POJ3177 redundant Paths "side double Unicom Component" "Tarjan"

Topic Links:

http://poj.org/problem?id=3177

Main topic:

Bessie's Farm has an F-block ranch, which is known to have at least one path connected to any of the two farms (not necessarily directly connected)

To move from one ranch to another, Bessie and her companions often need to go through the decaying woods. The cows were particularly

Disgusted by the bad way, so Bessie decided to build a few more roads on the farm, so that when going to a place can always have two

A completely independent way to choose. So here's the question: F-Block Ranch, R-Road, ask at least a few more roads to make the farm

There are at least two independent paths between any of the two ranches.

Ideas:

In order to have at least two independent paths between any of the two pastures in the farm. That is, the F-Ranch as a point, R-Bar.

The road is considered to be a two-way street between two farms. Then the problem is converted to: add at least a few edges, you can make the graph does not exist cutting edge.

That is, the same problem as POJ3352: add at least a few edges, so that any two edge connected components are connected to each other.

Specific Practice Reference Blog: http://blog.csdn.net/lianai911/article/details/43867789

Note: This problem is a bit more than the POJ3352, need to judge the weight ... So I used the map[][] array.

AC Code:

#include <iostream> #include <algorithm> #include <cstdio> #include <cstring>using namespace    std;const int MAXN = 5050;const int maxm = 50500;bool map[maxn][maxn];struct edgenode{int to; int next;} Edges[maxm];int head[maxn];int dfn[maxn],low[maxn],stack[maxn],degree[maxn];int m,id,scc,lay,M,N;void AddEdges (int    U,int v) {edges[id].to = v;    Edges[id].next = Head[u]; Head[u] = id++;}    void Tarjanbfs (int pos,int from) {//stack[m++] = pos;    Dfn[pos] = Low[pos] = ++lay;        for (int i = Head[pos]; I! =-1; i = edges[i].next) {if (edges[i].to = = from) continue;            if (!dfn[edges[i].to]) {TARJANBFS (edges[i].to,pos);                Low[pos] = min (low[pos],low[edges[i].to])//if (Dfn[pos] < low[edges[i].to])//{// while (stack[m--]! = edges[i].to);/}} else Low[pos] = min (low[pos],low[edges[i].t    O]);    }}int Main () {int u,v; while (~SCANF ("%d%d ", &n,&m)) {memset (head,-1,sizeof (Head));        memset (dfn,0,sizeof (DFN));        memset (low,0,sizeof (Low));        memset (degree,0,sizeof (degree));        m = Lay = SCC = id = 0;        memset (map,0,sizeof (MAP));            for (int i = 0; i < M; ++i) {scanf ("%d%d", &u,&v);                if (map[u][v] = = 0) {addedges (u,v);                Addedges (V,u);            MAP[U][V] = Map[v][u] = 1;        }} tarjanbfs (1,-1);                for (int i = 1, i <= N; ++i) {for (int j = head[i]; J! =-1; j = edges[j].next) {            if (low[i]! = low[edges[j].to]) degree[low[i]]++;        }} int ans = 0;        for (int i = 1; I <= N; ++i) if (degree[i] = = 1) ans++;    printf ("%d\n", (ans+1)/2); } return 0;}

POJ3177 redundant Paths "side double Unicom Component" "Tarjan"