hdu 3338 Kakuro Extension(最大流)

題目：

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.
　　　　　　　　 　　　　　
　　　　　　　　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

 

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

HDOJ Monthly Contest – 2010.03.06

 

Recommend

lcy

題意：有n*m的格子，黑格子左邊的數字代表它所在這一列的連續的白格子的值的和，右邊的數字代表它所在這一行連續的白格子的和，求每個白格子要填的數字。

思路：可以把這個圖想象成行進列出的網絡流，每個黑格子可以拆分成行的值和源點連邊，列的值和匯點連邊，然後每個白格子與所在行的黑格子，所在列的白格子連邊，每條邊的容量限制本來是1-9，由於有下限不好處理，就所有值統一減一變爲0-8，然後最後的結果都加一即可。每個白格子連的邊上的流量即爲填的數字。

代碼：

#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()) == '-')				//判斷正負
		flag = 1;

	else if(ch >= '0' && ch <= '9')			//得到完整的數
		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;//點數的最大值
const int MAXM=880010;//邊數的最大值
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 :說明殘留網絡中dep[i]==x的個數爲y

//n是總的點的個數，包括源點和匯點

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)//出現斷層，無增廣路
          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;
}