Seeding

題目描述

It is spring time and farmers have to plant seeds in the field. Tom has a nice field, which is a rectangle with n * m squares. There are big stones in some of the squares.
Tom has a seeding-machine. At the beginning, the machine lies in the top left corner of the field. After the machine finishes one square, Tom drives it into an adjacent square, and continues seeding. In order to protect the machine, Tom will not drive it into a square that contains stones. It is not allowed to drive the machine into a square that been seeded before, either.

Tom wants to seed all the squares that do not contain stones. Is it possible?

輸入

The first line of each test case contains two integers n and m that denote the size of the field. (1 < n, m < 7) The next n lines give the field, each of which contains m characters. ‘S’ is a square with stones, and ‘.’ is a square without stones.

Input is terminated with two 0’s. This case is not to be processed.

輸出

For each test case, print “YES” if Tom can make it, or “NO” otherwise.

樣例

Sample Input4 4 .S.. .S.. .... .... 4 4 .... ...S .... ...S 0 0Sample OutputYES NO

題意

一個N*M的矩陣 里面有“S”和“.”，從左上方0 0點開始能否將.走一邊
直接回溯就行了， 非常裸的一道題，就是考察個回溯

AC代碼

#pragma comment(linker, "/STACK:1024000000,1024000000") #include <cstdio> #include <iostream> #include <cstdlib> #include <cmath> #include <cctype> #include <string> #include <cstring> #include <algorithm> #include <stack> #include <queue> #include <set> #include <map> #include <ctime> #include <vector> #include <fstream> #include <list> #include <iomanip> #include <numeric> using namespace std;#define ls st<<1 #define rs st<<1|1 #define fst first #define snd second #define MP make_pair #define PB push_back #define LL long long #define PII pair<int,int> #define VI vector<int> #define CLR(a,b) memset(a, (b), sizeof(a)) #define ALL(x) x.begin(),x.end() #define rep(i,s,e) for(int i=(s); i<=(e); i++) #define tep(i,s,e) for(int i=(s); i>=(e); i--)const int INF = 0x3f3f3f3f; const int MAXN = 2e2+10; const int mod = 1e9+7; const double eps = 1e-8;void fe() {#ifndef ONLINE_JUDGEfreopen("in.txt", "r", stdin);freopen("out.txt","w",stdout);#endif } LL read() {LL x=0,f=1;char ch=getchar();while (ch<'0'||ch>'9') {if (ch=='-') f=-1;ch=getchar();}while (ch>='0'&&ch<='9') x=(x<<1)+(x<<3)+ch-'0',ch=getchar();return x*f; } int n, m; int k, kk; bool flag; int dx[] = {0, 0, 1, -1}; int dy[] = {1, -1, 0, 0}; char mps[MAXN][MAXN]; bool vis[MAXN][MAXN]; bool check(int x, int y) {return x>=0&&x<n&&y>=0&&y<m&&mps[x][y]!='S'&&(!vis[x][y]); } void dfs(int x, int y) {if(kk == n*m-k) {flag = true;return ;}for(int i = 0; i < 4; i++) {int xx = dx[i]+x;int yy = dy[i]+y;if(check(xx, yy)) {vis[xx][yy] = true;kk++;dfs(xx, yy);kk--;vis[xx][yy] = false;}}} int main(int argc, char const *argv[]) {while(cin >> n >> m,n|m) {CLR(vis, false);CLR(mps, false);flag = false;k = 0, kk = 1;for(int i = 0; i < n; i++) cin >> mps[i];for(int i = 0; i < n; i++) {for(int j = 0; j < m; j++) {if(mps[i][j] == 'S')k++;}}vis[0][0] = true;dfs(0,0);if(flag) cout << "YES\n";elsecout << "NO\n";}return 0; }

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