wlh1998的博客

#include#include#include#includeusing namespace std;const int maxn = 1e6+5;struct node{ int jav; int food; double ave; bool operator< (const node&I)const { //按照ave由da到xiao排列 return I.av

//函数返回值写成int，函数内外用了ans，调试只能显示其中一个，//垃圾dev半天没调出来 #include#include#include#includeusing namespace std;const int maxn = 1e3+10;const int maxm = 1e6+10;const int INF = 0x3f3f3f3f;int fa[maxn]

//重边//双向图//if 忘记大括号 #include#include#includeusing namespace std; const int maxn = 2e2+3;const int INF = 0x3f3f3f3f;int mp[maxn][maxn];int vis[maxn];int dis[maxn];int main (){ int m,n;

#include#include#includeconst int INF = 0x3f3f3f3f;const int maxn = 1e3;double mp[maxn][maxn]; // 存图； int vis[maxn]; //是否在新图内； double dis[maxn]; //第i个顶点到新图最短距离； double prim(int n);struct

#include#includeconst int INF = 0x3f3f3f3f;const int maxn = 1e3;int fa[maxn];int mp[maxn][maxn]; // 存图； int vis[maxn]; //是否在新图内； int dis[maxn]; //第i个顶点到新图最短距离； int prim(int n);int find(

#include#includeconst int INF = 0x3f3f3f3f;const int maxn = 1e3;int mp[maxn][maxn]; // 存图； int vis[maxn]; //是否在新图内； int dis[maxn]; //第i个顶点到新图最短距离； int prim(int n);int n,m;int main (){ i

#include#include const int maxn = 1e5;int mp[maxn];int vis[maxn];int ans[maxn];int prime[maxn],res;bool isn_prime [maxn];void get_prime (int n);int n;void dfs(int k){ if (k == n + 1){ i

//奇偶剪枝//01010101//10101010//01010101//10101010//1奇数步走到0,1偶数步走到1. #include#include#includeint n,m,t,ans = 0;const int maxn = 1e3;char a[maxn][maxn];int vis[maxn][maxn];int dx[10] = {0,0,

//搜索注意已搜索标记，第一次忘了判断，wa了// 第二次判断写错， #include#include #include using namespace std;struct node{ int pos; int step; node(int _pos, int _step):pos (_pos) ,step(_step) { }};const int maxn = 1e

#include#include#includeusing namespace std;const int maxn = 1e3 + 5;const int INF = 0x3f3f3f3f;char mp[maxn][maxn];int vis[maxn][maxn];int dx[10] = {0,0,-1,1,1,1,-1,-1};int dy[10] = {1,-

#include#includeconst int INF = 0x3f3f3f3f;const int maxn = 22;int mps[maxn][maxn]; // 存图； int mp[maxn][maxn];int pw[maxn]; // 点权 int vis[maxn]; //是否在新图内； int dis[maxn]; //第i个顶点到新图最短距离；