題目鏈接

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” cell2.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 _

AC

• 根據(jù)游戲規(guī)則，每行每列的數(shù)字之和只受到相應(yīng)的數(shù)字要求，這樣可以用最大流建邊寫
• 建邊：
  
• 假設(shè)流量是從上流入，然后從左流出
• 如果只有下邊的數(shù)字，這個(gè)數(shù)字是它下面所有的白色格子的和，所以將這個(gè)格子和它下面所有的白格子建邊，權(quán)值為8（最大流的出的可能有0的流量），將這個(gè)格子和源點(diǎn)相連（默認(rèn)流量從上流入）權(quán)值為數(shù)字 - 下方白色格子的個(gè)數(shù)（因?yàn)槊總€(gè)白格子建邊的時(shí)候都減1）
• 同理只有右邊的數(shù)字，就讓右邊的格子和這個(gè)格子建邊（注意建邊方向，流量從左流出），權(quán)值為8，并將這個(gè)格子和匯點(diǎn)建邊，權(quán)值為數(shù)字 - 右邊白色格子的數(shù)量
• 如果兩個(gè)數(shù)字同時(shí)存在，就拆點(diǎn)建邊，建邊過(guò)程同上
• 因?yàn)樾枰袛嗝總€(gè)格子的信息，所以可以用一個(gè)結(jié)構(gòu)體來(lái)存放每個(gè)格子的信息
  
• 類型（黑色， 白色， 有數(shù)字）
• 數(shù)字的大小
• 100 * 100 的圖，所以最多可以建100 * 100 * 2 * 2 條邊，（建邊的時(shí)候默認(rèn)兩條）
• head數(shù)組應(yīng)該開100 * 100 * 2
• 最后跑一邊Dinic

#include <iostream> #include <stdio.h> #include <map> #include <vector> #include <queue> #include <algorithm> #include <cmath> #define N 20010 #include <cstring> #define ll long long #define P pair<int, int> #define mk make_pair using namespace std; struct ac{int v, c, pre; }edge[40001]; int head[N], dis[N], curedge[N], cnt; int inf = 0x3f3f3f3f; void addedge(int u, int v, int c) {edge[cnt].v = v;edge[cnt].c = c;edge[cnt].pre = head[u];head[u] = cnt++;swap(u, v);edge[cnt].v = v;edge[cnt].c = 0;edge[cnt].pre = head[u];head[u] = cnt++; } bool bfs(int s, int e) {queue<int> que;que.push(s);memset(dis, 0, sizeof(dis));dis[s] = 1;while (!que.empty()) {int t = que.front();que.pop();for (int i = head[t]; i != -1; i = edge[i].pre) {if (dis[edge[i].v] || edge[i].c == 0) continue;dis[edge[i].v] = dis[t] + 1;que.push(edge[i].v);} }return dis[e] != 0; } int dfs(int s, int e, int flow) {if (s == e) return flow;for (int &i = curedge[s]; i != -1; i = edge[i].pre) {if (dis[edge[i].v] == dis[s] + 1 && edge[i].c) {int d = dfs(edge[i].v, e, min(flow, edge[i].c));if (d > 0) {edge[i].c -= d;edge[i ^ 1].c += d;return d; }}}return 0; } int solve(int s, int e) {int sum = 0;while (bfs(s, e)) {for (int i = 0; i <= e; ++i) {curedge[i] = head[i];}int d;while (d = dfs(s, e, inf)) {sum += d;}}return sum; } // 每個(gè)格子的信息 struct point{// type 標(biāo)記各自的類型// 黑色：-1， 白色0， 只有下1， 只有右2， 兩個(gè)都有3 // r 右邊數(shù)字的和 // d 下方數(shù)字的和 int type, r, d; }cell[120][120]; int main() { #ifndef ONLINE_JUDGEfreopen("in.txt", "r", stdin); #endif // ios::sync_with_stdio(false);int n, m;while (cin >> n >> m) {memset(head, -1, sizeof(head));cnt = 0;string s;// 讀入方格信息 for (int i = 1; i <= n; ++i) {for (int j = 1; j <= m; ++j) {cin >> s;if (s == "XXXXXXX") {cell[i][j].type = -1;}else if (s == ".......") {cell[i][j].type = 0;}else {int num1 = 0, num2 = 0;if (s[0] != 'X') {for (int k = 0; k < 3; ++k) {num1 = num1 * 10 + s[k] - '0';}}if (s[4] != 'X') {for (int k = 4; k <= 6; ++k) {num2 = num2 * 10 + s[k] - '0';}}if (num1 && num2) {cell[i][j].type = 3;cell[i][j].r = num2;cell[i][j].d = num1;}else if (num1) {cell[i][j].type = 1;cell[i][j].d = num1;}else {cell[i][j].type = 2;cell[i][j].r = num2;}}}}// 定義源點(diǎn)和匯點(diǎn) int start = 0, end = n * m * 2 + 1;for (int i = 1; i <= n; ++i) {for (int j = 1; j <= m; ++j) {int type = cell[i][j].type;if (type == -1 || type == 0) continue;if (type == 1) {int sum = 0;for (int k = i + 1; k <= n; ++k) {if (cell[k][j].type != 0) break;sum++;addedge(i * m - m + j, k * m - m + j, 8);}addedge(start, i * m - m + j, cell[i][j].d - sum);}else if (type == 2) {int sum = 0;for (int k = j + 1; k <= m; ++k) {if (cell[i][k].type != 0) break;sum++;addedge(i * m - m + k, i * m - m + j, 8);}addedge(i * m - m + j, end, cell[i][j].r - sum);}else if (type == 3) {// 拆點(diǎn) int sum;// 向下 sum = 0;for (int k = i + 1; k <= n; ++k) {if (cell[k][j].type != 0) break;sum++;addedge(i * m - m + j, k * m - m + j, 8);}addedge(start, i * m - m + j, cell[i][j].d - sum);// 向右 sum = 0;for (int k = j + 1; k <= m; ++k) {if (cell[i][k].type != 0) break;sum++;addedge(i * m - m + k, i * m - m + j + n * m, 8);}addedge(i * m - m + j + n * m, end, cell[i][j].r - sum);}}}solve(start, end);for (int i = 1; i <= n; ++i) {for (int j = 1; j <= m; ++j) {if (cell[i][j].type != 0) cout << "_";else {int sum = 0;cout << 8 - edge[head[i * m - m + j]].c + 1;}if (j == m) cout << endl; else cout << " ";}}}return 0; }

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