Vasya has a very beautiful country garden that can be represented as an n × m rectangular field divided into n·m squares. One beautiful day Vasya remembered that he needs to pave roads between k important squares that contain buildings. To pave a road, he can cover some squares of his garden with concrete.
For each garden square we know number aij that represents the number of flowers that grow in the square with coordinates (i, j). When a square is covered with concrete, all flowers that grow in the square die.
Vasya wants to cover some squares with concrete so that the following conditions were fulfilled:
- all k important squares should necessarily be covered with concrete
- from each important square there should be a way to any other important square. The way should go be paved with concrete-covered squares considering that neighboring squares are squares that have a common side
- the total number of dead plants should be minimum
As Vasya has a rather large garden, he asks you to help him.
Input
The first input line contains three integers n, m and k (1 ≤ n, m ≤ 100, n·m ≤ 200, 1 ≤ k ≤ min(n·m, 7)) — the garden‘s sizes and the number of the important squares. Each of the next n lines contains m numbers aij (1 ≤ aij ≤ 1000) — the numbers of flowers in the squares. Next k lines contain coordinates of important squares written as "x y" (without quotes) (1 ≤ x ≤ n, 1 ≤ y ≤ m). The numbers written on one line are separated by spaces. It is guaranteed that all k important squares have different coordinates.
Output
In the first line print the single integer — the minimum number of plants that die during the road construction. Then print n lines each containing m characters — the garden‘s plan. In this plan use character "X" (uppercase Latin letter X) to represent a concrete-covered square and use character "." (dot) for a square that isn‘t covered with concrete. If there are multiple solutions, print any of them.
Sample test(s)
input
3 3 21 2 31 2 31 2 31 23 3
output
9.X..X..XX
input
4 5 41 4 5 1 22 2 2 2 72 4 1 4 53 2 1 7 11 11 54 14 4
output
26X..XXXXXX.X.X..X.XX.原博文：http://blog.csdn.net/gzh1992n/article/details/9119543 不知道最小生成树能不能搞出来
/*
题意：把k个点连起来权值最小，并且输出图
令dp[i][j] 为当前为第i个点，已经连了j状态所需连得点的最小已连接权值
所以就有两个动态转移方程
1：根据边 dp[i][j] = min(dp[k][j] + maze[k][i]) 相同的状态，连接点转移
2：根据状态 dp[i][j] = min(dp[i][j1] + dp[i][j2] - maze[i/m][i%m]) j1,j2为j的子集
dfs找路径
*/
#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <vector>
#include <queue>
#include <cmath>
#include <string>
#include <map>
using namespace std;
const int inf = 1 << 29;
const int MAX = 200 + 10;
int dp[MAX][1<<7], pre[MAX][1<<7];
int n, m, k, nn, mm;
int hash1[MAX];
int maz[MAX][MAX];
char g[MAX][MAX];
bool visit[MAX][1<<7];
int dx[] = {0, 0, -1, 1};
int dy[] = {-1, 1, 0, 0};
struct Node {
int u, st;
Node(int _u, int _st){
u = _u, st = _st;//为了类型转换
}
};
queue<Node> que;
bool check(int x, int y) {
if(x >= 0 && x < n && y >= 0 && y < m) return true;
return false;
}
//update(u, hash[u], maz[a][b], -1);
void update(int u, int st, int w, int fa) {
if(dp[u][st] > w) {
dp[u][st] = w;
pre[u][st] = fa;//标记的点
if(!visit[u][st]) {
que.push(Node(u, st));
visit[u][st] = true;
}
}
}
void dfs(int u, int st) {
int x = u / m, y = u % m;
g[x][y] = ‘X‘;
if(pre[u][st] == -1) return;//如果到了标记的点就停止
else {
int v = pre[u][st] / 1000, stt = pre[u][st] % 1000;
dfs(v, stt);
if(stt - st) dfs(v, st - stt);//拆分当前状态
}
}
void solve() {//spfa
while(!que.empty()) {
Node now = que.front();
que.pop();
int u = now.u, x = now.u / m, y = now.u % m, st = now.st;
visit[u][st] = false;//把状态拆分
for(int i = 0; i < 4; i++) {
int xx = x + dx[i], yy = y + dy[i];
if(!check(xx, yy)) continue;
int v = xx * m + yy;
update(v, st, dp[u][st] + maz[xx][yy], u * 1000 + st);//点的动态方程
}
int t = mm - 1 - st;
for(int i = t; i; i = (i-1) & t) {
update(u, i | st, dp[u][i] + dp[u][st] - maz[x][y], u * 1000 + st);//状态的动态方程，因为是同一个点要减去一个重
}
}
int ans = inf, u;
for(int i = 0; i < nn; i++) {
if(ans > dp[i][mm-1]) {
ans = dp[i][mm-1];
u = i;
}
}//找点中最小值
dfs(u, mm - 1);//DFS找路径
cout << ans << endl;
for(int i = 0; i < n; i++) {
for(int j = 0; j < m; j++)
cout << g[i][j];
cout << endl;
}
}
int main() {
while(cin >> n >> m >> k) {
while(!que.empty()) que.pop();
for(int i = 0; i < n; i++) {
for(int j = 0; j < m; j++) {
cin >> maz[i][j];
g[i][j] = ‘.‘;//初始化图
}
}
nn = n * m, mm = 1 << k;//把图点化
memset(hash1, 0, sizeof(hash1));
memset(visit, false, sizeof(visit));
for(int i = 0; i < nn; i++)
for(int j = 0; j < mm; j++)
dp[i][j] = inf;//总点的编号，需要连得点的编号
for(int i = 0, a, b; i < k; i++) {
cin >> a >> b;
a--, b--;
int u = a * m + b;
hash1[u] = 1 << i;//记录状态
update(u, hash1[u], maz[a][b], -1);
}
solve();
}
return 0;
}