官方题解：http://discuss.acmcoder.com/topic/58dd094d6f37d3610507b06e
1. 直接从左往右扫，上升而且长度至少为1，然后再扫，下降的长度至少为1，更下答案，从这个位置重新开始。
1 /* 2 ID: y1197771 3 PROG: test 4 LANG: C++ 5 */ 6 #include<bits/stdc++.h> 7 #define pb push_back 8 #define FOR(i, n) for (int i = 0; i < (int)n; ++i) 9 #define dbg(x) cout << #x << " at line " << __LINE__ << " is: " << x << endl 10 typedef long long ll; 11 using namespace std; 12 typedef pair<int, int> pii; 13 const int maxn = 1e7 + 10; 14 int n; 15 int a[maxn]; 16 void solve() { 17 while(cin >> n){ 18 for (int i = 0; i < n; i++) { 19 cin >> a[i]; 20 } 21 int res = 0; 22 int tleft, tright; tleft = tright = -1; 23 int i = 0; 24 while(i < n) { 25 int tl = i; 26 while(i + 1 < n && a[i + 1] > a[i]) { 27 i++; 28 } 29 if(i == tl) { 30 i++; 31 continue; 32 } 33 int tm = i; 34 while(i + 1 < n && a[i + 1] < a[i]) { 35 i++; 36 } 37 if(i == tm) { 38 i++; 39 continue; 40 } 41 int tlen = i - tl + 1; 42 if(tlen > res) { 43 res = tlen; 44 tleft = tl; 45 tright = i; 46 } 47 } 48 cout << tleft << " " << tright << endl; 49 } 50 } 51 int main() { 52 freopen("test.in", "r", stdin); 53 //freopen("test.out", "w", stdout); 54 ios::sync_with_stdio(0); 55 cin.tie(0); cout.tie(0); 56 solve(); 57 return 0; 58 }
2. 直接按照要求，每一个句子搞一个set，然后对测试的每一个句子拆分单词，找最多的匹配，码的代码比较多，题意挺直接的。
1 /* 2 ID: y1197771 3 PROG: test 4 LANG: C++ 5 */ 6 #include<bits/stdc++.h> 7 #define pb push_back 8 #define FOR(i, n) for (int i = 0; i < (int)n; ++i) 9 #define dbg(x) cout << #x << " at line " << __LINE__ << " is: " << x << endl 10 typedef long long ll; 11 using namespace std; 12 typedef pair<int, int> pii; 13 const int maxn = 1e3 + 10; 14 int n, m; 15 set<string> se[603]; 16 string a[603]; 17 void solve() { 18 cin >> n >> m; 19 string s; 20 getchar(); 21 for (int i = 0; i < n; i++) { 22 getline(cin, s); 23 //cout << s << endl; 24 a[i] = s; 25 s += " "; 26 int tn = s.size(); 27 int p = 0; 28 while(p < tn) { 29 int t = p; 30 while(p < tn && s[p] != ' ') p++; 31 string st = s.substr(t, p - t); 32 se[i].insert(st); 33 p++; 34 } 35 } 36 set<string> v; 37 for (int i = 0; i < m; i++) { 38 getline(cin, s); 39 v.clear(); 40 s += " "; 41 int tn = s.size(); 42 int p = 0; 43 while(p < tn) { 44 int t = p; 45 while(p < tn && s[p] != ' ') p++; 46 string st = s.substr(t, p - t); 47 v.insert(st); 48 p++; 49 } 50 int res = 0, id = -1; 51 for (int j = 0; j < n; j++) { 52 int sc = 0; 53 for (string t : v) { 54 if(se[j].count(t)) { 55 sc++; 56 } 57 } 58 if(sc > res) { 59 res = sc; 60 id = j; 61 } 62 } 63 cout << a[id] << endl; 64 } 65 } 66 int main() { 67 freopen("test.in", "r", stdin); 68 //freopen("test.out", "w", stdout); 69 //ios::sync_with_stdio(0); 70 //cin.tie(0); cout.tie(0); 71 solve(); 72 return 0; 73 }
stringstream该可以按照空格拆分字符串，利用这个可以简单的拆分单词。
3. 题意挺明显的，先找最深层次，这关系到最后的输出长度，然后逐个扫描输出。
我不清楚到底要不要输出空格。一直是30%。
1 /* 2 ID: y1197771 3 PROG: test 4 LANG: C++ 5 */ 6 #include<bits/stdc++.h> 7 #define pb push_back 8 #define FOR(i, n) for (int i = 0; i < (int)n; ++i) 9 #define dbg(x) cout << #x << " at line " << __LINE__ << " is: " << x << endl 10 typedef long long ll; 11 using namespace std; 12 typedef pair<int, int> pii; 13 const int maxn = 1e3 + 10; 14 string s; 15 16 void solve() { 17 while(cin >> s) { 18 int n = s.size(); 19 int len = 0; 20 int x, y; 21 x = 0; 22 for (char t : s) { 23 if(t == '[') x++; 24 else x--; 25 len = max(len, x); 26 } 27 int m = 0; 28 int cur = 0; 29 for (int i = 0; i < n; i++) { 30 if(s[i] == '[') { 31 cur++; 32 if(cur == 1) { 33 m = 2 * len - 1; 34 35 cout << "+"; 36 for(int j = 0; j < m; j++) 37 cout << '-'; 38 cout << "+" << endl; 39 40 if(i + 1 < n && s[i + 1] == ']') { 41 cout << "|"; 42 for(int j = 0; j < m; j++) 43 cout << ' '; 44 cout << "|" << endl; 45 46 for (int j = 0; j < 2 * len + 1; j++) 47 cout << " "; 48 cout << endl; 49 50 cout << "|"; 51 for(int j = 0; j < m; j++) 52 cout << ' '; 53 cout << "|" << endl; 54 } 55 } else { 56 m = 2 * (len - cur + 1) - 1; 57 58 for (int j = 0; j < cur - 2; j++) 59 cout << " "; 60 cout << "|"; 61 cout << "+"; 62 for(int j = 0; j < m; j++) 63 cout << '-'; 64 cout << "+"; 65 cout << "|"; 66 for (int j = 0; j < cur - 2; j++) 67 cout << " "; 68 cout << endl; 69 70 71 if(i + 1 < n && s[i + 1] == ']') { 72 73 for (int j = 0; j < cur - 1; j++) 74 cout << " "; 75 cout << "|"; 76 for(int j = 0; j < m; j++) 77 cout << ' '; 78 cout << "|" ; 79 for (int j = 0; j < cur - 1; j++) 80 cout << " "; 81 cout << endl; 82 83 84 for (int j = 0; j < 2 * len + 1; j++) 85 cout << " "; 86 cout << endl; 87 88 89 for (int j = 0; j < cur - 1; j++) 90 cout << " "; 91 cout << "|"; 92 for(int j = 0; j < m; j++) 93 cout << ' '; 94 cout << "|"; 95 for (int j = 0; j < cur - 1; j++) 96 cout << " "; 97 cout << endl; 98 } 99 } 100 101 } else { 102 103 if(cur == 1) { 104 m = 2 * len - 1; 105 cout << "+"; 106 for(int j = 0; j < m; j++) 107 cout << '-'; 108 cout << "+" << endl; 109 110 } else { 111 m = 2 * (len - cur + 1) - 1; 112 for (int j = 0; j < cur - 2; j++) 113 cout << " "; 114 cout << "|"; 115 cout << "+"; 116 for(int j = 0; j < m; j++) 117 cout << '-'; 118 cout << "+"; 119 cout << "|"; 120 for (int j = 0; j < cur - 2; j++) 121 cout << " "; 122 cout << endl; 123 124 } 125 cur--; 126 } 127 } 128 } 129 } 130 int main() { 131 freopen("test.in", "r", stdin); 132 //freopen("test.out", "w", stdout); 133 ios::sync_with_stdio(0); 134 cin.tie(0); cout.tie(0); 135 solve(); 136 return 0; 137 }
4. 显然暴力是不行的，先写个暴力的，拿到30%的分数。接下来分析有没有什么好办法，考虑到题目数目范围[1， 1e5],很容的就可以开1e5大小的数组，然后就是桶排序。观察到i的顺序无关紧要。不能在线一个一个做的话，就离线，按照x的值进行排序，固定第一个值，然后动态的更新第二个数，而且第二个数还要求点更新和区间查询，显然树状数组是满足要求的，（当然线段树也是可以的，什么更高深的算法更是不在话下吧，学学莫队算法，treap，splay等）。然后就很直接了。
注意，即使使用优化的cin，cout也会超时tle，我猜是题目的数据太多了的缘故，（而且输出不是很大），必须换成scanf，printf或者开挂（采用read读取，write写出）。
1 /* 2 ID: y1197771 3 PROG: test 4 LANG: C++ 5 */ 6 #include<bits/stdc++.h> 7 #define pb push_back 8 #define FOR(i, n) for (int i = 0; i < (int)n; ++i) 9 #define dbg(x) cout << #x << " at line " << __LINE__ << " is: " << x << endl 10 typedef long long ll; 11 using namespace std; 12 typedef pair<int, int> pii; 13 const int maxn = 1e5 + 10; 14 int n, q; 15 vector<pii> e[maxn]; 16 pii a[maxn]; 17 int res[maxn]; 18 int f[maxn]; 19 int lb(int x) { 20 return x & -x; 21 } 22 int ask(int t) { 23 int r = 0; 24 while(t > 0) { 25 r += f[t]; 26 t -= lb(t); 27 } 28 return r; 29 } 30 void upd(int x) { 31 while(x <= 100000) { 32 f[x]++; 33 x += lb(x); 34 } 35 } 36 37 void solve() { 38 cin >> n >> q; 39 for (int i = 0; i < n; i++) { 40 //cin >> a[i].first; 41 scanf("%d", &a[i].first); 42 } 43 for (int i = 0; i < n; i++) { 44 //cin >> a[i].second; 45 scanf("%d", &a[i].second); 46 } 47 int x, y; 48 for (int i = 0; i < q; i++) { 49 //cin >> x >> y; 50 scanf("%d%d", &x, &y); 51 e[x].pb({y, i}); 52 } 53 sort(a, a + n); 54 int p = n - 1; 55 for (int i = 100000; i >=1; i--) { 56 while(p >= 0 && a[p].first >= i) { 57 int t = a[p].second; 58 upd(t); 59 p--; 60 //cout << i << " " << p << endl; 61 } 62 63 for (int j = 0; j < e[i].size(); j++) { 64 int t = e[i][j].first; 65 int t2 = e[i][j].second; 66 res[t2] = (n - 1 - p) - ask(t - 1); 67 } 68 69 } 70 for (int i = 0; i < q; i++) 71 printf("%d ", res[i]); 72 73 } 74 int main() { 75 freopen("test.in", "r", stdin); 76 //freopen("test.out", "w", stdout); 77 //ios::sync_with_stdio(0); 78 //cin.tie(0); cout.tie(0); 79 solve(); 80 return 0; 81 }