Case of Computer Network

CodeForces - 555E

Andrewid the Android is a galaxy-known detective. Now he is preparing a defense against a possible attack by hackers on a major computer network.

In this network are n vertices, some pairs of vertices are connected by m undirected channels. It is planned to transfer q important messages via this network, the i-th of which must be sent from vertex si to vertex di via one or more channels, perhaps through some intermediate vertices.

To protect against attacks a special algorithm was developed. Unfortunately it can be applied only to the network containing directed channels. Therefore, as new channels can't be created, it was decided for each of the existing undirected channels to enable them to transmit data only in one of the two directions.

Your task is to determine whether it is possible so to choose the direction for each channel so that each of the q messages could be successfully transmitted.

Input

The first line contains three integers nm and q (1 ≤ n, m, q ≤ 2·105) — the number of nodes, channels and important messages.

Next m lines contain two integers each, vi and ui (1 ≤ vi, ui ≤ nvi ≠ ui), that means that between nodes vi and ui is a channel. Between a pair of nodes can exist more than one channel.

Next q lines contain two integers si and di (1 ≤ si, di ≤ nsi ≠ di) — the numbers of the nodes of the source and destination of the corresponding message.

It is not guaranteed that in it initially possible to transmit all the messages.

Output

If a solution exists, print on a single line "Yes" (without the quotes). Otherwise, print "No" (without the quotes).

Examples

Input
4 4 2
1 2
1 3
2 3
3 4
1 3
4 2
Output
Yes
Input
3 2 2
1 2
3 2
1 3
2 1
Output
No
Input
3 3 2
1 2
1 2
3 2
1 3
2 1
Output
Yes

Note

In the first sample test you can assign directions, for example, as follows: 1 → 2, 1 → 3, 3 → 2, 4 → 3. Then the path for for the first message will be 1 → 3, and for the second one — 4 → 3 → 2.

In the third sample test you can assign directions, for example, as follows: 1 → 2, 2 → 1, 2 → 3. Then the path for the first message will be 1 → 2 → 3, and for the second one — 2 → 1.

简要题意:给出一个无向图,给出q个询问S,T表示从S走到T。
问能否给这张图的边定向,使得满足q个询问

sol:因为是无向图,所以在同一个强联通分量中是两两可以随意到达,然后可以搞成一棵树,树上可以差分一下,两个数组p0,p1,S位置p0++,T位置p1++,Lca处p0--,p1--,看看是否有一个点既有p0又有p1就No了

/*
¼òÒªÌâÒ⣺¸ø³öÒ»¸öÎÞÏòͼ£¬¸ø³öq¸öѯÎÊS£¬T±íʾ´ÓS×ßµ½T¡£
ÎÊÄÜ·ñ¸øÕâÕÅͼµÄ±ß¶¨Ïò£¬Ê¹µÃÂú×ãq¸öѯÎÊ
*/
#include <bits/stdc++.h>
using namespace std;
typedef int ll;
inline ll read()
{
ll s=; bool f=; char ch=' ';
while(!isdigit(ch)) {f|=(ch=='-'); ch=getchar();}
while(isdigit(ch)) {s=(s<<)+(s<<)+(ch^); ch=getchar();}
return (f)?(-s):(s);
}
#define R(x) x=read()
inline void write(ll x)
{
if(x<) {putchar('-'); x=-x;}
if(x<) {putchar(x+''); return;}
write(x/); putchar((x%)+'');
}
#define W(x) write(x),putchar(' ')
#define Wl(x) write(x),putchar('\n')
const int N=,M=;
int n,m,Q;
int tot=,Next[M],to[M],head[N];
struct Ques{int S,T;}qq[N];
inline void Link(int x,int y)
{
Next[++tot]=head[x]; to[tot]=y; head[x]=tot;
}
int cnt=,dfn[N],low[N],top=,Sta[N],now=,Bel[N];
bool ins[N];
inline void tarjan(int x,int fat)
{
int i;
bool bo=;
dfn[x]=low[x]=++cnt; Sta[++top]=x; ins[x]=;
for(i=head[x];i;i=Next[i])
{
if(to[i]==fat&&bo) {bo=; continue;}
if(!dfn[to[i]])
{
tarjan(to[i],x); low[x]=min(low[x],low[to[i]]);
}
else if(ins[to[i]]) low[x]=min(low[x],dfn[to[i]]);
}
if(dfn[x]==low[x])
{
int oo=; now++;
while(oo!=x)
{
oo=Sta[top--]; Bel[oo]=now; ins[oo]=;
}
}
}
vector<int>E[N];
#define PB push_back
int fa[N];
inline int gf(int x)
{
return (fa[x]==x)?(x):(fa[x]=gf(fa[x]));
}
int Dep[N],ff[N][];
inline void dfs(int x,int fat)
{
int i;
for(i=;i<E[x].size();i++) if(E[x][i]!=fat)
{
Dep[E[x][i]]=Dep[x]+; ff[E[x][i]][]=x; dfs(E[x][i],x);
}
}
inline int ask_lca(int x,int y)
{
int i;
if(Dep[x]<Dep[y]) swap(x,y);
for(i=;~i;i--) if(Dep[ff[x][i]]>=Dep[y]) x=ff[x][i];
if(x==y) return x;
for(i=;~i;i--) if(ff[x][i]!=ff[y][i]) x=ff[x][i],y=ff[y][i];
return ff[x][];
}
int path[N][];
bool Vis[N];
inline void dfss(int x,int fat)
{
int i;
Vis[x]=;
for(i=;i<E[x].size();i++)
{
int V=E[x][i]; if(V==fat) continue;
dfss(V,x);
path[x][]+=path[V][]; path[x][]+=path[V][];
}
}
int main()
{
// freopen("codeforces555E.in","r",stdin);
int i,j,x,y;
R(n); R(m); R(Q);
for(i=;i<=m;i++)
{
R(x); R(y); Link(x,y); Link(y,x);
}
for(i=;i<=Q;i++) {R(qq[i].S); R(qq[i].T);}
for(i=;i<=n;i++) if(!dfn[i]) tarjan(i,);
// cout<<"now="<<now<<endl;
// for(i=1;i<=n;i++) cout<<i<<' '<<Bel[i]<<endl;
// puts("");
for(i=;i<=now;i++) fa[i]=i;
for(i=;i<=n;i++)
{
for(j=head[i];j;j=Next[j])
{
int o1=gf(Bel[i]),o2=gf(Bel[to[j]]);
if(o1==o2) continue;
E[Bel[i]].PB(Bel[to[j]]); E[Bel[to[j]]].PB(Bel[i]); fa[o1]=o2;
}
}
for(i=;i<=now;i++) if(!Dep[i]) {Dep[i]=; dfs(i,);}
// for(i=1;i<=now;i++) cout<<i<<' '<<Dep[i]<<endl;
// puts("");
for(i=;i<=;i++) for(j=;j<=now;j++) ff[j][i]=ff[ff[j][i-]][i-];
// for(i=1;i<=now;i++) cout<<i<<' '<<ff[i][0]<<endl;
// puts("");
for(i=;i<=Q;i++)
{
int S=Bel[qq[i].S],T=Bel[qq[i].T];
// cout<<S<<' '<<gf(S)<<' '<<T<<' '<<gf(T)<<endl;
if(gf(S)!=gf(T)) return puts("No"),;
int lca=ask_lca(S,T);
// cout<<S<<' '<<T<<' '<<"lca="<<lca<<endl;
path[S][]++; path[lca][]--; path[T][]++; path[lca][]--;
}
// puts("");
// for(i=1;i<=now;i++) cout<<i<<' '<<path[i][0]<<' '<<path[i][1]<<endl;
for(i=;i<=now;i++) if(!Vis[i]) dfss(i,);
for(i=;i<=Q;i++)
{
int S=Bel[qq[i].S],T=Bel[qq[i].T]; if(S==T) continue;
if((path[S][]&&path[S][])||(path[T][]&&path[T][])) return puts("No"),;
}
puts("Yes");
return ;
}
/*
Input
4 4 2
1 2
1 3
2 3
3 4
1 3
4 2
Output
Yes Input
3 2 2
1 2
3 2
1 3
2 1
Output
No Input
3 3 2
1 2
1 2
3 2
1 3
2 1
Output
Yes
*/

codeforces555E的更多相关文章

随机推荐

  1. 装饰器中functools的用处

    定义一个最简单的装饰器 def user_login_data(f): def wrapper(*args, **kwargs): return f(*args, **kwargs) return w ...

  2. [Lua]LuaAPI整理

    ref :https://blog.csdn.net/ouyangshima/article/details/43339571   LUA和C/C++的沟通桥梁——栈 Lua生来就是为了和C交互的,因 ...

  3. 怎样监听xhr.readyState值的变化

    可以使用 xhr.onreadystatechange 属性指向的函数去监听 xhr.readyState 值的变化. 示例如下: var xhr = new XMLHttpRequest(); xh ...

  4. javascript——onsubmit和onreset事件 和开发中常用的方式

    <head> <meta charset="UTF-8"> <title></title> <script> funct ...

  5. springcloud必知功能使用教程

    springcloud Spring Cloud是一系列框架的有序集合.它利用Spring Boot的开发便利性巧妙地简化了分布式系统基础设施的开发,如服务发现注册.配置中心.消息总线.负载均衡.断路 ...

  6. BRD——>MRD——>PRD,产品经理三大文档概念详解及前后逻辑

    转自:https://blog.csdn.net/neikutaixiao/article/details/40819445 商业需求文档Business Requirement DocumentBR ...

  7. SQL学习——LIKE运算符

    原文链接 LIKE 作用 在WHERE子句中使用LIKE运算符来搜索列中的指定模式. 有两个通配符与LIKE运算符一起使用: % - 百分号表示零个,一个或多个字符 _ - 下划线表示单个字符 注意: ...

  8. 离线文档 real

    mac 上用 dash windows 用 real 上官网下载安装 tool-> 安装需要的文档 就可以使用. 没有难度,记录一下.

  9. Shell脚本相关

    cat /proc/17616/cmdline 17616代表进程号 用这个可以完整打印出当前的进程的全名 当前shell的进程号.你可以使用ps -A 看你自己shell 的pid.是内置变量. $ ...

  10. 前端知识体系:JavaScript基础-原型和原型链-instanceof的底层实现原理

    instanceof的底层实现原理(参考文档) instanceof的实现实际上是调用JS的内部函数 [[HasInstance]] 来实现的,其实现原理是:只要右边变量的prototype在左边变量 ...