?

大意：求割頂的數量以及刪除割頂之后子圖的數量。

思路：Tarjan算法求割頂，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; //直接將根節點賦值為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)?//處理根節點。?
????????????{
????????????????if(++rootson?>?1)?sub[u]++;
????????????}
????????????Tarjan(v,?u);
????????????low[u]?=?min(low[u],?low[v]);
????????????if(u?!=?root?&&?dfn[u]?<=?low[v])?sub[u]++;?//根節點已經被處理過。
????????}
????????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;
}

?

?

轉載于:https://www.cnblogs.com/g0feng/archive/2012/11/03/2752321.html

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