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;
}