图论刷题1

582div3 G. Path Queries

题意

给定一颗$n$个点的加权树，以及$m$次询问，每次询问输出存在简单路径中边权不大于$x$的顶点对数

( $1\le n,m\le 2\cdot 10^5$ )——树中的顶点数和查询数。$x\le 2\cdot 10^5$

思路

可以发现$x$越大则满足的点对数越多，所以考虑离线按$x$从小到大处理，每次将所有边权小于$x$的边加入，使用并查集维护互相可达的顶点数即可

代码

#include<bits/stdc++.h>#define ull unsigned long long 
#define ll long long
#define inf 1e9
#define INF 1e18
#define lc p<<1
#define rc p<<1|1
#define endl '\n'
#define all(a) a.begin()+1,a.end()
#define all0(a) a.begin(),a.end()
#define lowbit(a) (a&-a)
#define fi first
#define se second
#define pb push_back
#define yes cout<<"YES"<<endl
#define no cout<<"NO"<<endlusing namespace std;
const double eps=1e-6;
typedef pair<int,int>PII;
typedef array<int,3>PIII;
mt19937_64 rnd(time(0));  const int N=2e5+10;struct edge{int u,v,w;bool operator <(const edge&t)const{return w<t.w;}
};struct query{int x,id;bool operator <(const query&t)const{return x<t.x;}
};int p[N];
int siz[N];int find(int x)
{if(x!=p[x]) p[x]=find(p[x]);return p[x];
}void merge(int x,int y)
{x=find(x);y=find(y);if(x!=y){p[x]=y;siz[y]+=siz[x];}
}void solve()
{int n,m;cin>>n>>m;for(int i=1;i<=n;i++) {p[i]=i;siz[i]=1;}vector<edge>e(n);for(int i=1;i<n;i++){int a,b,c;cin>>a>>b>>c;e[i]={a,b,c};}sort(all(e));vector<query>q(m+1);for(int i=1;i<=m;i++) {cin>>q[i].x;q[i].id=i;}sort(all(q));vector<ll>ans(m+1);ll res=0;int j=1;for(int i=1;i<=m;i++){while(j<n && e[j].w<=q[i].x){int u=e[j].u;int v=e[j].v;u=find(u);v=find(v);if(u!=v){res+=1ll*siz[u]*siz[v];merge(u,v);}j++;}ans[q[i].id]=res;}for(int i=1;i<=m;i++) cout<<ans[i]<<" ";
}int main()
{ios::sync_with_stdio(0),cin.tie(0),cout.tie(0);solve();return 0;
}

516div1 B. Labyrinth

题意

在一个迷宫中，上下移动不受限制，但左右移动次数最多为$x,y$，求最多可到达的单元格数

思路

一开始四维枚举所有状态，不出所料的t了，其实是$01BFS$板子题，对于边权为$0$的操作(上下移动)将其加到队头，边权为1(左右移动)则加入到队尾即可，因为这样保证一定是距离短的点先到

代码

#include<bits/stdc++.h>#define ull unsigned long long 
#define ll long long
#define inf 1e9
#define INF 1e18
#define lc p<<1
#define rc p<<1|1
#define endl '\n'
#define all(a) a.begin()+1,a.end()
#define all0(a) a.begin(),a.end()
#define lowbit(a) (a&-a)
#define fi first
#define se second
#define pb push_back
#define yes cout<<"YES"<<endl
#define no cout<<"NO"<<endlusing namespace std;
const double eps=1e-6;
typedef pair<int,int>PII;
typedef array<int,3>PIII;
mt19937_64 rnd(time(0));  int dx[]={0,1,0,-1};
int dy[]={-1,0,1,0};void solve()
{int n,m,r,c,x,y;cin>>n>>m>>r>>c>>x>>y;vector<vector<char>>g(n+1,vector<char>(m+1));for(int i=1;i<=n;i++){for(int j=1;j<=m;j++) cin>>g[i][j];}deque<array<int,4>>q;vector<vector<int>>vis(n+1,vector<int>(m+1));q.push_back({r,c,x,y});int ans=0;while(!q.empty()){auto [x,y,ra,rb]=q.front();q.pop_front();if(vis[x][y] || ra<0 || rb<0)continue;vis[x][y]=1;ans++;for(int i=0;i<4;i++){int a=x+dx[i];int b=y+dy[i];if(a<1 || a>n || b<1 || b>m) continue;if(vis[a][b]) continue;if(g[a][b]=='*') continue;if(i==1 || i==3) {q.push_front({a,b,ra,rb});continue;}if(i==0) {q.push_back({a,b,ra-1,rb});continue;}if(i==2) q.push_back({a,b,ra,rb-1});}}cout<<ans<<endl;
}int main()
{ios::sync_with_stdio(0),cin.tie(0),cout.tie(0);solve();return 0;
}

div1 599 B. 0-1 MST

题意

给定$n$个点的完全图$n\le 10^5$,以及$m$条边，除了这$m$条边的边权为1，其余边的边权都为0,求最小生成树边权

思路

因为是完全图，直接建图跑最小生成树显然不可取，不难发现答案其实为补图的连通块个数-1，因为补图的边权都为0，我们跑出补图的连通块个数即可(时间复杂度有点妙？)

代码

#include<bits/stdc++.h>#define ull unsigned long long 
#define ll long long
#define inf 1e9
#define INF 1e18
#define lc p<<1
#define rc p<<1|1
#define endl '\n'
#define all(a) a.begin()+1,a.end()
#define all0(a) a.begin(),a.end()
#define lowbit(a) (a&-a)
#define fi first
#define se second
#define pb push_back
#define yes cout<<"YES"<<endl
#define no cout<<"NO"<<endlusing namespace std;
const double eps=1e-6;
typedef pair<int,int>PII;
typedef array<int,3>PIII;
mt19937_64 rnd(time(0));  const int N=1e5+10;
set<int>e[N];void solve()
{int n,m;cin>>n>>m;set<int>s;for(int i=1;i<=n;i++) s.insert(i);for(int i=1;i<=m;i++){int a,b;cin>>a>>b;e[a].insert(b);e[b].insert(a);}int ans=0;auto dfs=[&](auto &&dfs,int u)->void{vector<int>v;for(auto t:s){if(!e[t].count(u)) v.pb(t);//补图的点}for(auto t:v) s.erase(t);for(auto t:v) dfs(dfs,t);};for(int i=1;i<=n;i++){if(s.count(i)) {ans++;dfs(dfs,i);}}cout<<ans-1<<endl;}int main()
{ios::sync_with_stdio(0),cin.tie(0),cout.tie(0);solve();return 0;
}

div3 656 E. Directing Edges

题意

给定$n$个点的图，包含有向边和无向边，你需要把所有无向边指定方向，使得图是一个有向无环图

思路

要使图中无环，则需要基于拓扑序来定向，即让拓扑序较小的边指向拓扑序较大的边即可，注意拓扑排序一定包含所有点，所以我们用有向图跑拓扑排序得到拓扑序列之后，根据拓扑序定向即可

代码

#include<bits/stdc++.h>#define ull unsigned long long 
#define ll long long
#define inf 1e9
#define INF 1e18
#define lc p<<1
#define rc p<<1|1
#define endl '\n'
#define all(a) a.begin()+1,a.end()
#define all0(a) a.begin(),a.end()
#define lowbit(a) (a&-a)
#define fi first
#define se second
#define pb push_back
#define yes cout<<"YES"<<endl
#define no cout<<"NO"<<endlusing namespace std;
const double eps=1e-6;
typedef pair<int,int>PII;
typedef array<int,3>PIII;
mt19937_64 rnd(time(0));  const int N=2e5+10;
vector<int>e[N];void solve()
{int n,m;cin>>n>>m;for(int i=1;i<=n;i++) e[i].clear();vector<int>din(n+1);vector<int>U(m+1),V(m+1);for(int i=1;i<=m;i++){int a,b,c;cin>>a>>b>>c;if(a) e[b].pb(c);U[i]=b;V[i]=c;}vector<int>top;//top序列auto topsort=[&](){for(int i=1;i<=n;i++){for(auto ed:e[i]) din[ed]++;}queue<int>q;for(int i=1;i<=n;i++) if(!din[i]) q.push(i);while(!q.empty()){auto t=q.front();q.pop();top.pb(t);for(auto ed:e[t]){din[ed]--;if(!din[ed]) q.push(ed);}}return top.size()==n;};if(topsort()){yes;vector<int>rk(n+1);for(int i=0;i<n;i++) rk[top[i]]=i;for(int i=1;i<=m;i++){if(rk[U[i]]>rk[V[i]]) swap(U[i],V[i]);}for(int i=1;i<=m;i++) cout<<U[i]<<" "<<V[i]<<endl;}else no;}int main()
{ios::sync_with_stdio(0),cin.tie(0),cout.tie(0);int t;cin>>t;while(t--)solve();return 0;
}