HDU3338Kakuro Extension(最大流+边的流量)

题目传送门：http://acm.hdu.edu.cn/showproblem.php?pid=2883

Kakuro Extension

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1986    Accepted Submission(s): 692
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 6XXXXXXX XXXXXXX 028\XXX 017\XXX 028\XXX XXXXXXXXXXXXXX 022\022 ....... ....... ....... 010\XXXXXX\034 ....... ....... ....... ....... .......XXX\014 ....... ....... 016\013 ....... .......XXX\022 ....... ....... ....... ....... XXXXXXXXXXXXXX XXX\016 ....... ....... XXXXXXX XXXXXXX5 8XXXXXXX 001\XXX 020\XXX 027\XXX 021\XXX 028\XXX 014\XXX 024\XXXXXX\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 _

题目大意：

原数谜是个很有趣的游戏，每一行或每一列空白称为一个回，每一回都对应着一个整数sum，sum就是这回的和。这些空白格里只能填入1—9这九个数字，每一回中可以重复。全黑色的格为空，有数字的格，左下角的表示列的和，右上角的表示行的和，则可以得到第二个图。

分析：

一个数等于若干个数的和，可以看做一条入流分解为若干条出流，入流量等于总的出流量。每个格子(i,j)可以由行和列两个坐标确定，所以可以建立行的点和列的点，代表i行的点向代表j列的点连接一条边（容量范围【1,9】）就是代表(i，j)这个格子。具有行总和的黑色格子，可以表示为流入行点的总流量，具有列总和的黑色格子，可以表示该列流出流量的和。某些行(列)可能有多个约束总和，我们可以将每个总和都看做单独一行（列），所以实际的行数和列数并不一定等于原图的。还要记录每个格子对应那一条边，最后流过那条边的总流量就是要填的数字。由于流量有上下限限制，可以给每个数都减掉1，对应的和也对应减去几，则填出来的数字范围为0—8, 就可以用单纯的网络流搞定了。求出来后再加上1就可以了，这样就没有下界需要处理了。

建图：

一共有四类点：

1. 构造源点S，汇点T

2. 有行和的格子，此类节点设为A

3. 空白格，设为B

4. 有列和的格子，设为C

则可以建边：

1. <S, A> 容量和行和

2. <A, B> 容量为8

3. <B, C> 容量为8

4. <C, T> 容量为列和

收获：

该题中对边需要求解实际流量的边进行记录，求出最大流后，则可以得到实际流量。

#include<iostream>#include<algorithm>#include<cstring>#include<cstdio>#include<vector>#include<queue>#include<set>#include<map>using namespace std;#define PI acos(-1.0)#define eps 1e-8#define ll long long#define MEM(a, b) memset(a, b, sizeof(a))#define pb push_back#define MII map<int,int>::iterator#define MLL map<LL,LL>::iterator#define pii pair<int,int>#define SI set<int>::iterator#define SL set<LL>::iterator#define dug printf("bug-------bug-------bug\n")const int maxn = 10005;const int inf = 0x3f3f3f3f;struct Edge{    int to, cap, nxt;    Edge(){}    Edge(int t, int c, int nx):to(t), cap(c), nxt(nx){}};int head[maxn], tol;Edge edge[100*maxn];int n, m;int S, T;struct Val{    int fir, sec;    int emp;};Val mp[105][105];int r[105][105], c[105][105], bel[105][105];int row[maxn], col[maxn];bool vis[maxn];void AddEdge(int u, int v, int cap){    edge[tol] = Edge(v, cap, head[u]);    head[u] = tol++;    edge[tol] = Edge(u, 0, head[v]);    head[v] = tol++;}Val turn(char s[]){    Val ret;    if(s[0] == 'X' || s[0] == '.')        ret.fir = -1;    else    {        int tmp = 0;        for(int i = 0; i < 3; i++)            tmp = tmp * 10 + s[i] - '0';        ret.fir = tmp;    }    if(s[4] == 'X' || s[4] == '.')        ret.sec = -1;    else    {        int tmp = 0;        for(int i = 0; i < 3; i++)            tmp = tmp * 10 + s[i+4] - '0';        ret.sec = tmp;    }    ret.emp = (s[0] == '.' && s[4] == '.');    return ret;}void init(){    tol = 0;    memset(head, -1, sizeof(head));    char s[8];    for(int i = 1; i <= n; i++)        for(int j = 1; j <= m; j++)        {            scanf("%s", s);            mp[i][j] = turn(s);        }    S = 0, T = n*m + 1;    int rowc = 1, colc = 1;    int cnt = 1;    for(int i = 1; i <= n; i++)        for(int j = 1; j <= m; j++)        {            Val cur = mp[i][j];            if(cur.emp)                continue;            if(cur.sec != -1)            {                int ed = j + 1;                while(ed <= m && mp[i][ed].emp)                    r[i][ed] = rowc, ed++;                row[rowc] = cnt++;                AddEdge(0, row[rowc++], cur.sec - (ed - j - 1));            }            if(cur.fir != -1)            {                int ed = i + 1;                while(ed <= n && mp[ed][j].emp)                    c[ed][j] = colc, ed++;                col[colc] = cnt++;                AddEdge(col[colc++], T, cur.fir - (ed - i - 1));            }        }    memset(bel, -1, sizeof(bel));    for(int i = 1; i <= n; i++)        for(int j = 1; j <= m; j++)        if(mp[i][j].emp)        {            AddEdge(row[r[i][j]], col[c[i][j]], 8);            bel[i][j] = tol - 2;        }}int d[maxn];bool BFS(){    queue<int> q;    memset(vis, false, sizeof(vis));    vis[S] = true;    d[S] = 0;    q.push(S);    while(!q.empty())    {        int u = q.front();        q.pop();        for(int i = head[u]; i != -1; i = edge[i].nxt)        {            int v = edge[i].to, c = edge[i].cap;            if(!vis[v] && c > 0)            {                vis[v] = true;                d[v] = d[u] + 1;                q.push(v);            }        }    }    return vis[T];}int cur[maxn];int dfs(int u, int T, int a){    if(u == T || a == 0)        return a;    int flow = 0, f;    for(int &i = cur[u]; i != -1; i = edge[i].nxt)    {        int v = edge[i].to, c = edge[i].cap, r = i^1;        if(d[v] == d[u] + 1 && c > 0 && (f = dfs(v, T, min(c, a))) > 0)        {            edge[i].cap -= f;            edge[r].cap += f;            flow += f;            a -= f;            if(a == 0)                break;        }    }    return flow;}int dinic(int S, int T){    int ret = 0;    while(BFS())    {        for(int i = 0; i <= T; i++)            cur[i] = head[i];        int f = inf;        //while((f = dfs(S, T, f)) > 0)        f = dfs(S, T, f);            ret += f;    }    return ret;}int ans[maxn][maxn];int main(){    while(cin >> n >> m)    {        init();        int t = dinic(S, T);        for(int i = 1; i <= n; i++)            for(int j = 1; j <= m; j++)            {                if(bel[i][j] == -1)                {                    ans[i][j] = 0;                    continue;                }                ans[i][j] = 9 - edge[bel[i][j]].cap;            }        for(int i = 1; i <= n; i++)            for(int j = 1; j <= m; j++)            {                if(!ans[i][j])                    putchar('_');                else                    printf("%d", ans[i][j]);                putchar(j == m ? '\n':' ');            }    }    return 0;}