UVA11882-Biggest Number(DFS+最优化剪枝)

Problem UVA11882-Biggest Number

Accept: 177    Submit: 3117
Time Limit: 1000 mSec    Memory Limit : 128MB

Problem Description

Input

There will be at most 25 test cases. Each test begins with two integers R and C (2 ≤ R,C ≤ 15, R∗C ≤ 30), the number of rows and columns of the maze. The next R rows represent the maze. Each line contains exactly C characters (without leading or trailing spaces), each of them will be either ‘#’ or one of the nine non-zero digits. There will be at least one non-obstacle squares (i.e. squares with a non-zero digit in it) in the maze. The input is terminated by a test case with R = C = 0, you should not process it.

 Output

For each test case, print the biggest number you can find, on a single line.

 

 Sample Input

3 7

##9784#

##123##

##45###

0 0

 

 Sample Output

791452384

题解：一看题目觉得是个大水题，dfs一下就行，然后果断TLE，这个题对效率要求还是比较高的，两个比较重要的剪枝：

1、如果剩余最长可走长度加上目前已有长度小于答案的长度，直接return.

2、如果剩余最长可走长度加上目前已有长度等于答案的长度，比较字典序，如果答案更优就retrun.

 #include <bits/stdc++.h>
 
 using namespace std;
 
 const int maxn = ;
 
 int n, m, cnt;
 int head, tail;
 pair<int,int> que[maxn*maxn];
 int dx[] = { ,-,, };
 int dy[] = { ,,-, };
 char gra[maxn][maxn];
 bool vis[maxn][maxn],vvis[maxn][maxn];
 string ans, tmp;
 
 int h(int x,int y) {
     head = tail = ;
     int cnt = ;
     que[tail++] = make_pair(x, y);
     memcpy(vvis, vis, sizeof(vis));
 
     while (head < tail) {
         pair<int, int> temp = que[head++];
         for (int i = ; i < ; i++) {
             int xx = temp.first + dx[i], yy = temp.second + dy[i];
             if ( <= xx &&  <= yy && xx < n && yy < m && !vvis[xx][yy] && gra[xx][yy] != '#') {
                 vvis[xx][yy] = true;
                 cnt++;
                 que[tail++] = make_pair(xx, yy);
             }
         }
     }
     return cnt;
 }
 
 void update(const string& s) {
     int len1 = s.size(), len2 = ans.size();
     if (len2 < len1 || (len2 == len1 && ans < s)) ans = s;
 }
 
 void dfs(int x,int y,string s,int deep) {
     int hh = h(x, y);
     int l = ans.size();
     if (deep + hh < l) return;
     if (deep + hh == l) {
         if (s + "z" < ans) return;
     }
 
     update(s);
 
     for (int i = ; i < ; i++) {
         int xx = x + dx[i], yy = y + dy[i];
         if ( <= xx &&  <= yy && xx < n && yy < m && !vis[xx][yy] && gra[xx][yy] != '#') {
             vis[xx][yy] = true;
             dfs(xx, yy, s + gra[xx][yy], deep + );
             vis[xx][yy] = false;
         }
     }
 }
 
 int main()
 {
     //freopen("input.txt", "r", stdin);
     while (~scanf("%d%d", &n, &m) && (n || m)) {
         ans = "";
         for (int i = ; i < n; i++) {
             scanf("%s", gra[i]);
         }
 
         for (int i = ; i < n; i++) {
             for (int j = ; j < m; j++) {
                 if (isdigit(gra[i][j])) {
                     tmp = "";
                     tmp += gra[i][j];
                     vis[i][j] = true;
                     dfs(i, j, tmp, );
                     vis[i][j] = false;
                 }
             }
         }
         cout << ans << endl;
     }
     return ;
 }