow<=8)&&(Mat1[x][y]==0))
{
//cout<<inpos(v)<<" "<<E[j].flow<<endl;
Ans[x][y]=8-E[j].flow;//得到答案，8减去残量就是流量，也就是这个格上的数
}
}
//cout<<endl;
}
for (int i=1;i<=n;i++)
{
for (int j=1;j<=m;j++)
if (Mat1[i][j]!=0)
printf("_ ");
else
printf("%d ",Ans[i][j]+1);//注意这里要+1，因为我们求的流量是[0,8]而实际求的数是[1,9]
printf("\n");
}
}
return 0;
}

{
cnt++;
E[cnt].u=u;
E[cnt].v=v;
E[cnt].flow=flow;

cnt++;
E[cnt].u=v;
E[cnt].v=u;
E[cnt].flow=0;
return;
}

bool bfs()
{
memset(depth,-1,sizeof(depth));
int h=1,t=0;
Q[1]=0;
depth[0]=1;
do
{
t++;
int u=Q[t];
{
int v=E[i].v;
if ((depth[v]==-1)&&(E[i].flow>0))
{
depth[v]=depth[u]+1;
h++;
Q[h]=v;
}
}
}
while (t!=h);
if (depth[n*m*2+1]==-1)
return 0;
return 1;
}

int dfs(int u,int flow)
{
if (u==n*m*2+1)
return flow;
for (int &i=cur[u];i!=-1;i=Next[i])
{
int v=E[i].v;
if ((depth[v]==depth[u]+1)&&(E[i].flow>0))
{
int di=dfs(v,min(flow,E[i].flow));
if (di>0)
{
E[i].flow-=di;
E[i^1].flow+=di;
return di;
}
}
}
return 0;
}
posted @ 2017-08-20 15:04 SYCstudio 阅读(...) 评论(...) 编辑 收藏