逃不掉的路 CH Round #24 - 三体杯 Round #1
题目描述
现代社会，路是必不可少的。任意两个城镇都有路相连，而且往往不止一条。但有些路连年被各种XXOO，走着很不爽。按理说条条大路通罗马，大不了绕行其他路呗——可小撸却发现：从a城到b城不管怎么走，总有一些逃不掉的必经之路。
他想请你计算一下，a到b的所有路径中，有几条路是逃不掉的？
输入格式
第一行是n和m，用空格隔开。接下来m行，每行两个整数x和y，用空格隔开，表示x城和y城之间有一条长为1的双向路。第m+2行是q。接下来q行，每行两个整数a和b，用空格隔开，表示一次询问。
输出格式
对于每次询问，输出一个正整数，表示a城到b城必须经过几条路。
样例输入
5 5
1 2
1 3
2 4
3 4
4 5
2
1 4
2 5
1 2
1 3
2 4
3 4
4 5
2
1 4
2 5
样例输出
0
1
1
样例解释
第1次询问，1到4的路径有 1--2--4 ，还有 1--3--4 。没有逃不掉的道路，所以答案是0。
第2次询问，2到5的路径有 2--4--5 ，还有 2--1--3--4--5 。必须走“4--5”这条路，所以答案是1。
数据约定与范围
共10组数据，每组10分。
有3组数据，n ≤ 100 , n ≤ m ≤ 200 , q ≤ 100。
另有2组数据，n ≤ 103 , n ≤ m ≤ 2 x 103 , 100 < q ≤ 105。
另有3组数据，103 < n ≤ 105 , m = n-1 , 100 < q ≤ 105。
另有2组数据，103 < n ≤ 105 , n ≤ m ≤ 2 x 105 , 100 < q ≤ 105。
对于全部的数据，1 ≤ x,y,a,b ≤ n；对于任意的道路，两端的城市编号之差不超过104；
任意两个城镇都有路径相连；同一条道路不会出现两次；道路的起终点不会相同;查询的两个城市不会相同。
题解
要求无向图的必经边，较为简单的做法是找出边双缩点，然后转化成树上距离问题。
#include<bits/stdc++.h>
#define rg register
#define il inline
#define co const
template<class T>il T read(){
rg T data=0,w=1;rg char ch=getchar();
for(;!isdigit(ch);ch=getchar())if(ch=='-') w=-w;
for(;isdigit(ch);ch=getchar()) data=data*10+ch-'0';
return data*w;
}
template<class T>il T read(rg T&x) {return x=read<T>();}
typedef long long ll;
using namespace std;
co int N=2e5+1;
int n,m;
int head[N],ver[N*2],next[N*2],tot=1;
int dfn[N],low[N],c[N],num,dcc;
bool bridge[N*2];
int hc[N],vc[N*2],nc[N*2],tc=1;
int d[N],f[N][18];
queue<int> q;
il void add(int x,int y){
ver[++tot]=y,::next[tot]=head[x],head[x]=tot;
}
il void add_c(int x,int y){
vc[++tc]=y,nc[tc]=hc[x],hc[x]=tc;
}
void tarjan(int x,int in_edge){
dfn[x]=low[x]=++num;
for(int i=head[x];i;i=::next[i]){
int y=ver[i];
if(!dfn[y]){
tarjan(y,i);
low[x]=min(low[x],low[y]);
if(low[y]>dfn[x]) bridge[i]=bridge[i^1]=1;
}
else if(i!=(in_edge^1)) low[x]=min(low[x],dfn[y]);
}
}
void dfs(int x){ // dye
c[x]=dcc;
for(int i=head[x];i;i=::next[i]){
int y=ver[i];
if(c[y]||bridge[i]) continue;
dfs(y);
}
}
void bfs(){
}
int lca(int x,int y){
if(d[x]<d[y]) swap(x,y);
for(int i=17;i>=0;--i)
if(d[f[x][i]]>=d[y]) x=f[x][i];
if(x==y) return x;
for(int i=17;i>=0;--i)
if(f[x][i]!=f[y][i]) x=f[x][i],y=f[y][i];
return f[x][0];
}
int main(){
read(n),read(m);
for(int x,y;m--;){
read(x),read(y);
add(x,y),add(y,x);
}
// 缩点建图
for(int i=1;i<=n;++i)
if(!dfn[i]) tarjan(i,0);
for(int i=1;i<=n;++i)
if(!c[i]) ++dcc,dfs(i);
for(int i=2;i<=tot;++i){
int x=ver[i^1],y=ver[i];
if(c[x]!=c[y]) add_c(c[x],c[y]);
}
// 倍增lca预处理
d[1]=1,q.push(1);
while(q.size()){
int x=q.front();q.pop();
for(int i=hc[x];i;i=nc[i]){
int y=vc[i];
if(d[y]) continue;
d[y]=d[x]+1,f[y][0]=x;
for(int j=1;j<18;++j) f[y][j]=f[f[y][j-1]][j-1];
q.push(y);
}
}
// 处理询问
read(m);
for(int x,y;m--;){
x=c[read<int>()],y=c[read<int>()];
printf("%d
",d[x]+d[y]-2*d[lca(x,y)]);
}
return 0;
}
Traffic Real Time Query System
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 3808 Accepted Submission(s): 759
Problem Description
City C is really a nightmare of all drivers for its traffic jams. To solve the traffic problem, the mayor plans to build a RTQS (Real Time Query System) to monitor all traffic situations. City C is made up of N crossings and M roads, and each road connects two crossings. All roads are bidirectional. One of the important tasks of RTQS is to answer some queries about route-choice problem. Specifically, the task is to find the crossings which a driver MUST pass when he is driving from one given road to another given road.
Input
There are multiple test cases.
For each test case:
The first line contains two integers N and M, representing the number of the crossings and roads.
The next M lines describe the roads. In those M lines, the ith line (i starts from 1)contains two integers Xi and Yi, representing that roadi connects crossing Xi and Yi (Xi≠Yi).
The following line contains a single integer Q, representing the number of RTQs.
Then Q lines follows, each describing a RTQ by two integers S and T(S≠T) meaning that a driver is now driving on the roads and he wants to reach roadt . It will be always at least one way from roads to roadt.
The input ends with a line of “0 0”.
Please note that: 0<N<=10000, 0<M<=100000, 0<Q<=10000, 0<Xi,Yi<=N, 0<S,T<=M
For each test case:
The first line contains two integers N and M, representing the number of the crossings and roads.
The next M lines describe the roads. In those M lines, the ith line (i starts from 1)contains two integers Xi and Yi, representing that roadi connects crossing Xi and Yi (Xi≠Yi).
The following line contains a single integer Q, representing the number of RTQs.
Then Q lines follows, each describing a RTQ by two integers S and T(S≠T) meaning that a driver is now driving on the roads and he wants to reach roadt . It will be always at least one way from roads to roadt.
The input ends with a line of “0 0”.
Please note that: 0<N<=10000, 0<M<=100000, 0<Q<=10000, 0<Xi,Yi<=N, 0<S,T<=M
Output
For each RTQ prints a line containing a single integer representing the number of crossings which the driver MUST pass.
Sample Input
5 6 1 2 1 3 2 3 3 4 4 5 3 5 2 2 3 2 4 0 0
Sample Output
0 1
Source
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给你一个无向图，询问边a和边b，问从边a到边b的路径中几个点是必须要经过的。
题解
求无向图的必经点，跟上题类似，求出点双后缩点，然后转化成查询路径上割点的个数。由于点双缩点的特殊性，所以算距离的时候写的不一样。
#include<iostream>
#include<cstring>
#include<queue>
#define rg register
#define il inline
#define co const
template<class T>il T read(){
rg T data=0,w=1;rg char ch=getchar();
for(;!isdigit(ch);ch=getchar())if(ch=='-') w=-w;
for(;isdigit(ch);ch=getchar()) data=data*10+ch-'0';
return data*w;
}
template<class T>il T read(rg T&x) {return x=read<T>();}
typedef long long ll;
using namespace std;
co int N=2e4+1,M=2e5+2;
int n,m,t;
int head[N],ver[M],next[M],tot;
int dfn[N],low[N],stack[N],new_id[N],c[N],belong[M],num,root,top,cnt;
int d[N],dist[N],f[N][16];
bool cut[N];
vector<int> dcc[N];
int hc[N],vc[M],nc[M],tc;
queue<int> q;
il void add(int x,int y){
ver[++tot]=y,::next[tot]=head[x],head[x]=tot;
}
il void add_c(int x,int y){
vc[++tc]=y,nc[tc]=hc[x],hc[x]=tc;
}
void tarjan(int x){
dfn[x]=low[x]=++num;
stack[++top]=x;
if(x==root&&head[x]==0){
dcc[++cnt].push_back(x);
return;
}
int flag=0;
for(int i=head[x];i;i=::next[i]){
int y=ver[i];
if(!dfn[y]){
tarjan(y);
low[x]=min(low[x],low[y]);
if(low[y]>=dfn[x]){
++flag;
if(x!=root||flag>1) cut[x]=1;
++cnt;
int z;
do{
z=stack[top--];
dcc[cnt].push_back(z);
}while(z!=y);
dcc[cnt].push_back(x);
}
}
else low[x]=min(low[x],dfn[y]);
}
}
void bfs(int s){
d[s]=1,dist[s]=s>cnt,q.push(s);
for(int i=0;i<16;++i) f[s][i]=0;
while(q.size()){
int x=q.front();q.pop();
for(int i=hc[x];i;i=nc[i]){
int y=vc[i];
if(d[y]) continue;
d[y]=d[x]+1,dist[y]=dist[x]+(y>cnt),f[y][0]=x;
for(int j=1;j<16;++j) f[y][j]=f[f[y][j-1]][j-1];
q.push(y);
}
}
}
int lca(int x,int y){
if(d[x]<d[y]) swap(x,y);
for(int i=15;i>=0;--i)
if(d[f[x][i]]>=d[y]) x=f[x][i];
if(x==y) return x;
for(int i=15;i>=0;--i)
if(f[x][i]!=f[y][i]) x=f[x][i],y=f[y][i];
return f[x][0];
}
int main(){
while(read(n)|read(m)){
memset(head,0,sizeof head);
memset(hc,0,sizeof hc);
memset(dfn,0,sizeof dfn);
memset(d,0,sizeof d);
memset(cut,0,sizeof cut);
memset(c,0,sizeof c);
for(int i=1;i<=n;++i) dcc[i].clear();
tot=1,num=cnt=top=0;
for(int x,y;m--;){
read(x),read(y);
add(x,y),add(y,x);
}
for(int i=1;i<=n;++i)
if(!dfn[i]) root=i,tarjan(i);
num=cnt;
for(int i=1;i<=n;++i)
if(cut[i]) new_id[i]=++num;
tc=1;
for(int i=1;i<=cnt;++i){
for(int j=0;j<dcc[i].size();++j){
int x=dcc[i][j];
if(cut[x]) add_c(i,new_id[x]),add_c(new_id[x],i);
c[x]=i;
}
for(int j=0;j<dcc[i].size()-1;++j){
int x=dcc[i][j];
for(int k=head[x];k;k=::next[k]){
int y=ver[k];
if(c[y]==i) belong[k/2]=i; // 输入的第k/2条边(x,y)处于第i个v-DCC内
}
}
}
// 编号 1~cnt 的为原图的v-DCC，编号 cnt+1~num 的为原图割点
for(int i=1;i<=num;++i)
if(!d[i]) bfs(i);
read(t);
while(t--){
int es=read<int>(),et=read<int>();
int x=belong[es],y=belong[et],z=lca(x,y);
printf("%d
",dist[x]+dist[y]-2*dist[z]+(z>cnt));
}
}
return 0;
}