题目：
Kakuro Extension
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 1412 Accepted Submission(s): 474
Special Judge
Problem 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.
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Picture of the first sample input Picture of the first sample output
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.
.......— "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
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 _
Author
NotOnlySuccess@HDU
Source
Recommend
lcy
题意：有n*m的格子，黑格子左边的数字代表它所在这一列的连续的白格子的值的和，右边的数字代表它所在这一行连续的白格子的和，求每个白格子要填的数字。
思路：可以把这个图想象成行进列出的网络流，每个黑格子可以拆分成行的值和源点连边，列的值和汇点连边，然后每个白格子与所在行的黑格子，所在列的白格子连边，每条边的容量限制本来是1-9，由于有下限不好处理，就所有值统一减一变为0-8，然后最后的结果都加一即可。每个白格子连的边上的流量即为填的数字。
代码：
#include <cstdlib>
#include <cctype>
#include <cstring>
#include <cstdio>
#include <cmath>
#include<climits>
#include <algorithm>
#include <vector>
#include <string>
#include <iostream>
#include <sstream>
#include <map>
#include <set>
#include <queue>
#include <stack>
#include <fstream>
#include <numeric>
#include <iomanip>
#include <bitset>
#include <list>
#include <stdexcept>
#include <functional>
#include <utility>
#include <ctime>
using namespace std;
#define PB push_back
#define MP make_pair
#define REP(i,x,n) for(int i=x;i<(n);++i)
#define FOR(i,l,h) for(int i=(l);i<=(h);++i)
#define FORD(i,h,l) for(int i=(h);i>=(l);--i)
#define SZ(X) ((int)(X).size())
#define ALL(X) (X).begin(), (X).end()
#define RI(X) scanf("%d", &(X))
#define RII(X, Y) scanf("%d%d", &(X), &(Y))
#define RIII(X, Y, Z) scanf("%d%d%d", &(X), &(Y), &(Z))
#define DRI(X) int (X); scanf("%d", &X)
#define DRII(X, Y) int X, Y; scanf("%d%d", &X, &Y)
#define DRIII(X, Y, Z) int X, Y, Z; scanf("%d%d%d", &X, &Y, &Z)
#define OI(X) printf("%d",X);
#define RS(X) scanf("%s", (X))
#define MS0(X) memset((X), 0, sizeof((X)))
#define MS1(X) memset((X), -1, sizeof((X)))
#define LEN(X) strlen(X)
#define F first
#define S second
#define Swap(a, b) (a ^= b, b ^= a, a ^= b)
#define Dpoint strcut node{int x,y}
#define cmpd int cmp(const int &a,const int &b){return a>b;}
/*#ifdef HOME
freopen("in.txt","r",stdin);
#endif*/
const int MOD = 1e9+7;
typedef vector<int> VI;
typedef vector<string> VS;
typedef vector<double> VD;
typedef long long LL;
typedef pair<int,int> PII;
//#define HOME
int Scan()
{
int res = 0, ch, flag = 0;
if((ch = getchar()) == '-') //判断正负
flag = 1;
else if(ch >= '0' && ch <= '9') //得到完整的数
res = ch - '0';
while((ch = getchar()) >= '0' && ch <= '9' )
res = res * 10 + ch - '0';
return flag ? -res : res;
}
/*----------------PLEASE-----DO-----NOT-----HACK-----ME--------------------*/
const int MAXN=20010;//点数的最大值
const int MAXM=880010;//边数的最大值
const int INF=0x3f3f3f3f;
struct Node
{
int from,to,next;
int cap;
}edge[MAXM];
int tol;
int head[MAXN];
int dep[MAXN];
int gap[MAXN];//gap[x]=y :说明残留网络中dep[i]==x的个数为y
//n是总的点的个数，包括源点和汇点
void init()
{
tol=0;
memset(head,-1,sizeof(head));
}
void addedge(int u,int v,int w)
{
edge[tol].from=u;
edge[tol].to=v;
edge[tol].cap=w;
edge[tol].next=head[u];
head[u]=tol++;
edge[tol].from=v;
edge[tol].to=u;
edge[tol].cap=0;
edge[tol].next=head[v];
head[v]=tol++;
}
void BFS(int start,int end)
{
memset(dep,-1,sizeof(dep));
memset(gap,0,sizeof(gap));
gap[0]=1;
int que[MAXN];
int front,rear;
front=rear=0;
dep[end]=0;
que[rear++]=end;
while(front!=rear)
{
int u=que[front++];
if(front==MAXN)front=0;
for(int i=head[u];i!=-1;i=edge[i].next)
{
int v=edge[i].to;
if(dep[v]!=-1)continue;
que[rear++]=v;
if(rear==MAXN)rear=0;
dep[v]=dep[u]+1;
++gap[dep[v]];
}
}
}
int SAP(int start,int end,int n)
{
int res=0;
BFS(start,end);
int cur[MAXN];
int S[MAXN];
int top=0;
memcpy(cur,head,sizeof(head));
int u=start;
int i;
while(dep[start]<n)
{
if(u==end)
{
int temp=INF;
int inser;
for(i=0;i<top;i++)
if(temp>edge[S[i]].cap)
{
temp=edge[S[i]].cap;
inser=i;
}
for(i=0;i<top;i++)
{
edge[S[i]].cap-=temp;
edge[S[i]^1].cap+=temp;
}
res+=temp;
top=inser;
u=edge[S[top]].from;
}
if(u!=end&&gap[dep[u]-1]==0)//出现断层，无增广路
break;
for(i=cur[u];i!=-1;i=edge[i].next)
if(edge[i].cap!=0&&dep[u]==dep[edge[i].to]+1)
break;
if(i!=-1)
{
cur[u]=i;
S[top++]=i;
u=edge[i].to;
}
else
{
int min=n;
for(i=head[u];i!=-1;i=edge[i].next)
{
if(edge[i].cap==0)continue;
if(min>dep[edge[i].to])
{
min=dep[edge[i].to];
cur[u]=i;
}
}
--gap[dep[u]];
dep[u]=min+1;
++gap[dep[u]];
if(u!=start)u=edge[S[--top]].from;
}
}
return res;
}
int emp=0,rownum=0,colnum=0;
struct A
{
int x;
int y;
int val;
}row[15000],col[15000];
int g[200][200];
int print(int tp)
{
int id=tp+rownum;
int ans=0;
for(int i=head[id];i!=-1;i=edge[i].next)
{
int v=edge[i].to;
if(v<=rownum)
{ans+=edge[i].cap;
break;}
}
return ans+1;
}
int main()
{
int n,m;
while(RII(n,m)!=EOF)
{MS0(g);
init();
char s[10];
emp=0,rownum=0,colnum=0;
for(int i=0;i<n;i++)
for(int j=0;j<m;j++)
{
scanf("%s",s);
if(s[0]=='.')
g[i][j]=++emp;
else
{if(s[4]!='X')
{ int v=(s[4]-'0')*100+(s[5]-'0')*10+s[6]-'0';
row[++rownum].x=i;
row[rownum].y=j;
row[rownum].val=v;
}
if(s[0]!='X')
{
int v=(s[0]-'0')*100+(s[1]-'0')*10+s[2]-'0';
col[++colnum].x=i;
col[colnum].y=j;
col[colnum].val=v;
}
}
}
int S=0,T=emp+rownum+colnum+1;
for(int i=1;i<=rownum;i++)
{
int x=row[i].x;
int y=row[i].y;
int v=row[i].val;
int cnt=0;
for(int j=y+1;j<m;j++)
{
if(g[x][j]!=0)
{
cnt++;
addedge(i,rownum+g[x][j],8);
}
else
break;
}
addedge(0,i,v-cnt);
}
for(int i=1;i<=colnum;i++)
{
int x=col[i].x;
int y=col[i].y;
int v=col[i].val;
int cnt=0;
for(int j=x+1;j<n;j++)
{
if(g[j][y])
{
cnt++;
addedge(rownum+g[j][y],rownum+emp+i,8);
}
else
break;
}
addedge(rownum+emp+i,T,v-cnt);
}
SAP(S,T,T+1);
for(int i=0;i<n;i++)
for(int j=0;j<m;j++)
{
if(g[i][j]==0)
printf("_");
else
{
printf("%d",print(g[i][j]));
}
if(j==m-1)
printf("\n");
else
printf(" ");
}
}
return 0;
}