bzoj 1023 仙人掌圖
來源:程序員人生 發(fā)布時(shí)間:2015-03-10 08:29:06 閱讀次數(shù):3163次
Description
求1個(gè)神仙掌圖的直徑
Solution
神仙掌圖有個(gè)性質(zhì),1條邊要末是割邊要末就是在環(huán)內(nèi),那末我們可以對(duì)它進(jìn)行Dp辣!
令f[u]表示以u為根的子樹最長(zhǎng)鏈長(zhǎng)度
如果u?v是橋的話轉(zhuǎn)移就是ans=max(ans,f[u]+f[v]+1),f[u]=max(f[u],f[v]+1),由于當(dāng)前f[u]都是由它的孩子更新來的
如果是環(huán)的話,變環(huán)為鏈,用單調(diào)隊(duì)列dp出ans,然后用環(huán)上的f值更新f[u]的值就能夠了,具體實(shí)現(xiàn)見代碼
Code
using namespace std;
const int N = 100005, M = N << 1;
int ans, ind, tot, cnt, fa[N], cir[N << 1], to[M << 1], nxt[M << 1], head[N], dfn[N], low[N], f[N];
inline int read(int &t) {
int f = 1;char c;
while (c = getchar(), c < '0' || c > '9') if (c == '-') f = -1;
t = c - '0';
while (c = getchar(), c >= '0' && c <= '9') t = t * 10 + c - '0';
t *= f;
}
struct data {
int p, w;
}q[N];
void add(int u, int v) {
to[tot] = v, nxt[tot] = head[u], head[u] = tot++;
to[tot] = u, nxt[tot] = head[v], head[v] = tot++;
}
void gao() {
int h = 1, r = 1;
for (int i = 1; i <= cnt; ++i) cir[cnt + i] = cir[i];
for (int i = 1; i <= (cnt << 1); ++i) {
while (h < r && i - q[h].p > cnt / 2) ++h;
while (h < r && q[r].w <= f[cir[i]] - i) --r;
q[++r].p = i, q[r].w = f[cir[i]] - i;
ans = max(ans, f[cir[i]] + i + q[h].w);
}
}
void dfs(int u) {
low[u] = dfn[u] = ++ind;
for (int i = head[u], v; ~i; i = nxt[i]) {
v = to[i];
if (fa[v] != 0 && v != fa[u]) low[u] = min(low[u], dfn[v]);
if (fa[v] == 0) {
fa[v] = u;
dfs(v);
low[u] = min(low[u], low[v]);
}
}
for (int i = head[u], v; ~i; i = nxt[i]) {
v = to[i];
if (fa[v] == u && low[v] > dfn[u]) { //bridge
ans = max(ans, f[u] + f[v] + 1);
f[u] = max(f[u], f[v] + 1);
}
if (fa[v] != u && dfn[u] < dfn[v]) { //circle
cnt = 0;
while (v != fa[u]) cir[++cnt] = v, v = fa[v];
gao();
for (int j = 1; j < cnt; ++j) f[u] = max(f[u], f[cir[j]] + min(j, cnt - j));
}
}
}
int main() {
int n, m;
memset(head, -1, sizeof(head));
read(n), read(m);
for (int i = 1, x, y, z; i <= m; ++i) {
read(x), read(y);
for (int j = 1; j < x; ++j) {
read(z);
add(y, z);
y = z;
}
}
fa[1] = -1;
dfs(1);
printf("%d
", ans);
return 0;
}
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