POJ3177 Redundant Paths

文章目錄

• • Description
  • 題意：
  • 題解：
  • 代碼：

Time Limit: 1000MS Memory Limit: 65536K Total Submissions: 21945 Accepted: 9056

Description

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

7 7 1 2 2 3 3 4 2 5 4 5 5 6 5 7

Sample Output

2

Hint

Explanation of the sample:

One visualization of the paths is: 1 2 3 ±–±--+
| |
| | 6 ±–±--+ 4
/ 5
/
/ 7 + Building new paths from 1 to 6 and from 4 to 7 satisfies the conditions. 1 2 3 ±–±--+ : | | : | |
6 ±–±--+ 4
/ 5 :
/ :
/ : 7 + - - - - 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.

題意：

n個(gè)點(diǎn)，m個(gè)邊，問再添加多少邊可以使得任意兩點(diǎn)有兩條路徑（且不可重復(fù)）

題解：

如果任意兩點(diǎn)至少存在兩條邊不重復(fù)路徑，則稱該圖為邊雙連通的。
我們可以用Tarjan來求出每個(gè)邊雙聯(lián)通分量，對(duì)于同一個(gè)邊雙連通分量的點(diǎn)之間都至少有兩條路徑，但是不同之間只會(huì)有一條路徑。
所以我們對(duì)每個(gè)邊雙連通分量進(jìn)行縮點(diǎn)，就可以得到一個(gè)樹，要使這個(gè)無根數(shù)變成邊雙聯(lián)通圖，我們要先看看樹中誰需要連線，沒錯(cuò)就是葉子節(jié)點(diǎn)，如果我們將所有葉子節(jié)點(diǎn)都消滅那不就行了，所以至少要添加（葉子節(jié)點(diǎn)數(shù)+1）/2條邊
可以結(jié)合圖分析一下（圖中為題目給的樣例）

代碼：

代碼參考

#include <iostream> #include <cstdio> #include <algorithm> #include <cstring> using namespace std; typedef long long LL; const int INF = 2e9; const LL LNF = 9e18; const int MOD = 1e9+7; const int MAXN = 5e3+10;struct Edge {int to, next;bool cut; }edge[MAXN<<2]; int head[MAXN], tot;int index, dfn[MAXN], low[MAXN]; int block, belong[MAXN]; int top, Stack[MAXN], instack[MAXN]; int degree[MAXN];void addedge(int u, int v) {edge[tot].to = v;edge[tot].cut = false;edge[tot].next = head[u];head[u] = tot++; }void Tarjan(int u, int pre) {low[u] = dfn[u] = ++index;Stack[top++] = u;instack[u] = true;for(int i = head[u]; i!=-1; i = edge[i].next){int v = edge[i].to;if(v==pre) continue;if(!dfn[v]){Tarjan(v, u);low[u] = min(low[u], low[v]);if(low[v]>dfn[u])//當(dāng)前邊i所連接的點(diǎn)為葉子節(jié)點(diǎn) {edge[i].cut = true;edge[i^1].cut = true;//標(biāo)記兩次 }}else if(instack[v])low[u] = min(low[u], dfn[v]);}if(low[u]==dfn[u]){block++;int v;do{v = Stack[--top];instack[v] = false;belong[v] = block;}while(v!=u);} }void init() {tot = 0;memset(head,-1,sizeof(head));index = 0;memset(dfn,0,sizeof(dfn));memset(low,0,sizeof(low));block = top = 0;memset(instack,0,sizeof(instack));memset(degree,0,sizeof(degree)); }int main() {int n, m;while(scanf("%d%d",&n,&m)!=EOF){init();for(int i = 1; i<=m; i++){int u, v;scanf("%d%d",&u,&v);addedge(u, v);addedge(v,u);}Tarjan(1, 1);for(int u = 1; u<=n; u++)for(int i = head[u]; i!=-1; i = edge[i].next)if(edge[i].cut) //不需要兩端都加，因?yàn)橐粭l割邊被標(biāo)記了兩次。一次正好對(duì)應(yīng)一個(gè)端點(diǎn)。degree[belong[u]]++;//求各點(diǎn)的度數(shù) int leaf = 0;for(int i = 1; i<=block; i++)if(degree[i]==1) leaf++;printf("%d\n", (leaf+1)/2);} }

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