UVAlive X-Plosives 思路:    “如果车上存在k个简单化合物,正好包含k种元素,那么他们将组成一个易爆的混合物”  如果将(a,b)看作一条边那么题意就是不能出现环,很容易联想到Kruskal算法中并查集的判环功能(新加入的边必须属于不同的两个集合否则出现环),因此本题可以用并查集实现.模拟装车过程即可. 代码: #include<cstdio> #include<cstring> #define FOR(a,b,c) for(int a=(b);a<(c…
UVAlive 3135 Argus Argus Time Limit: 3000MS   Memory Limit: Unknown   64bit IO Format: %lld & %llu Submit Status Description A data stream is a real-time, continuous, ordered sequence of items. Some examples include sensor data, Internet traffic, fin…
UVAlive 3026 Period 题目: Period   Time Limit: 3000MS   Memory Limit: Unknown   64bit IO Format: %lld & %llu Submit Status Description For each prefix of a given string S with N characters (each character has an ASCII code between 97 and 126, inclusive…
UVAlive 3942 Remember the Word 题目: Remember the Word   Time Limit: 3000MS   Memory Limit: Unknown   64bit IO Format: %lld & %llu Submit Status Description Neal is very curious about combinatorial problems, and now here comes a problem about words. Kn…
UVAlive 4329 Ping pong 题目: Ping pong Time Limit: 3000MS   Memory Limit: Unknown   64bit IO Format: %lld & %llu Submit Status Description N(3N20000) ping pong players live along a west-east street(consider the street as a line segment). Each player ha…
UVAlive 3027 Corporative Network 题目:   Corporative Network Time Limit: 3000MS   Memory Limit: 30000K Total Submissions: 3450   Accepted: 1259 Description A very big corporation is developing its corporative network. In the beginning each of the N ent…
UVAlive 4670 Dominating Patterns 题目:   Dominating Patterns   Time Limit: 3000MS   Memory Limit: Unknown   64bit IO Format: %lld & %llu Submit Status Description The archaeologists are going to decipher a very mysterious ``language". Now, they kno…
UVa11991 Easy Problem from Rujia Liu?  思路:  构造数组data,使满足data[v][k]为第k个v的下标.因为不是每一个整数都会出现因此用到map,又因为每个数出现次数不等可能相差很大,因此用到vector. 注意:对于数据的清空与判空不要忘记,而map在调用之前必须有map.count的检查. 代码: #include<cstdio> #include<map> #include<vector> using namespac…
UVa11995  I Can Guess the Data Structure! 思路:边读边模拟,注意empty的判断! 代码如下: #include<iostream> #include<queue> #include<stack> using namespace std; int main(){ queue<int> q; priority_queue<int> pri_q; stack<int> sta; int n; wh…
范围最小值问题: 提供操作: Query(L,R):计算min{AL ~ AR } Sparse-Table算法: 定义d[i][j]为从i开始长度为2j的一段元素的最小值.所以可以用递推的方法表示. 预处理RMQ_init如下(感觉像区间DP): int RMQ_init(const vector<int>& A){ int n=A.size(); ;i<n;i++) d[i][] = A[i]; ;(<<j)<=n;j++) ;i+(<<j)-&…