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
InputThe 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.OutputPrint 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
Sample Output
_ _ _ _ _ _
_ _ 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 _题意：
给定横着的和和竖着的和，输出可行解．
思路:
将横着的限制看成一个点，竖着的限制看成一个点，白色方块在中间即可．
白块限制流量１～9,本来应该是上下界网络流，但是因为每一条的边的下界是一样的，所以通过减一处理即可转换为最大流．
#include<iostream>
#include<algorithm>
#include<vector>
#include<stack>
#include<queue>
#include<map>
#include<set>
#include<cstdio>
#include<cstring>
#include<cmath>
#include<ctime>#define fuck(x) cerr<<#x<<" = "<<x<<endl;
#define debug(a, x) cerr<<#a<<"["<<x<<"] = "<<a[x]<<endl;
#define ls (t<<1)
#define rs ((t<<1)|1)
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
const int loveisblue = ;
const int maxn = ;
const int maxm = ;
const int inf = 0x3f3f3f3f;
const ll Inf = ;
const int mod = ;
const double eps = 1e-;
const double pi = acos(-);int Head[maxn],cnt;
struct edge{
int Next,v,w;
}e[maxm];
void add_edge(int u,int v,int w){
// cout<<u<<" "<<v<<" "<<w<<endl;
e[cnt].Next=Head[u];
e[cnt].v=v;
e[cnt].w=w;
Head[u]=cnt++; e[cnt].Next=Head[v];
e[cnt].v=u;
e[cnt].w=;
Head[v]=cnt++;
}int D_vis[maxn],D_num[maxn];
int source,meeting;
bool bfs()
{ memset(D_vis,,sizeof(D_vis));
for(int i=;i<=meeting;i++){//注意要覆盖所有点
D_num[i]=Head[i];
}
D_vis[source]=;
queue<int>q;
q.push(source);
int r=;
while(!q.empty()){
int u=q.front();
q.pop();
int k=Head[u];
while(k!=-){
if(!D_vis[e[k].v]&&e[k].w){
D_vis[e[k].v]=D_vis[u]+;
q.push(e[k].v);
}
k=e[k].Next;
}
}
// fuck(meeting)
return D_vis[meeting];
}
int dfs(int u,int f)
{
if(u==meeting){return f;}
int &k=D_num[u];
while(k!=-){
if(D_vis[e[k].v]==D_vis[u]+&&e[k].w){
int d=dfs(e[k].v,min(f,e[k].w));
if(d>){
e[k].w-=d;
e[k^].w+=d;
return d;
}
}
k=e[k].Next;
}
return ;
}
int Dinic()
{
int ans=;
while(bfs()){
int f;
while((f=dfs(source,inf))>){
ans+=f;
}
}
return ans;
}char s[][][];
int mp1[][];
int mp2[][];
int mph[][];
int mps[][];
int mpk[][];
int cal(char a,char b,char c){
return (a-)*+(b-)*+c-;
}int main() {
// ios::sync_with_stdio(false);
// freopen("in.txt", "r", stdin); int n,m;
while (scanf("%d%d",&n,&m)!=EOF){
memset(Head,-,sizeof(Head));
memset(mp1,,sizeof(mp1));
memset(mp2,,sizeof(mp2));
cnt = ;
int cur = ;
for(int i=;i<=n;i++){
for(int j=;j<=m;j++){
scanf("%s",s[i][j]);
if(s[i][j][]=='.'){
cur++;
mp1[i][j]=cur;
cur++;
mp2[i][j]=cur;
mpk[i][j]=cnt;
add_edge(mp1[i][j],mp2[i][j],);
}
}
}
source = ;
meeting = ; for(int i=;i<=n;i++){
for(int j=;j<=m;j++){
if(s[i][j][]!='.'&&s[i][j][]!='X'){
cur++;
mps[i][j]=cur;
int sum = cal(s[i][j][],s[i][j][],s[i][j][]);
for(int k=i+;k<=n;k++){
if(!mp1[k][j]){ break;}
add_edge(cur,mp1[k][j],inf);
sum--;
}
add_edge(source,cur,sum);
}if(s[i][j][]!='.'&&s[i][j][]!='X'){
cur++;
mph[i][j]=cur;
int sum = cal(s[i][j][],s[i][j][],s[i][j][]);
for(int k=j+;k<=m;k++){
if(!mp2[i][k]){ break;}
add_edge(mp2[i][k],cur,inf);
sum--;
}
add_edge(cur,meeting,sum);
}
}
}
int ans = Dinic();
// fuck(ans)
// fuck("????")
for(int i=;i<=n;i++){
for(int j=;j<m;j++){
if(mp1[i][j]==){
printf("_ ");
}else{
printf("%d ",-e[mpk[i][j]].w+);
}
}
if(mp1[i][m]==){
printf("_\n");
}else{
printf("%d\n",-e[mpk[i][m]].w+);
}
} } return ;
}
