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 = 16;

int n, m, cnt;
int head, tail;
pair<int,int> que[maxn*maxn];
int dx[] = { 1,-1,0,0 };
int dy[] = { 0,0,-1,1 };
char gra[maxn][maxn];
bool vis[maxn][maxn],vvis[maxn][maxn];
string ans, tmp;

int h(int x,int y) {
    head = tail = 0;
    int cnt = 0;
    que[tail++] = make_pair(x, y);
    memcpy(vvis, vis, sizeof(vis));

    while (head < tail) {
        pair<int, int> temp = que[head++];
        for (int i = 0; i < 4; i++) {
            int xx = temp.first + dx[i], yy = temp.second + dy[i];
            if (0 <= xx && 0 <= 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 = 0; i < 4; i++) {
        int xx = x + dx[i], yy = y + dy[i];
        if (0 <= xx && 0 <= yy && xx < n && yy < m && !vis[xx][yy] && gra[xx][yy] != '#') {
            vis[xx][yy] = true;
            dfs(xx, yy, s + gra[xx][yy], deep + 1);
            vis[xx][yy] = false;
        }
    }
}

int main()
{
    //freopen("input.txt", "r", stdin);
    while (~scanf("%d%d", &n, &m) && (n || m)) {
        ans = "";
        for (int i = 0; i < n; i++) {
            scanf("%s", gra[i]);
        }
        
        for (int i = 0; i < n; i++) {
            for (int j = 0; j < m; j++) {
                if (isdigit(gra[i][j])) {
                    tmp = "";
                    tmp += gra[i][j];
                    vis[i][j] = true;
                    dfs(i, j, tmp, 1);
                    vis[i][j] = false;
                }
            }
        }
        cout << ans << endl;
    }
    return 0;
}