A题。。暴力枚举在每个位置添加字符，然后检查一下是不是回文串
1 #include <iostream>
2 #include <cstdio>
3 #include <cstring>
4 #include <algorithm>
5 #include <cmath>
6 #include <vector>
7
8 using namespace std;
9
10 #define LL long long
11 #define eps 1e-8
12 #define inf 0x3f3f3f3f
13 #define lson l, m, rt << 1
14 #define rson m+1, r, rt << 1 | 1
15 #define mnx 31000
16
17 char s[20], ch[20];
18 bool check(){
19 int n = strlen( ch );
20 for( int i = 0, j = n-1; i <= j; ++i, --j ){
21 if( ch[i] != ch[j] ) return false;
22 }
23 return true;
24 }
25 int main(){
26 scanf( "%s", &s );
27 int n = strlen( s );
28 for( int i = 0; i <= n; ++i ){
29 for( int j = ‘a‘; j <= ‘z‘; ++j ){
30 for( int k = 0, m = 0; k <= n; ++k ){
31 if( i == k )
32 ch[k] = (char)j;
33 else
34 ch[k] = s[m++];
35 }
36 ch[n+1] = ‘\0‘;
37 if( check() ){
38 printf( "%s\n", ch );
39 return 0;
40 }
41 }
42 }
43 puts( "NA" );
44 return 0;
45 }
B题。。也是暴力枚举每种颜色，dfs算一下就好
1 #include <iostream>
2 #include <cstdio>
3 #include <cstring>
4 #include <algorithm>
5 #include <cmath>
6 #include <vector>
7
8 using namespace std;
9
10 #define LL long long
11 #define eps 1e-8
12 #define inf 0x3f3f3f3f
13 #define lson l, m, rt << 1
14 #define rson m+1, r, rt << 1 | 1
15 #define mnx 500
16
17 int fst[mnx], nxt[mnx], vv[mnx], col[mnx], e, ans;
18 void add( int u, int v, int c ){
19 vv[e] = v, col[e] = c, nxt[e] = fst[u], fst[u] = e++;
20 }
21 bool vis[mnx], ok;
22 void dfs( int u, int gg, int vol ){
23 if( u == gg ){
24 ans++; ok = 1; return ;
25 }
26 if( vis[u] ) return ;
27 vis[u] = 1;
28 for( int i = fst[u]; i != -1; i = nxt[i] ){
29 if( ok ) return ;
30 int v = vv[i], c = col[i];
31 if( c != vol ) continue;
32 dfs( v, gg, vol );
33 }
34 }
35 int main(){
36 int n, m, q;
37 scanf( "%d %d", &n, &m );
38 memset( fst, -1, sizeof( fst ) );
39 for( int i = 0; i < m; ++i ){
40 int u, v, c;
41 scanf( "%d %d %d", &u, &v, &c );
42 add( u, v, c );
43 add( v, u, c );
44 }
45 scanf( "%d", &q );
46 while( q-- ){
47 int u, v;
48 scanf( "%d%d", &u, &v );
49 ans = 0;
50 for( int i = 1; i <= m; ++i ){
51 memset( vis, 0, sizeof( vis ) );
52 ok = 0;
53 dfs( u, v, i );
54 }
55 cout << ans << endl;
56 }
57 return 0;
58 }
C题。。比赛的时候想不出来，看了题解才知道 第二维的状态最多不超过500。。因为你1+2+...+250 > 3w，这样每次步数减一或者加一的总的状态不会超过500，所以dp[30000][600]就够了， 把第二维300当做第一次的步数d，然后每次有可能走 d+j-n - 1, d+j-n, d+j-n+1步，感觉看代码容易理解一些。。
1 #include <iostream>
2 #include <cstdio>
3 #include <cstring>
4 #include <algorithm>
5 #include <cmath>
6 #include <vector>
7
8 using namespace std;
9
10 #define LL long long
11 #define eps 1e-8
12 #define inf 0x3f3f3f3f
13 #define lson l, m, rt << 1
14 #define rson m+1, r, rt << 1 | 1
15 #define mnx 31000
16
17 const int N = 300;
18 int dp[mnx][700], val[mnx];
19 int main(){
20 int n, m;
21 while( scanf( "%d%d", &n, &m ) != EOF ){
22 memset( val, 0, sizeof val );
23 memset( dp, 0, sizeof dp );
24 for( int i = 0; i < n; ++i ){
25 int x;
26 scanf( "%d", &x );
27 val[x]++;
28 }
29 dp[m][N] = val[m] + val[0] + 1;
30 int ans = 0;
31 for( int i = m; i < mnx; ++i ){
32 for( int j = 0; j < 600; ++j ){
33 if( !dp[i][j] ) continue;
34 int l = m + j - N;
35 ans = max( dp[i][j] - 1, ans );
36 if( i + l >= mnx ) continue;
37 if( l > 0 )
38 dp[i+l][j] = max( dp[i+l][j], dp[i][j] + val[i+l] );
39 if( l > 1 )
40 dp[i+l-1][j-1] = max( dp[i+l-1][j-1], dp[i][j] + val[i+l-1] );
41 if( l >= 0 )
42 dp[i+l+1][j+1] = max( dp[i+l+1][j+1], dp[i][j] + val[i+l+1] );
43 }
44 }
45 cout << ans << endl;
46 }
47 return 0;
48 }
D题。。好像是B的加强版，用并查集搞。。明天再看看
时间： 2024-10-10 13:53:28