yyg的专栏

思路：最终添加完边的图，肯定可以分成两个部X和Y，其中只有X到Y的边没有Y到X的边，那么要使得边数尽可能的多，则X部肯定是一个完全图，Y部也是，同时X部中每个点到Y部的每个点都有一条边，假设X部有x个点，Y部有y个点，有x+y=n，同时边数F=x*y+x*(x-1)+y*(y-1)，整理得：F=N*N-N-x*y，当x+y为定值时，二者越接近，x*y越大，所以要使得边数最多，那么X部和Y部

#include#include#include#includeusing namespace std;const int inf = 0x3f3f3f3f;int n , m , a , b , c;double map[105][105];double dist[105] , ans1 , ans2;int vis[105] , use[105][105] , p[20

use标记当前边是否属于最小生成树。d[i][j]代表从i->j的瓶颈路。#include#include#includeusing namespace std;const int inf = 0x3f3f3f3f;int n , m , a , b , c;int map[105][105];int dist[105] , vis[105] , ans1 , an

复习了一下次小生成树。。。。#include#include#include#includeusing namespace std;const int inf = 0x3f3f3f3f;int n , m , a , b , c;double map[1005][1005];double dist[1005] , ans1 , ans2;int vis[1005]

#include#include#include#includeusing namespace std;const int maxn = 100500;const int inf = 0x3f3f3f3f;int t , n , m , c , vis[maxn * 4] , cnt[maxn * 4] , dist[maxn * 4] , cc , a , b;struct

#include#include#includeconst int maxn = 70010;const int inf = 0x3f3f3f3f;using namespace std;int vis[maxn] , cnt[maxn] , dist[maxn] , n , m , c , a , b;struct G{ int head[maxn] , e;

#include#include#includeconst int maxn = 15005;int n , m , fa[maxn] , rank[maxn];struct T{ int u , v , c;}edge[maxn * 2] , edge1[maxn * 2];void init(){ for(int i = 1 ; i <= n ; i ++

#include#include#includeconst int maxn = 255003;int fa[maxn] , rank[maxn] , n , m , k , p , q , c , tot , head[maxn] , ans , cnt;struct T{ int u , v , c , next;}edge[maxn * 2];int cmp(

#include#include#includeusing namespace std;const int maxn = 5008;int head[maxn] , low[maxn] , pre[maxn] , depth , iscut[maxn] , instack[maxn];int n , m , e , a , b , ret , del;struct T{

思路：首先用tarjan算法算出每个点属于哪个连通块，再缩点后求出各个连通块的入度出度。    入度为0的点的个数就是问一的解。            入度和出度为0的点个数多的就是问二的解。代码：#include#include#include#includeusing namespace std;const int maxn = 119;int n , in[ma

莫名其妙的 runtime error。代码：#include#include#include#define _n printf("\n");using namespace std;struct T{ int x; int y; int step;};int dir[5][5] = {{1 , 0} , {-1 , 0} , {0 , 1} , {0 , -1}};

简单BFS#include#include#include#include#includeusing namespace std;const int maxn = 1005;int n , ans , e , m;int head[maxn] , vis[maxn];bool flag;struct T{ int u , v , next;}edge[m