POJ 1523 SPF

大意：求割顶的数量以及删除割顶之后子图的数量。

思路：Tarjan算法求割顶，同POJ 1144 NetWork.

CODE1：

#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<algorithm>
using namespace std;

const int MAXN = 100010;
const int MAXM = 500010;

struct Edge
{
int v, next;
}edge[MAXM];

int dfn[MAXN], low[MAXN], sub[MAXN];
int first[MAXN];
int N, M;
int cnt, tot;
int u, v;

void read_graph(int u, int v)
{
edge[cnt].v = v;
edge[cnt].next = first[u], first[u] = cnt++;
}

void init()
{
cnt = 0;
tot = 0;
memset(first, -1, sizeof(first));
memset(dfn, 0, sizeof(dfn));
}

void read_graph2()
{
scanf("%d", &v);
read_graph(u, v);
read_graph(v, u);
N = max(N, max(u, v));
while(scanf("%d", &u) && u)
{
scanf("%d", &v);
read_graph(u, v);
read_graph(v, u);
N = max(N, max(u, v));
}
}

void tarjan(int u)
{
dfn[u] = low[u] = ++tot;
for(int e = first[u]; e != -1; e = edge[e].next)
{
int v = edge[e].v;
if(!dfn[v])
{
tarjan(v);
low[u] = min(low[u], low[v]);
if(dfn[u] <= low[v]) sub[u]++;
}
low[u] = min(low[u], dfn[v]);
}
}

void solve(int root)
{
for(int i = 1; i <= N; i++) sub[i] = (i == root)? 0:1; //直接将根节点赋值为0
tarjan(root);
int flag = 0;
for(int i = 1; i <= N; i++)
{
if(sub[i] > 1)
{
flag = 1;
printf(" SPF node %d leaves %d subnets\n", i, sub[i]);
}
}
if(!flag)
printf(" No SPF nodes\n");
}

int main()
{
int times = 0;
while(scanf("%d", &u) && u)
{
init();
read_graph2();
if(times) printf("\n");
printf("Network #%d\n", ++times);
solve(1);
}
return 0;
}

CODE2：

#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<algorithm>
using namespace std;

const int MAXN = 100010;
const int MAXM = 500010;

struct Edge
{
int v, next;
}edge[MAXM];

int dfn[MAXN], low[MAXN], sub[MAXN];
int first[MAXN];
int N, M;
int cnt, tot;
int u, v;
int root = 1;

inline void read_graph(int u, int v)
{
edge[cnt].v = v;
edge[cnt].next = first[u], first[u] = cnt++;
}

inline void init()
{
cnt = 0;
tot = 0;
memset(first, -1, sizeof(first));
memset(dfn, 0, sizeof(dfn));
}

inline void read_graph2()
{
scanf("%d", &v);
read_graph(u, v);
read_graph(v, u);
N = max(N, max(u, v));
while(scanf("%d", &u) && u)
{
scanf("%d", &v);
read_graph(u, v);
read_graph(v, u);
N = max(N, max(u, v));
}
}

inline void Tarjan(int u, int fa)
{
int rootson = 0;
low[u] = dfn[u] = ++tot;
for(int e = first[u]; e != -1; e = edge[e].next)
{
int v = edge[e].v;
if(!dfn[v])
{
if(u == root) //处理根节点。
{
if(++rootson > 1) sub[u]++;
}
Tarjan(v, u);
low[u] = min(low[u], low[v]);
if(u != root && dfn[u] <= low[v]) sub[u]++; //根节点已经被处理过。
}
low[u] = min(low[u], dfn[v]);
}
}

inline void solve()
{
int flag = 0;
for(int i = 1; i <= N; i++) sub[i] = 1;
Tarjan(root, -1);
for(int i = 1; i <= N; i++)
{
if(sub[i] > 1)
{
flag = 1;
printf(" SPF node %d leaves %d subnets\n", i, sub[i]);
}
}
if(!flag) printf(" No SPF nodes\n");
}

int main()
{
int times = 0;
while(scanf("%d", &u) && u)
{
init();
read_graph2();
if(times) printf("\n");
printf("Network #%d\n", ++times);
solve();
}
return 0;
}