做了一道题，对了，但是还是掉分了。
第二道题也做了，但是没有交上，不知道对错。
后来交上以后发现少判断了一个条件，改过之后就对了。
第一道题爆搜的，有点麻烦了，其实几行代码就行。
250贴代码：
1 #include <iostream> 2 #include <cstring> 3 #include <queue> 4 #include <cmath> 5 #include <cstdio> 6 #include <algorithm> 7 #include <vector> 8 #define LL long long 9 using namespace std; 10 11 class BishopMove 12 { 13 public: 14 struct node 15 { 16 int x, y, step; 17 } next, pos; 18 int howManyMoves(int r1, int c1, int r2, int c2) 19 { 20 int vis[10][10], i, a, b; 21 memset(vis, 0, sizeof(vis)); 22 if(r1==r2 && c1==c2) 23 return 0; 24 25 queue<node>q; 26 vis[r1][c1] = 1; 27 next.x = r1; 28 next.y = c1; 29 next.step = 0; 30 q.push(next); 31 while(!q.empty()) 32 { 33 pos = q.front(); 34 if(pos.x == r2 && pos.y == c2) 35 return pos.step; 36 37 q.pop(); 38 for(i = 1; i <= 7; i ++) 39 { 40 a = pos.x+i; 41 b = pos.y+i; 42 if(a>=0&&a<=7 && b>=0 && b<=7 && vis[a][b]==0) 43 { 44 vis[a][b] = 1; 45 next.x = a; 46 next.y = b; 47 next.step = pos.step+1; 48 q.push(next); 49 } 50 a = pos.x+i; 51 b = pos.y-i; 52 if(a>=0&&a<=7 && b>=0 && b<=7 && vis[a][b]==0) 53 { 54 vis[a][b] = 1; 55 next.x = a; 56 next.y = b; 57 next.step = pos.step+1; 58 q.push(next); 59 } 60 a = pos.x-i; 61 b = pos.y+i; 62 if(a>=0&&a<=7 && b>=0 && b<=7 && vis[a][b]==0) 63 { 64 vis[a][b] = 1; 65 next.x = a; 66 next.y = b; 67 next.step = pos.step+1; 68 q.push(next); 69 } 70 a = pos.x-i; 71 b = pos.y-i; 72 if(a>=0&&a<=7 && b>=0 && b<=7 && vis[a][b]==0) 73 { 74 vis[a][b] = 1; 75 next.x = a; 76 next.y = b; 77 next.step = pos.step+1; 78 q.push(next); 79 } 80 } 81 } 82 return -1; 83 } 84 }; 85 86 int main() 87 { 88 int r1, c1, r2, c2; 89 while(1) 90 { 91 cin>>r1>>c1>>r2>>c2; 92 BishopMove y; 93 cout<<y.howManyMoves(r1, c1, r2, c2)<<endl; 94 } 95 return 0; 96 }
题意：有三种括号 和 x，x能变成任意的括号，求能否通过变化x使得给的字符串符合括号匹配
500贴代码：
AC代码：
1 #include <iostream> 2 #include <cstring> 3 #include <cstdlib> 4 #include <cmath> 5 #include <cstdio> 6 #include <vector> 7 #define LL long long 8 using namespace std; 9 10 class BracketExpressions 11 { 12 public: 13 int _max(int c, int d) 14 { 15 return c > d?c:d; 16 } 17 int match(char a, char b) 18 { 19 if(a=='(' && b==')') 20 return 1; 21 if(a=='{' && b=='}') 22 return 1; 23 if(a=='[' && b==']') 24 return 1; 25 if(a=='X' &&(b==']'||b=='}'||b==')')) 26 return 1; 27 if(b=='X' && (a=='['||a=='{'||a=='(')) 28 return 1; 29 if(a=='X' && b=='X') 30 return 1; 31 return 0; 32 } 33 string ifPossible(string expression) 34 { 35 int i, j, k, g, len; 36 int d[100][100]; 37 string s = expression; 38 len = s.size(); 39 memset(d, 0, sizeof(d)); 40 for(i = 0; i < len-1; i++) 41 if(match(s[i], s[i+1])) 42 d[i][i+1] = 1; 43 for(k = 2; k < len; k++) 44 { 45 for(i = 0; i < len-k; i++) 46 { 47 j = i+k; 48 if(match(s[i], s[j])) d[i][j] = d[i+1][j-1] + 1; 49 for(g = 0; g < k; g++) 50 d[i][j] = _max(d[i][i+g]+d[i+g+1][j], d[i][j]); 51 } 52 } 53 if(2*d[0][len-1]!=len) 54 return "impossible"; 55 else 56 return "possible"; 57 } 58 };