[BZOJ1023]静态仙人掌 Tarjan+dp+单调队列

来源：互联网发布：崩坏学园2淘宝代练编辑：程序博客网时间：2024/05/23 13:46

诶写完写完代码以后才发现天黑了，赶快溜出去找东西吃

求仙人掌的直径，不能直接两边DFS找最长路，有这样一个反例

绿色为真实的直径，可是如果第一遍找到最长路是红色路径的话，那么就找不到绿色路径

还是用最传统的方法，dp求从该节点出发的最长连

引用一篇写的虽然有点长但是写的很好的博客

http://z55250825.blog.163.com/blog/static/150230809201412793151890/

#include <iostream>#include <cstdio>#define N 100005#define M 20000050using namespace std;void ut(int &x,int y) { x = max(x,y); }inline int read() {    int x=0,f=1;char ch=getchar();    while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}    while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}    return x*f;}struct Edge{ int b,next; }E[M];int h[N],D[N],fa[N],cnt;int dfn[N],low[N],tme;int a[N],F[N],q[N],n,m,ans;void add(int a,int b) {E[++cnt] = (Edge){b,h[a]};h[a] = cnt; return ;}void dp(int rt,int x) {//将树上的链取下来 int tot = D[x] - D[rt] + 1;for (int i=x;i!=rt;i=fa[i]) a[tot--] = F[i]; a[tot] = F[rt];//倍长处理环 tot = D[x] - D[rt] + 1;for (int i=1;i<=tot;i++) a[i+tot] = a[i];q[1] = 1; int l = 1 , r = 1;for (int i=2;i<=2*tot;i++) {while (l <= r && i-q[l] > tot/2) l++;ut(ans , a[i]+a[q[l]]+ i-q[l] );while (l <= r && a[q[r]]-q[r] <= a[i]-i) r--;q[++r] = i;}for (int i=2;i<=tot;i++)ut(F[rt] , a[i] + min(i-1,tot-i+1));}void Tarjan(int u) {#define v E[i].bdfn[u] = low[u] = ++tme;//sta.push(u); ins[u] = 1;for (int i=h[u];i;i=E[i].next) {//int v = E[i].b;if (v == fa[u]) continue; if (!dfn[v]) {fa[v] = u;D[v] = D[u] + 1;Tarjan(v);low[u] = min(low[u] , low[v]);} else low[u] = min(low[u] , dfn[v]);if (dfn[u] < low[v]) {ut(ans,F[u]+F[v]+1);ut(F[u],F[v]+1);}}//环 for (int i=h[u];i;i=E[i].next)if (fa[v] != u && dfn[u] < dfn[v]) dp(u,v);return ;}int main() {n = read(); m = read(); ans = -1;for (int _=1;_<=m;_++) {int k = read() , a = read() ,b;for (int i=1;i<=k-1;i++) b = read() , add(a,b) , add(b,a) , a = b;}Tarjan(1);printf("%d\n",ans);return 0;}

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