A题逗比了，没有看到All ai are distinct. All bi are distinct. 其实很水的。。
1 #include <iostream>
2 #include <cstdio>
3 #include <cstring>
4 #include <algorithm>
5 using namespace std;
6
7 #define mnx 100002
8
9
10 struct latop{
11 int p, q;
12 bool operator < ( const latop & b ) const {
13 return p < b.p;
14 }
15 }a[mnx];
16 int main(){
17 int n;
18 scanf( "%d", &n );
19 for( int i = 0; i < n; i++ ){
20 scanf( "%d %d", &a[i].p, &a[i].q );
21 }
22 sort( a, a + n );
23 bool flag = 0;
24 for( int i = 1; i < n; i++ ){
25 if( a[i].q < a[i-1].q ) flag = 1;
26 }
27 if( flag ) puts( "Happy Alex" );
28 else puts( "Poor Alex" );
29 return 0;
30 }
B题找循环节。。循环节是4，然后输入再长的数，也只需要看最后两位，因为100的倍数肯定能够整除4，就看最后两位。。结果又逗比了，数组开小了，最后re了。。太粗心了。。
1 #include<string>
2 #include<cmath>
3 #include<map>
4 #include<queue>
5
6 using namespace std;
7
8 #define mnx 1000000
9 #define PI acos(-1.0)
10 #define ll long long
11 #define ull unsigned long long
12 #define inf 0x3f3f3f3f
13 #define eps 1e-8
14 #define MP make_pair
15 #define lson l, m, rt << 1
16 #define rson m+1, r, rt << 1 | 1
17 #define mod 2333333
18
19 int a[4] = { 4, 0, 0, 0 };
20 char ch[mnx];
21 int main(){
22 scanf( "%s", &ch );
23 int n = strlen( ch );
24 int k = 0;
25 if( n <= 2 ){
26 for( int i = 0; i < n; i++ ){
27 k = k * 10 + ch[i] - ‘0‘;
28 }
29 // cout<<k<<endl;
30 k %= 4;
31 printf( "%d\n", a[k] );
32 }
33 else{
34 for( int i = n-2; i < n; i++ ){
35 k = k * 10 + ch[i] - ‘0‘;
36 }
37 // cout<<k<<endl;
38 k %= 4;
39 printf( "%d\n", a[k] );
40 }
41 return 0;
42 }
C题先排序，然后把相同的弄掉，if( b[i] == b[i-1] + 1 ) dp[i] = max( dp[i-2] + sum[i], dp[i-1] ),要么就是选当前这个b[i],然后dp[i]的值就是dp[i-2] + sum[i],要么就选b[i-1]，然后dp[i]的值就是dp[i-1]；else dp[i] = dp[i] = dp[i-1] + sum[i]; 最后答案就是 dp[cnt]; 具体看代码。。当时写的比较稳妥，多写了一些。。结果还是因为longlong跪了几发，这是后来写的代码
1 #include<iostream>
2 #include<cstring>
3 #include<cstdio>
4 #include<algorithm>
5 #include<string>
6 #include<cmath>
7 #include<map>
8 #include<queue>
9
10 using namespace std;
11
12 #define mnx 200000
13 #define PI acos(-1.0)
14 #define ll long long
15 #define ull unsigned long long
16 #define inf 0x3f3f3f3f
17 #define eps 1e-8
18 #define MP make_pair
19 #define lson l, m, rt << 1
20 #define rson m+1, r, rt << 1 | 1
21 #define mod 2333333
22
23 ll a[mnx], b[mnx], sum[mnx], dp[mnx];
24 int main(){
25 int n;
26 scanf( "%d", &n );
27 for( int i = 1; i <= n; i++ ){
28 scanf( "%I64d", &a[i] );
29 }
30 sort( a + 1, a + 1 + n );
31 int cnt = 0;
32 for( int i = 1; i <= n; i++ ){
33 if( a[i] != b[cnt] ){
34 ++cnt;
35 b[cnt] = a[i];
36 sum[cnt] = a[i];
37 }
38 else sum[cnt] += a[i];
39 }
40 dp[1] = sum[1];
41 for( int i = 2; i <= cnt; i++ ){
42 if( b[i] == b[i-1] + 1 ){
43 dp[i] = max( dp[i-2] + sum[i], dp[i-1] );
44 }
45 else{
46 dp[i] = dp[i-1] + sum[i];
47 }
48 }
49 cout<<dp[cnt]<<endl;
50 return 0;
51 }
D题trie树，不会做。。赛后问了思路，弄了很久才明白。。原本的想法是 trie的根就是必败，然后dfs，第一层就是必胜，遍历到叶子节点，看叶子节点是必败还是必胜，如果叶子节点全是必胜的，说明先手的每次必胜，如果叶子节点全是必败的，说明先手的每次必败,如果叶子节点既有必胜又有必败的，说明先手的每次既能必胜也能必败。。后来发现这样想是错的，如果是abcs，abcab，先手走到c，接下来先手的输赢全部是由后手决定，后手走s，先手输，后手走a，先手赢。。这道题输赢的状态必须由叶子节点开始向根的方向推，看了超哥的代码，开两个bool数组，win[], can[],win数组表示下一步先手开始走，能不能赢，can数组表示下一步先手开始走，能不能输。。所以叶子节点就是win[u] = 0(先手下一步不能走，就不能赢)，can[u] = 1(先手下一步不能走，肯定输)。。当一个节点有多个子节点，win[u] = 1就看 win[son[u][i]]里面有没有为0的，否则就是0；can[u] = 1就看can[son[u][i]]里面有没有为0的，否则就是0。。推到根节点的状态。。这样讲有点抽象，可以手动模拟一下 abcs， abcab这种情况。。
1 #include <iostream>
2 #include <cstdio>
3 #include <cstring>
4 #include <algorithm>
5 using namespace std;
6
7 #define mnx 100002
8
9 int son[mnx][26], tot;
10 bool win[mnx], can[mnx];
11 void insert(char *s) {
12 int t = 0;
13 for (int i = 0; s[i]; ++i) {
14 int c = s[i] - ‘a‘;
15 if (!son[t][c]) son[t][c] = ++tot;
16 t = son[t][c];
17 }
18 }
19 void dfs( int u ){
20 win[u] = 0, can[u] = 1;
21 bool hav = 0;
22 for( int i = 0; i < 26; i++ ){
23 if( son[u][i] ){
24 dfs( son[u][i] ); hav = 1;
25 }
26 }
27 if( !hav ) return ;
28 can[u] = 0;
29 for( int i = 0; i < 26; i++ ){
30 if( son[u][i] && !win[son[u][i]] ) win[u] = 1;
31 if( son[u][i] && !can[son[u][i]] ) can[u] = 1;
32 }
33 }
34 char ch[mnx];
35 int main(){
36 int n, k;
37 scanf( "%d%d", &n, &k );
38 for( int i = 0; i < n; i++ ){
39 scanf( "%s", ch );
40 insert( ch );
41 }
42 dfs( 0 );
43 if( win[0] == false ) puts( "Second" );
44 else{
45 if( can[0] == true ) puts( "First" );
46 else puts( (k & 1) ? "First" : "Second" );
47 }
48 return 0;
49 }
E题没时间看，赛后看了一下，发现也不是很难，树的直径和并查集。。比较难处理的就是connect them by a road so as to minimize the length of the longest path in the resulting region 使联通后的区域的直径最小，那么就应该从两条直径的中间点连一条边，使他们联通。。假设要联通两个区域为u，v，直径为val[u], val[v]，则联通后的直径应该是 max( val[u], val[v], 1 + ( val[u] - val[u] / 2 ) + ( val[v] - val[v] / 2 );
1 #include <iostream>
2 #include <cstdio>
3 #include <cstring>
4 #include <algorithm>
5 using namespace std;
6
7 #define mnx 600005
8 #define inf 0x3f3f3f3f
9
10 int dis[2][mnx], fa[mnx], val[mnx];
11 int vv[mnx], first[mnx], nxt[mnx], e;
12 bool vis[mnx], in[mnx];
13 int n, m, query, tot;
14 void init(){
15 memset( first, -1, sizeof(first) );
16 memset( dis, 0x3f, sizeof(dis) );
17 e = 0;
18 }
19 void add( int u, int v ){
20 vv[e] = v, nxt[e] = first[u], first[u] = e++;
21 }
22 int q[mnx];
23 int bfs( int s, int *dis ){
24 int head = 0, tail = 0;
25 q[tail++] = s, dis[s] = 0;
26 while( head < tail ){
27 int u = q[head++];
28 vis[u] = 1;
29 for( int i = first[u]; i != -1; i = nxt[i] ){
30 int v = vv[i];
31 if( dis[v] > dis[u] + 1 ){
32 dis[v] = dis[u] + 1;
33 q[tail++] = v;
34 }
35 }
36 }
37 int id = s, d = dis[s];
38 for( int i = 0; i < tail; i++ ){
39 if( dis[q[i]] > d ){
40 d = dis[q[i]], id = q[i];
41 }
42 }
43 return id;
44 }
45 int find( int s ){
46 if( s != fa[s] ){
47 fa[s] = find( fa[s] );
48 }
49 return fa[s];
50 }
51 void treeD( int s ){
52 int v = bfs( s, dis[0] );
53 int u = bfs( v, dis[1] );
54 val[find(u)] = dis[1][u];
55 }
56 int main(){
57 scanf( "%d %d %d", &n, &m, &query );
58 tot = n;
59 init();
60 for( int i = 0; i <= n + query; i++ ){
61 fa[i] = i;
62 }
63 while( m-- ){
64 int u, v;
65 scanf( "%d %d", &u, &v );
66 add( u, v ), add( v, u );
67 int fu = find( u ), fv = find( v );
68 fa[fu] = fv;
69 }
70 for( int i = 1; i <= n; i++ ){
71 if( !vis[i] ){
72 treeD( i );
73 }
74 }
75 while( query-- ){
76 int tye, u, v;
77 scanf( "%d", &tye );
78 if( tye == 1 ){
79 scanf( "%d", &u );
80 printf( "%d\n", val[find(u)] );
81 }
82 else{
83 scanf( "%d%d", &u, &v );
84 int fu = find( u ), fv = find( v );
85 if( fu == fv ) continue;
86 ++tot;
87 fa[fu] = fa[fv] = tot;
88 val[tot] = 1 + ( val[fu] - val[fu] / 2 ) + ( val[fv] - val[fv] / 2 );
89 val[tot] = max( val[tot], val[fu] );
90 val[tot] = max( val[tot], val[fv] );
91 }
92 }
93 return 0;
94 }
Codeforces Round #260 (Div. 2) ABCDE
时间： 2024-11-05 13:28:42