任务说明：搜索可以穷举各种情况。很多题目都可以用搜索完成。就算不能，搜索也是骗分神器。
P1219 八皇后
直接dfs。对角线怎么判断：同一条对角线的横纵坐标的和或者差相同。
1 #include <iostream> 2 #include <vector> 3 #include <cstdio> 4 #include <cstdlib> 5 #include <cmath> 6 7 using namespace std; 8 9 static int total = 0; 10 vector<vector<int>> ans; 11 12 void print(int t, vector<vector<bool>>& chess) { 13 printf("the %dth answer ", t); 14 for(int i = 0; i < chess.size(); ++i) { 15 for(int j = 0; j < chess[i].size(); ++j) { 16 //printf("%d", chess[i][j]); 17 cout << chess[i][j]; 18 } 19 printf(" "); 20 } 21 } 22 23 void print(int t, vector<int>& current_ans) { 24 //printf("the %dth answer ", t); 25 printf("%d", current_ans[0]); 26 for(int i = 1; i < current_ans.size(); ++i) { 27 printf(" %d", current_ans[i]); 28 } 29 printf(" "); 30 } 31 32 33 void queuens(vector<vector<bool>>& chess, int i, const int N, 34 vector<int>& cols, vector<int>& diagnals, vector<int>& anti_diagnals, 35 vector<int>& current_answer) { 36 if (i == N) { 37 total++; 38 if (total <= 3) { 39 print(total, current_answer); 40 } 41 return; 42 } 43 for(int col = 0; col < N; ++col) { 44 if(cols[col] == 0 && anti_diagnals[i-col+N] == 0 && diagnals[i+col] == 0) { 45 cols[col] = 1; anti_diagnals[i-col+N] = 1 ; diagnals[i+col] = 1; 46 current_answer[i] = col+1; 47 queuens(chess, i+1, N, cols, diagnals, anti_diagnals, current_answer); 48 cols[col] = 0; anti_diagnals[i-col+N] = 0 ; diagnals[i+col] = 0; 49 current_answer[i] = 0; 50 } 51 } 52 } 53 54 55 int main() { 56 int N; 57 cin >> N; 58 vector<vector<bool>> chess(N, vector<bool>(N, false)); 59 vector<int> diagnals(2*N-1, 0); 60 vector<int> anti_diagnals(2*N-1, 0); 61 vector<int> cols(N, 0); 62 vector<int> current_answer(N, 0); 63 queuens(chess, 0, N, cols, diagnals, anti_diagnals, current_answer); 64 printf("%d ", total); 65 return 0; 66 }
P1019 单词接龙
做的时候没写题解，结果现在忘记了。。。orz
1 #include <iostream> 2 #include <cstdio> 3 #include <cstdlib> 4 #include <utility> 5 #include <string> 6 #include <cstring> 7 #include <vector> 8 using namespace std; 9 10 typedef pair<string, int> P; 11 static int maxans = 0; 12 13 14 void print (vector<string>& vecStr) { 15 printf("vector dragon "); 16 for(auto ele : vecStr) { 17 printf("%s ", ele.c_str()); 18 } 19 printf(" "); 20 } 21 22 void dfs(int level, vector<P>& vecWord, int& ans, vector<string>& dragon) 23 { 24 string lastWord = dragon[level-1]; 25 //printf("info for debug: level[%d] ", level); 26 //printf("lastWord[%s]", lastWord.c_str()); 27 //print(dragon); 28 29 for (int i = 0; i < vecWord.size(); ++i) { 30 if(vecWord[i].second < 2) { 31 string curWord = vecWord[i].first; 32 int startIdx; 33 if (level == 1) { startIdx = 0; } 34 else {startIdx = 1;} 35 for (int j = lastWord.size() - 1; j >= startIdx; --j) { 36 string lssubstr = lastWord.substr(j); 37 if (curWord.find(lssubstr) != 0 || curWord == lssubstr) { continue; } 38 vecWord[i].second++; 39 dragon[level] = curWord; 40 ans += curWord.size() - lssubstr.size(); 41 if (ans > maxans) {maxans = ans;} 42 dfs(level+1, vecWord, ans, dragon); 43 dragon[level] = ""; 44 vecWord[i].second--; 45 ans -= curWord.size() - lssubstr.size(); 46 } 47 } 48 } 49 } 50 51 int main() { 52 int n; 53 cin >> n; 54 vector<P> vecWord(n); 55 for(int i = 0; i < n; ++i) { 56 string str; 57 cin >> str; 58 P p; 59 p.first = str; 60 p.second = 0; 61 vecWord[i] = p; 62 } 63 vector<string> vecDragon(2*n+1); 64 string strDragon; 65 cin >> strDragon; 66 67 vecDragon[0] = strDragon; 68 int ans = 1; 69 dfs(1, vecWord, ans, vecDragon); 70 cout << maxans << endl; 71 return 0; 72 }
P1101 单词方阵
就是在一个字母方阵中画出 “yizhong” 这个单词，八种方向都ok，但是一旦选定了方向，就不能换了。
如图：
解法：我的解法就是用一个相同的布尔方阵去决定这个字母是否显示，dfs决定use方阵的01。
1 #include <iostream> 2 #include <cstdio> 3 #include <cstdlib> 4 #include <utility> 5 #include <string> 6 #include <cstring> 7 #include <vector> 8 using namespace std; 9 10 string pattern = "yizhong"; 11 12 void print (vector<vector<char>>& chess, vector<vector<bool>>& use) { 13 for(int i = 0; i < chess.size(); ++i) { 14 for (int j = 0; j < chess.size(); ++j) { 15 if (use[i][j]) { printf("%c", chess[i][j]);} 16 else { printf("*"); } 17 } 18 printf(" "); 19 } 20 } 21 22 bool dfs(vector<vector<char>>& chess, vector<vector<bool>>& use, int level, const int addx, const int addy, int x, int y) { 23 if (level == 7) { 24 return true; 25 } 26 if (x+addx >= 0 && x+addx < chess.size() && y+addy >= 0 && y+addy < chess.size()) { 27 if (pattern[level] == chess[x+addx][y+addy]) { 28 bool ans = dfs(chess, use, level+1, addx, addy, x+addx, y+addy); 29 if (ans) { 30 use[x+addx][y+addy] = ans; 31 } 32 return ans; 33 } 34 else { 35 return false; 36 } 37 } else { 38 return false; 39 } 40 } 41 42 void findword(vector<vector<char>>& chess, vector<vector<bool>>& use, int x, int y) { 43 for(int addx = -1; addx <= 1; ++addx) { 44 for(int addy = -1; addy <= 1; ++addy) { 45 if (addx == 0 && addy == 0) { continue; } 46 if (x+addx >= 0 && x+addx < chess.size() && y+addy >= 0 && y+addy < chess.size()) { 47 bool ans = dfs(chess, use, 1, addx, addy, x, y); 48 if (ans) { 49 use[x][y] = ans; 50 } 51 } 52 } 53 } 54 55 } 56 57 58 int main() { 59 int n; 60 cin >> n; 61 vector<vector<char>> chess(n, vector<char>(n)); 62 63 int x = -1, y = -1; 64 for(int i = 0; i < n; ++i) { 65 for (int j = 0; j < n; ++j) { 66 cin >> chess[i][j]; 67 if (chess[i][j] == 'y' && x == -1 && y == -1) { 68 x = i, y = j; 69 } 70 } 71 } 72 vector<vector<bool>> use(n, vector<bool>(n, false)); 73 if (x == -1 && y == -1) { 74 print(chess, use); 75 return 0; 76 } 77 78 for(int i = 0; i < chess.size(); ++i) { 79 for(int j = 0; j < chess[0].size(); ++j) { 80 if (chess[i][j] == 'y') { 81 findword(chess, use, i, j); 82 } 83 } 84 } 85 print(chess, use); 86 87 88 return 0; 89 }
P1605 迷宫
给定一个N*M方格的迷宫，迷宫里有T处障碍，障碍处不可通过。给定起点坐标和终点坐标，问: 每个方格最多经过1次，有多少种从起点坐标到终点坐标的方案。在迷宫中移动有上下左右四种方式，每次只能移动一个方格。数据保证起点上没有障碍。
解答： 我是直接dfs， 爆搜出来的...题面只给了坐标，需要用坐标复现出棋盘的样子，然后爆搜就好
第一次提交70分， 原因在于 迷宫的起点也不能重复通过。
//initialize chess,
//state = 0 can go
//state = -1 obstacle
//state = 1 visited
1 //1605 2 //第一次提交70分， 问题在于迷宫起点也不能重复通过 3 4 #include <iostream> 5 #include <vector> 6 #include <cstdio> 7 #include <cstdlib> 8 #include <cmath> 9 10 using namespace std; 11 12 static int ans = 0; 13 int n, m, t; 14 int sx, sy, dx, dy; 15 16 int addx[4] = {0, -1, 0, 1}; 17 int addy[4] = {-1, 0, 1, 0}; 18 19 void print(const vector<vector<int>>& chess) { 20 printf("======begin debug======= "); 21 for(int i = 0; i < chess.size(); ++i) { 22 for(int j = 0; j < chess[0].size(); ++j) { 23 printf("%d ", chess[i][j]); 24 } 25 printf(" "); 26 } 27 printf("=======end============== "); 28 } 29 30 31 void dfs(vector<vector<int>>& chess, int x, int y) { 32 if (x == dx && y == dy) { 33 ++ans; 34 return; 35 } 36 for(int i = 0; i < 4; ++i) { 37 int newx = x + addx[i], newy = y + addy[i]; 38 if(newx > 0 && newx <= n && newy > 0 && newy <= m && chess[newx][newy] != -1 && chess[newx][newy] != 1) { 39 chess[newx][newy] = 1; 40 dfs(chess, newx, newy); 41 chess[newx][newy] = 0; 42 } 43 } 44 } 45 46 47 int main() { 48 49 cin >> n >> m >> t; 50 cin >> sx >> sy >> dx >> dy; 51 //initialize chess, 52 //state = 0 can go 53 //state = -1 obstacle 54 //state = 1 visited 55 56 vector<vector<int>> chess(n+1, vector<int>(m+1, 0)); 57 for(int i = 0; i < m+1; ++i) { 58 chess[0][i] = -1; 59 } 60 for (int i = 0; i < n+1; ++i) { 61 chess[i][0] = -1; 62 } 63 while(t--) { 64 int x, y; 65 cin >> x >> y; 66 chess[x][y] = -1; 67 } 68 //print(chess); 69 chess[sx][sy] = 1; //start point can not repeat pass 70 dfs(chess, sx, sy); 71 72 cout << ans << endl; 73 return 0; 74 }