weixin_44566432的博客

#include <iostream> #include <cstdio> using namespace std; const int MAXN = 25; int height[MAXN]; int dp[MAXN]; int main(int argc, char** argv) { int n; while(scanf("%d", &n) != EOF){ for(int i = 0; i < n; i++){ scanf("%d", .

二维矩阵的最大子矩阵和，从i到j行的最大矩阵 当i=j时，即为最大连续子序列和 当i != j时，把从i到j行的所有行元素加起来，得到只有一行的一维数组，转变为求一维数组的最大连续子序列和，即为最大子矩阵和。 注意，用辅助二维矩阵记录原始矩阵从上到下加起来的累加矩阵，求i到j行的一维数组只需要将辅助矩阵进行对行的减法即可。 #include <iostream> #include <cstdio> using namespace std; const int MAX.

寻找状态转移方程 dp[i] 是以A[i]作为末尾的连续序列最大和 #include <iostream> #include <cstdio> using namespace std; const int MAXN = 1000000; long long arr[MAXN]; long long dp[MAXN]; long long MaxSubsequence(int n){ long long maximum = 0; for(int i = 0; i &.

数列递推关系式 #include <iostream> #include <cstdio> using namespace std; const int MAXN = 91; long long dp[MAXN]; int main(int argc, char** argv) { dp[0] = 0; dp[1] = 1; for(int i = 2; i < MAXN; i++){ dp[i] = dp[i - 1] + dp[i - 2]; } .

#include <iostream> #include <cstdio> #include <queue> #include <vector> #include <cstring> #include <climits> using namespace std; const int MAXN = 1e5 + 7; const int INF = INT_MAX; const int MOD = 1E9 + 7; vector&l.

emmm还有点疑问，待更新完善。 #include <iostream> #include <cstdio> #include <queue> #include <vector> #include <cstring> #include <climits> using namespace std; const int MAXN = 1001; const int INF = INT_MAX; struct Edge{ int.

努力加油！棒棒的！ #include <iostream> #include <cstdio> #include <cstring> #include <queue> using namespace std; const int MAXN = 501; vector<int> graph[MAXN]; int inDegree[MAXN]; vector<int> TopologicalSort(int n){ vect.

判断某图是否是无向图（图中是否存在环）时，即求拓扑排序。 一个图存在符合拓扑次序的顶点序列，即为有向无环图。反之，非有向无环图 #include <iostream> #include <cstdio> #include <cstring> #include <queue> using namespace std; const int MAXN = 500; vector<int> graph[MAXN]; int inDegree[.

#include <iostream> #include <cstdio> #include <vector> #include <cstring> #include <queue> #include <climits> using namespace std; const int MAXN = 1001; const int INF = INT_MAX; struct Edge{ int to; int length; .

#include <iostream> #include <cstdio> #include <vector> #include <cstring> #include <queue> #include <climits> using namespace std; const int MAXN = 200; const int INF = INT_MAX; struct Edge{ int to; int length; E.

#include <iostream> #include <cstdio> #include <algorithm> using namespace std; const int MAXN = 100; struct Edge{ int from; int to; int length; bool operator< (const Edge& e) const{ return length < e.length; } }; Ed.

#include <iostream> #include <cstdio> #include <algorithm> using namespace std; const int MAXN = 100; struct Edge{ int from; int to; int length; bool operator< (const Edge& e) const{ return length < e.length; } }; .

#include <iostream> #include <cstdio> using namespace std; const int MAXN = 1000; int father[MAXN]; int height[MAXN]; int inDegree[MAXN]; bool visit[MAXN]; void Initial(){ for(int i = 0; i <= MAXN; i++){ father[i] = i; height[i] .

连通分量判断连通图。 不断合并图中边相连的两个点所属集合，最后计算集合个数便是图的连通分量个数。如果连通分量等于1，即为连通图。 #include <iostream> #include <cstdio> using namespace std; const int MAXN = 1000; int father[MAXN]; int height[MAXN]; void Initial(int n){ for(int i = 0; i <= n; i++){ .

#include <iostream> #include <cstdio> using namespace std; const int MAXN = 1000; int father[MAXN]; int height[MAXN]; void Initial(int n){ for(int i = 0; i <= n; i++){ father[i] = i; //父亲结点 height[i] = 0; //结点高度 } } int Find(i.

利用映射，将子串当做关键字，将其出现次数当做映射值 #include <iostream> #include <cstdio> #include <string> #include <map> using namespace std; int main(int argc, char** argv) { string str; while(cin >> str){ map<string, int> number; f.

注意一点：双向映射 #include <iostream> #include <cstdio> #include <string> #include <map> using namespace std; map<string, string> dictionary; int main(int argc, char** argv) { string str; while(getline(cin, str)){ if(str == .

#include <iostream> #include <cstdio> #include <map> #include <string> using namespace std; map<string, string> student; int main(int argc, char** argv) { int n; scanf("%d", &n); getchar(); for(int i = 0; i < n;.

#include <iostream> #include <cstdio> #include <map> using namespace std; map<string, int> mmp; int main(int argc, char** argv) { mmp["Emma"] = 67; mmp["Benedict"] = 100; mmp.insert(pair<string, int>("Bob", 72)); mmp.i.

#include <iostream> #include <cstdio> #include <queue> using namespace std; int main(int argc, char** argv) { int n; while(scanf("%d", &n) != EOF){ priority_queue<int, vector<int>, greater<int> > mpq; while(.

#include <iostream> #include <queue> #include <cstdio> using namespace std; struct TreeNode{ int data; TreeNode *leftChild; TreeNode *rightChild; }; //前序遍历 void PreOrder(TreeNode* root){ if(root == NULL){ .

#include <iostream> #include <cstdio> #include <cstring> #include <queue> using namespace std; const int MAXN = 100001; struct Status{ int n; int t; //到n 所消耗的时间 Status(int n, int t): n(n), t(t){ } }; bool visit[MAXN]; //记.

#include <iostream> #include <cstdio> using namespace std; int CountNodes(int m, int n){ if(m > n){ //递归出口，以m为根节点的树为空 return 0; }else{ return 1 + CountNodes(m*2, n) + CountNodes(m * 2 + 1, n); } } int main(int argc, char** arg.

#include <iostream> #include <cstdio> using namespace std; long long Function(int n){ if(n == 1){ return 2; }else{ return 3 * Function(n-1) + 2; //重点在这！ } } int main(int argc, char** argv) { int n; while(scanf("%d", &n) != EOF.

#include <iostream> #include <cstdio> #include <algorithm> #include <cstring> #include <queue> using namespace std; const int MAXN = 200001; struct Island{ long long left; //岛屿左端点 long long right; //岛屿右端点 }; struct .

简单贪心策略，先对数组排序（无排序，不贪心），然后找出最优方案即可 #include <iostream> #include <cstdio> #include <algorithm> using namespace std; struct Program{ int startTime; int endTime; }; const int MAXN = 1000; Program arr[MAXN]; bool Compare(Program x, P.

#include <iostream> #include <cstdio> #include <algorithm> using namespace std; const int MAXN = 1000; long long gun[MAXN]; long long monster[MAXN]; bool Compare(long long x, long long y){ return x > y; //从大到小排序 } int main(int.

策略：对weight和cost进行性价比计算排序，之后根据性价比依次装入 #include <iostream> #include <cstdio> #include <algorithm> using namespace std; const int MAXN = 1000; struct JavaBean{ double weight; double cost; }; JavaBean arr[MAXN]; //比较性价比 bool Comp.

#include <iostream> #include <cstdio> #include <string> #include <cstring> using namespace std; const int MAXN = 10000; struct BigInteger{ int digit[MAXN]; int length; BigInteger(); BigInteger(int x); BigInteger(string s.

#include <iostream> #include <cstdio> using namespace std; struct Matrix{ int matrix[3][3]; int row, col; Matrix(int r, int c): row(r), col(c){ } }; //矩阵乘法 Matrix Multiply(Matrix x, Matrix y){ Matrix answer(x.row, y.col); for(int i.

#include <iostream> #include <cstdio> using namespace std; struct Matrix{ int matrix[3][3]; int row, col; Matrix(int r, int c): row(r), col(c){ } }; Matrix Multiply(Matrix x, Matrix y){ Matrix answer(x.row, y.col); for(int i = 0; i &.

int FastExponentiation(int a, int b, int mod){ int answer = 1; while(b != 0){ //不断将b 转换成二进制数 if(b % 2 == 1){ answer *= a; answer %= mod; //只求后三位 } b /= 2; a *= a; //a 不断平方 a %= mod; } return answer; } #include <iost..

#include <iostream> #include <cstdio> #include <string> #include <vector> using namespace std; char IntToChar(int x){ //数字转字符 if(x < 10){ return x + '0'; }else{ return x - 10 + 'a'; } } int CharToInt(char c){ //字符.