hcCuriosity的博客

前向星void add_edge(int u, int v, int w) { //加边，u起点，v终点，w边权 edge[cnt].to = v; //终点 edge[cnt].w = w; //权值 edge[cnt].next = head[u];//以u为起点上一条边的编号，也就是与这个边起点相同的上一条边的编号 head[u] = cnt++;//更新以u为起点上一条边的编号}弗洛伊德（floyd）void Floyd(MGraph *mGraph, int **iArrPath

大数乘以正整数string Multiply(string s,int x) //大数乘以整形数{ reverse(s.begin(),s.end()); int cmp=0; for(int i=0;i<s.size();i++) { cmp=(s[i]-'0')*x+cmp; s[i]=(cmp%10+'0'); cmp/=10; } while(cmp) { s+=(cm

HDU Parentheses Matching 2020 第三场 1009杭电OJ6799Problem DescriptionGiven a string P consisting of only parentheses and asterisk characters (i.e. “(”, “)” and “"), you are asked to replace all the asterisk characters in order to get a balanced parenthesi