題目傳送門
Problem Description
There are some locations in a park, and some of them are connected by roads. The park manger needs to build some railways along the roads, and he would like to arrange tourist routes to each circuit. If a railway belongs to more than one tourist routes, there might be clash on it, and if a railway belongs to none tourist route, it doesn’t need to build.
Now we know the plan, and can you tell us how many railways are no need to build and how many railways where clash might happen.
Input
The Input consists of multiple test cases. The first line of each test case contains two integers, n (0 < n <= 10000), m (0 <= m <= 100000), which are the number of locations and the number of the railways. The next m lines, each line contains two integers, u, v (0 <= u, v < n), which means the manger plans to build a railway on the road between u and v.
You can assume that there is no loop and no multiple edges.
The last test case is followed by two zeros on a single line, which means the end of the input.
Output
Output the number of railways that are no need to build, and the number of railways where clash might happen. Please follow the format as the sample.
Sample Input
8 10
0 1
1 2
2 3
3 0
3 4
4 5
5 6
6 7
7 4
5 7
0 0
Sample Output
1 5
思路
不衝突路也就是橋邊，這個直接一個模板就完事了。對於有n個節點和n條邊的點雙連通分量,這種分量只有一個大環,不存在其他任何環了,所以這種分量中的邊都不是衝突邊。而一個點雙連通分量(大環)有小的點雙連通分量(小環)，也就是大環是由多個小環組成的這種點雙連通分量，那麼這個大環所有的邊都是衝突邊。根據這個原理 並且 圖沒有重邊那麼同時跑割邊和點雙連通分量。用堆棧記錄邊，用Set對點判重。
#include <iostream>
#include <cstring>
#include <cstdio>
#include <algorithm>
#include <vector>
#include <stack>
#include <set>
using namespace std;
const int maxn = 10005;
struct edge{
int from;
int to;
int next;
}e[maxn*20];
int head[maxn];
int dfn[maxn];
int low[maxn];
stack<int>st;
set<int>s;
int tot,cnt;
int res,sum;
void clear_set()
{
tot = cnt = sum = res = 0;
memset(head,-1,sizeof(head));
memset(low,0,sizeof(low));
memset(dfn,0,sizeof(dfn));
while(!st.empty()) st.pop();
}
void addedge(int x,int y)
{
e[tot].from = x;
e[tot].to = y;
e[tot].next = head[x];
head[x] = tot++;
}
void tarjan(int x,int fx)
{
dfn[x] = low[x] = ++cnt;
int k = 0;
for(int i = head[x];~i;i = e[i].next){
int y = e[i].to;
if(y == fx) continue;
if(!dfn[y]){
st.push(i);
tarjan(y,x);
low[x] = min(low[x],low[y]);
if(low[y] > dfn[x]){ //橋邊
res++;
}
if(low[y] >= dfn[x]){ //點雙連通分量
int num = 0;s.clear();
while(true){
int t = st.top();
num++;
st.pop();
s.insert(e[t].from);
s.insert(e[t].to);
if(e[t].from == x && e[t].to == y) break;
}
if(num > s.size()) sum += num; //邊數多於點數則衝突
}
}
else if(dfn[y] < dfn[x]){
st.push(i);
low[x] = min(low[x],dfn[y]);
}
}
}
int main()
{
int n,m;
while(~scanf("%d%d",&n,&m)){
if(n == 0 && m == 0) break;
clear_set();
int x,y;
for(int i = 0;i < m;i++){
scanf("%d%d",&x,&y);
addedge(x,y); addedge(y,x);
}
for(int i = 0;i < n;i++){
if(!dfn[i]){
tarjan(i,-1);
}
}
printf("%d %d\n",res,sum);
}
return 0;
}
願你走出半生，歸來仍是少年~