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;

}