玩了二十多次，终于杀到了div1，纪念一下
1、水题，求最大最小值的问题。
2、话说找工作的时候有道类似的题，也是在图上遍历，然后将符合性质的点归为一类，将种类数计为n，且最少有一个，那么就是
2^n - 1种。
1 #include <iostream> 2 #include <string> 3 #include <vector> 4 #include <cstdlib> 5 #include <cmath> 6 #include <map> 7 #include <stack> 8 #include <algorithm> 9 #include <list> 10 #include <ctime> 11 #include <set> 12 #include <queue> 13 typedef long long ll; 14 using namespace std; 15 #define CLR(arr, what) memset(arr, what, sizeof(arr)) 16 int pos[100][100]; 17 int r, c; 18 void dfs(int x, int y, int cur, int dist) { 19 if (x < 0 || x >= r || y < 0 || y >= c) 20 return; 21 22 if (pos[x][y] == 1) { 23 pos[x][y] = cur; 24 for (int i = x - dist; i < x + dist + 1; i++) { 25 for (int j = y - dist; j < y + dist + 1; j++) { 26 int tmp = abs(i - x) + abs(j - y); 27 if (tmp <= dist) { 28 dfs(i, j, cur, dist); 29 } 30 } 31 } 32 } else if (pos[x][y] == 0) { 33 return; 34 } else { 35 return; 36 } 37 } 38 class GooseInZooDivTwo { 39 public: 40 int count(vector<string> field, int dist) { 41 r = field.size(); 42 c = field[0].size(); 43 for (int i = 0; i < r; i++) { 44 for (int j = 0; j < c; j++) { 45 pos[i][j] = (field[i][j] == 'v'); 46 } 47 } 48 int tag = 1; 49 for (int i = 0; i < r; i++) { 50 for (int j = 0; j < c; j++) { 51 if (pos[i][j] == 1) { 52 tag++; 53 dfs(i, j, tag, dist); 54 } 55 } 56 } 57 if (1 == tag) 58 return 0; 59 int res = 1; 60 for (int i = 1; i < tag; i++) { 61 res = (res * 2) % 1000000007; 62 } 63 res = res - 1; 64 return res; 65 } 66 };
3、第三题没思路，伤不起。
DIV-1
1、和div-2的第二题类似，多了一个限制条件。
1 #include <iostream> 2 #include <string> 3 #include <vector> 4 #include <cstdlib> 5 #include <cmath> 6 #include <map> 7 #include <stack> 8 #include <algorithm> 9 #include <list> 10 #include <ctime> 11 #include <set> 12 #include <queue> 13 typedef long long ll; 14 using namespace std; 15 #define CLR(arr, what) memset(arr, what, sizeof(arr)) 16 int pos[100][100]; 17 int r, c; 18 int dfs(int x, int y, int cur, int dist) { 19 if (x < 0 || x >= r || y < 0 || y >= c) 20 return 0; 21 22 if (pos[x][y] == 1) { 23 int res = 1; 24 pos[x][y] = cur; 25 for (int i = x - dist; i < x + dist + 1; i++) { 26 for (int j = y - dist; j < y + dist + 1; j++) { 27 int tmp = abs(i - x) + abs(j - y); 28 if (tmp <= dist) { 29 res = res + dfs(i, j, cur, dist); 30 } 31 } 32 } 33 return res; 34 } else if (pos[x][y] == 0) { 35 return 0; 36 } else { 37 return 0; 38 } 39 } 40 class GooseInZooDivOne { 41 public: 42 int count(vector<string> field, int dist) { 43 r = field.size(); 44 c = field[0].size(); 45 for (int i = 0; i < r; i++) { 46 for (int j = 0; j < c; j++) { 47 pos[i][j] = (field[i][j] == 'v'); 48 } 49 } 50 int tag = 1; 51 int odd = 0; 52 int even = 0; 53 for (int i = 0; i < r; i++) { 54 for (int j = 0; j < c; j++) { 55 if (pos[i][j] == 1) { 56 tag++; 57 int tmp = dfs(i, j, tag, dist); 58 if ((tmp % 2) == 0) { 59 even++; 60 } else { 61 odd++; 62 } 63 } 64 } 65 } 66 if (1 == tag) 67 return 0; 68 int res = 1; 69 for (int i = 0; i < (even); i++) { 70 res = (res * 2) % 1000000007; 71 } 72 73 for (int i = 0; i < (odd - 1); i++) { 74 res = (res * 2) % 1000000007; 75 } 76 res = res - 1; 77 return res; 78 } 79 };