Uva 10765 Doves and bombs (点双联通分量 + Block Forest Data Structure)
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题意:给定一个n个点的连通无向图,一个点的鸽子值定义为将它从图中删去后的连通块个数。求每个点的鸽子值。
分析:删除后影响图连通性的点一定是割点,在求出所有双联通分量后重新构图,每个联通分量添加一个虚点,把连通分量的每个点向虚点连边,然后我们就得到了一个和原图连通性一样的森林(Block Forest Data Structure),Block Forest Data Structure森林有个特性,虚点和割点一定是非叶子节点,叶子节点删去一定不会影响图的连通性。本题每个割点连接的虚点的个数就是删去此割点后的连通块个数。
#include<iostream>#include<string>#include<algorithm>#include<cstdlib>#include<cstdio>#include<set>#include<map>#include<vector>#include<cstring>#include<stack>#include<cmath>#include<queue>#define INF 0x3f3f3f3f#define eps 1e-9#define MAXN 10005using namespace std;int n,m,x,y,dfs_clock,bcc_cnt,pre[MAXN],iscut[MAXN],bccno[MAXN],low[MAXN],lt[MAXN],ans[MAXN];vector <int> G[MAXN],G2[MAXN],bcc[MAXN];bool camp(int x,int y){if(lt[x] != lt[y]) return lt[x] > lt[y];return x < y;}struct Edge{int u,v;Edge() { }Edge(int x,int y){u = x;v = y;}};stack<Edge> S;int dfs(int u,int fa){low[u] = pre[u] = ++dfs_clock;int child = 0;for(int v : G[u]){Edge e = Edge(u,v);if(!pre[v]){S.push(e);child++;low[u] = min(low[u],dfs(v,u));if(low[v] >= pre[u]){iscut[u] = true;bcc[++bcc_cnt].clear();for(;;){Edge x = S.top();S.pop();if(bccno[x.u] != bcc_cnt){bcc[bcc_cnt].push_back(x.u);bccno[x.u] = bcc_cnt;}if(bccno[x.v] != bcc_cnt){bcc[bcc_cnt].push_back(x.v);bccno[x.v] = bcc_cnt;}if(x.u == u && x.v == v) break;}}}else if(pre[v] < pre[u] && v != fa) { S.push(e); low[u] = min(low[u],pre[v]); }}if(fa < 0 && child == 1) iscut[u] = 0;return low[u]; }void got_bcc(){memset(pre,0,sizeof(pre));memset(iscut,0,sizeof(iscut));memset(bccno,0,sizeof(bccno));bcc_cnt = 0;for(int i = 1;i <= n;i++) if(!pre[i]) { dfs_clock = 0; dfs(i,-1); } }int main(){while(~scanf("%d%d",&n,&m) && n+m){memset(lt,0,sizeof(lt));for(int i = 1;i <= n;i++) G[i].clear(); while(~scanf("%d%d",&x,&y) && x+y != -2){x++;y++;G[x].push_back(y);G[y].push_back(x);}got_bcc();for(int i = 1;i <= bcc_cnt;i++) for(int v : bcc[i]) lt[v]++;for(int i = 1;i <= n;i++) ans[i] = i;sort(ans+1,ans+1+n,camp);for(int i = 1;i <= m;i++) cout<<ans[i]-1<<" "<<lt[ans[i]]<<endl;cout<<endl;}}
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