SPF
Description Consider the two networks shown below. Assuming that data moves around these networks only between directly connected nodes on a peer-to-peer basis, a failure of a single node, 3, in the network on the left would prevent some of the still available nodes from communicating with each other. Nodes 1 and 2 could still communicate with each other as could nodes 4 and 5, but communication between any other pairs of nodes would no longer be possible. Input The input will contain the description of several networks. A network description will consist of pairs of integers, one pair per line, that identify connected nodes. Ordering of the pairs is irrelevant; 1 2 and 2 1 specify the same connection. All node numbers will range from 1 to 1000. A line containing a single zero ends the list of connected nodes. An empty network description flags the end of the input. Blank lines in the input file should be ignored. Output For each network in the input, you will output its number in the file, followed by a list of any SPF nodes that exist.
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题意
给你一个图，让你求图种的割点（删去后破坏图的连通性的点），以及去掉割点后原图被分成几个子图
解题思路
先求割点，再用染色法求子图数。
ACCode
#include <iostream>
#include <cstdio>
#include <vector>
#include <cstring>
#include <queue>
using namespace std;
vector<int > vec[1001];
bool flag[1001];
int num[1001],low[1001],root,index,n;
int _max(int x,int y,int z)
{
return max(x,max(y,z));
}
void init()
{
for(int i=1;i<=1000;i++) {
vec[i].clear();
num[i] = 0;
low[i] = 0;
flag[i] = false;
}
index = 0;
root = 1;
n = 0;
return ;
}
void dfs(int cur,int f)
{
int child=0;
index++;
num[cur] = index;
low[cur] = index;
int to;
for(int i=0;i<vec[cur].size();i++) {
to = vec[cur][i];
if(num[to] == 0) {
child++;
dfs(to,cur);
low[cur] = min(low[cur],low[to]);
if(cur != root && low[to] >= num[cur]) flag[cur] = true;
if(cur == root && child == 2) flag[cur] = true;
}
else if(to != f) { //&&num[to] != 0
low[cur] = min(low[cur],num[to]);
}
}
return ;
}
void bfs(int u,int col)
{
num[u] = col;
queue<int > q;
q.push(u);
while(!q.empty()) {
int h = q.front();
for(int i=0;i<vec[h].size();i++) {
int t = vec[h][i];
if(num[t] == 0) {
num[t] = col;
q.push(t);
}
}
q.pop();
}
}
int main()
{
int u,v,Network=0;
while(scanf("%d",&u) && u != 0) {
scanf("%d",&v);
init();
n = _max(n,u,v);
vec[u].push_back(v);
vec[v].push_back(u);
while(scanf("%d",&u) && u != 0) {
scanf("%d",&v);
n = _max(n,u,v);
vec[u].push_back(v);
vec[v].push_back(u);
}
dfs(1,0);
printf("Network #%d\n",++Network);
bool flag2 = false;
for(int i=1;i<=n;i++) {
if(flag[i]) {
flag2 = true ; //有割点
for(int j=1;j<=n;j++)
num[j] = 0; //用作color
num[i] = -1; //保证不会走割点
int tot = 0;
for(int j=1;j<=n;j++) {
if(num[j] == 0 && j != i) {
tot++;
bfs(j,tot);
}
}
printf(" SPF node %d leaves %d subnets\n",i,tot);
}
}
if(!flag2) printf(" No SPF nodes\n");
printf("\n");
}
return 0;
}