[POJ 3177] Redundant Paths【Tarjan边双连通分量】
Problem:
| Time Limit: 1000MS | Memory Limit: 65536K |
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
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.
Source
Solution:
这道题的思路很明确，但是处理起来有一定的难度。
首先题目的要求是加最少的边使整张图构成一个边双连通分量。
那么我们很容易想到先将原图的边双缩点，那么就可以得到一棵树，并且这棵树的所有边都是桥（即割边）。
观察这棵树可以发现，所有的非叶子节点（如果根只有一棵子树，我们也将其看作“叶子结点”）都和至少 2 个其他的点相连，所以我们只需要保持叶子结点的连通性即可。所以在每 2 个叶子结点上连一条边，我们就可以保证得到的图是边双连通的。
注意数据中可能出现重边，所以判回边时不能直接判当前边的终点与父节点是否相等，而是要判当前边是否为前驱边的反向边。具体见代码。
Code: O(F+R) [192K, 0MS]
#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<cmath>
#include<iostream>
#include<algorithm>
using namespace std;
int F, R;
struct Edge{
int np;
Edge *nxt, *rev;
};
struct Graph{
Edge *V[5002], E[20002];
int tope;
inline void addedge(int u, int v){
E[tope].np = v, E[tope].nxt = V[u], V[u] = &E[tope];
E[tope].rev = &E[tope ^ 1], tope++;
}
} G, C;
int dfn[5002], dfstime = 0;
int low[5002];
int Stack[5002], tops = 0;
int Status[5002];
int Color[5002], topc = 0;
inline void Tarjan(int u, Edge *prev){
dfn[u] = low[u] = ++dfstime;
Stack[++tops] = u, Status[u] = 1;
for(register Edge *ne = G.V[u]; ne; ne = ne->nxt){
if(prev && ne == prev->rev) continue;
// Not to backtrack
// Can't use "ne->np == fa" to find reversed edge, because there may be repeated paths
if(Status[ne->np] == 0){
Tarjan(ne->np, ne);
low[u] = min(low[u], low[ne->np]);
}
else if(Status[ne->np] == 1) low[u] = min(low[u], dfn[ne->np]);
}
if(low[u] == dfn[u]){
topc++;
while(Stack[tops + 1] != u){
Color[Stack[tops]] = topc;
Status[Stack[tops--]] = -1;
}
}
}
int rootchild = 0;
bool isLeaf[5002];
bool vis[5002];
inline void dfs(int u){
vis[u] = 1;
for(register Edge *ne = C.V[u]; ne; ne = ne->nxt){
if(vis[ne->np]) continue;
if(u == 1){
if(++rootchild <= 1) isLeaf[u] = 1;
else isLeaf[u] = 0;
} // The root is a leaf, unless it has at least 2 sub-trees
else isLeaf[u] = 0; // The non-root is a leaf, unless it has sub-trees
dfs(ne->np);
}
}
int main(){
scanf("%d%d", &F, &R);
for(register int i = 1; i <= R; i++){
int u, v;
scanf("%d%d", &u, &v);
G.addedge(u, v), G.addedge(v, u);
}
Tarjan(1, NULL);
for(register int i = 1; i <= F; i++)
for(register Edge *ne = G.V[i]; ne; ne = ne->nxt)
if(Color[i] != Color[ne->np]) C.addedge(Color[i], Color[ne->np]);
if(topc == 1){
printf("0\n"); // Already bi-connected
return 0;
}
memset(isLeaf, 1, sizeof(isLeaf));
dfs(1); // Find the leaf point in the shrunk graph
int leaf = 0;
for(register int i = 1; i <= topc; i++)
if(isLeaf[i]) leaf++;
printf("%d\n", leaf + 1 >> 1);
return 0;
}
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