本文主要是介绍【Redundant Paths】【无向图】【双连通分量】【缩点】，希望对大家解决编程问题提供一定的参考价值，需要的开发者们随着小编来一起学习吧！
Redundant Paths
Time Limit: 3000/1000MS (Java/Others) Memory Limit: 65535/65535KB (Java/Others)
In order to get from one of the F(1≤F≤5,000) grazing fields (which are numbered 1⋯F ) to another field, Bessie and the rest of the herd are forced to cross near the Tree of Rotten Apples. The cows are now tired of often being forced to take a particular path and want to build some new paths so that they will always have a choice of at least two separate routes between any pair of fields. They currently have at least one route between each pair of fields and want to have at least two. Of course, they can only travel on Official Paths when they move from one field to another.
Given a description of the current set of R(F−1≤R≤10,000) paths that each connect exactly two different fields, determine the minimum number of new paths (each of which connects exactly two fields) that must be built so that there are at least two separate routes between any pair of fields. Routes are considered separate if they use none of the same paths, even if they visit the same intermediate field along the way.
There might already be more than one paths between the same pair of fields, and you may also build a new path that connects the same fields as some other path.
Input
Line 1 : Two space-separated integers: F and R
Lines 2⋯R+1 : Each line contains two space-separated integers which are the fields at the endpoints of some path.
Output
Line 1 : A single integer that is the number of new paths that must be built.
Sample input and output
| Sample Input | Sample Output |
|---|---|
7 7 1 2 2 3 3 4 2 5 4 5 5 6 5 7 | 2 |
Hint
Explanation of the sample:
One visualization of the paths is:
Building new paths from 1 to 6 and from 4 to 7 satisfies the conditions.
Check some of the routes:
- 1–2:1–>2 and 1–>6–>5–>2
- 1–4:1–>2–>3–>4 and 1–>6–>5–>4
- 3–7:3–>4–>7 and 3–>2–>5–>7 Every pair of fields is, in fact, connected by two routes.
It's possible that adding some other path will also solve the problem (like one from 6 to 7 ). Adding two paths, however, is the minimum
#include <iostream>
#include <cstring>
#include <cmath>
#include <queue>
#include <stack>
#include <list>
#include <map>
#include <set>
#include <string>
#include <cstdlib>
#include <cstdio>
#include <algorithm>
using namespace std;#define rep(i,a,n) for (int i=a;i<n;i++)
#define per(i,a,n) for (int i=n-1;i>=a;i--)
#define mp push_backconst int MAXN = 50010;
const int MAXM = 100010;
int n,m;
struct Edge
{int to,next,id;
}edge[MAXM];
int vis[MAXN];
int low[MAXN];
int myst[MAXN];
int inst[MAXN];
int myblock[MAXN];
int in_de[MAXN];
int out_de[MAXN];
int H[MAXN];
int top;
int len;
int times;
int sccnum;void addedge(int from,int to,int id)
{edge[len].id = id;edge[len].to = to;edge[len].next = H[from];H[from] = len ++;
}void tarjan(int u,int x)
{myst[top++] = u;low[u] = vis[u] = ++times;inst[u] = true;for(int i=H[u];~i;i=edge[i].next){int v = edge[i].to;if(x == edge[i].id) continue;if(vis[v] == -1){tarjan(v,edge[i].id);low[u] = min(low[u],low[v]);}else {low[u] = min(low[u],vis[v]);}}int t;if(low[u] == vis[u]){sccnum ++;do{t = myst[--top];inst[t] = false;myblock[t] = sccnum;} while(t != u);}
}void solve(int n)
{for (int i = 1; i <= n; ++i){if(vis[i] == -1)tarjan(i,-1);}for (int u = 1; u <= n; ++u){for (int j = H[u]; ~j; j = edge[j].next){int v = edge[j].to;if(myblock[u] != myblock[v]){out_de[myblock[v]] ++;out_de[myblock[u]] ++;}}}int cnt = 0;for(int i=1;i<=sccnum;i++){if(out_de[i]/2 == 1){++cnt;}}printf("%d\n", (cnt + 1) >> 1);}
int main()
{freopen("in.txt","r",stdin);while(scanf("%d%d",&n,&m) != EOF){len = 0;times = 0;top = 0;sccnum = 0;memset(vis,-1,sizeof(vis));memset(H,-1,sizeof(H));memset(inst,0,sizeof(inst));memset(myst,0,sizeof(myst));memset(myblock,0,sizeof(myblock));memset(in_de,0,sizeof(in_de));memset(out_de,0,sizeof(out_de));for(int i=1;i<=m;i++){int u,v;scanf("%d%d",&u,&v);addedge(u,v,i);addedge(v,u,i);}solve(n);}}
这篇关于【Redundant Paths】【无向图】【双连通分量】【缩点】的文章就介绍到这儿，希望我们推荐的文章对编程师们有所帮助！
