ZROI17普及23-A.如烟题解--技巧枚举

• 题目链接

  因版权原因不予提供

• 分析

  别看这是普及模拟赛,其实基本上是提高难度...像这题做NOIpT1的话也说的过去

  有个很显然的暴力思路就是枚举c,a,b,时间复杂度\(O(N^3)\)，

  然后正解其实就是改变枚举顺序,我们先枚举a点,然后将所有可作为c点的点存起来,再从那些c点遍历得到可行b点统计答案,这样就不会重复且符合题意

  不过这道题需要仔细读题，像我这种菜B一开始就理解错题意了

• 代码

include

include

include

include

include

include

include

include

include

define ll long long

define ri register int

define ull unsigned long long

using namespace std;
template inline void read(T &x){
x=0;int ne=0;char c;
while(!isdigit(c=getchar()))ne=c'-';
x=c-48;
while(isdigit(c=getchar()))x=(x<<3)+(x<<1)+c-48;
x=ne?-x:x;return ;
}
const int maxn=3005;
const int inf=0x7fffffff;
struct Edge{
int ne,to;
}edge[maxn<<1],_edge[maxn<<1];
int h[maxn],num_edge=1;
inline void add_edge(int f,int to){
edge[++num_edge].ne=h[f];
edge[num_edge].to=to;
h[f]=num_edge;
}
int _h[maxn],_num_edge=1;//反向边
inline void _add_edge(int f,int to){
_edge[++_num_edge].ne=_h[f];
_edge[_num_edge].to=to;
_h[f]=_num_edge;
}
int n,m,sc,sb;
bool vis[maxn];
vector g;
void _dfs(int now,int fa){
int v;if(vis[now])return ;
vis[now]=1;g.push_back(now);
for(ri i=_h[now];i;i=_edge[i].ne){
v=_edge[i].to;
if(vfa)continue;
_dfs(v,now);
}
return ;
}
void dfs(int now,int fa){
int v;if(vis[now])return ;
sb++;vis[now]=1;
for(ri i=h[now];i;i=edge[i].ne){
v=edge[i].to;
if(v==fa)continue;
dfs(v,now);
}
return ;
}
int main(){
int T,x,y,z;
srand(19260817);//闷声发大财 预祝长者大寿 Long Live Jiang !!!
read(T);
while(T--){
ll ans=0;
read(n),read(m);
memset(h,0,sizeof(h));
memset(_h,0,sizeof(_h));
num_edge=_num_edge=1;
for(ri i=1;i<=m;i++){
read(x),read(y);
add_edge(x,y);
_add_edge(y,x);
}
int ss=(n+1)*sizeof(bool);
for(ri i=1;i<=n;i++){
memset(vis,0,ss);
g.clear();
_dfs(i,0);
memset(vis,0,ss);
for(ri j=0;j<g.size();j++){
sb=0;
dfs(g[j],0);
ans+=sb;
//printf("--%d %d %d\n",g[j],i,sb);
}
}
printf("%lld\n",ans);
}
return 0;
}