BZOJ1023 [SHOI2008]cactus仙人掌图

滚回第一页去了。。。

好吧，看了题解蒟蒻也写不粗来，怎么办捏？

看这个吧：Orz YDC巨巨；但是巨巨写的程序又不优美，于是程序Orz hzwer

其实这题的重点在于tarjan和单调队列dp里"f的更新"和"ans的更新"的先后顺序。。。

蒟蒻我研究半天才略懂还写不出题解，真是弱到不行了我去 ≥v≤ ~~~

 

  1 /**************************************************************
  2     Problem: 1023
  3     User: rausen
  4     Language: C++
  5     Result: Accepted
  6     Time:176 ms
  7     Memory:161276 kb
  8 ****************************************************************/
  9  
 10 #include <cstdio>
 11 #include <algorithm>
 12  
 13 using namespace std;
 14 const int N = 50005;
 15 const int M = 20000005;
 16  
 17 struct edges{
 18     int next, to;
 19 }e[M];
 20 int first[N], dep[N], f[N];
 21 int dfn[N], low[N], fa[N];
 22 int a[N * 2], q[N * 2], l, r;
 23 int n, m, tot, cnt, ans;
 24  
 25 inline int read(){
 26     int x = 0;
 27     char ch = getchar();
 28     while (ch < '0' || ch > '9')
 29         ch = getchar();
 30  
 31     while (ch >= '0' && ch <= '9'){
 32         x = x * 10 + ch - '0';
 33         ch = getchar();
 34     }
 35     return x;
 36 }
 37  
 38 inline void add_edge(int X, int Y){
 39     e[++tot].next = first[X];
 40     first[X] = tot;
 41     e[tot].to = Y;
 42 }
 43  
 44 void add_Edges(int x, int y){
 45     add_edge(x, y);
 46     add_edge(y, x);
 47 }
 48  
 49 void DP(int root, int p){
 50     int T = dep[p] - dep[root] + 1;
 51     for (int i = p; i != root; i = fa[i])
 52         a[T--] = f[i];
 53     a[T] = f[root];
 54     T = dep[p] - dep[root] + 1;
 55     for (int i = 1; i <= T; ++i)
 56         a[i + T] = a[i];
 57     q[1] = l = r = 1;
 58     for (int i = 2; i <= 2 * T; ++i){
 59         while (l <= r && i - q[l] > T / 2) ++l;
 60         ans = max(ans, a[i] + a[q[l]] + i - q[l]);
 61         while (l <= r && a[q[r]] - q[r] <= a[i] - i) --r;
 62         q[++r] = i;
 63     }
 64     for (int i = 2; i <= T; ++i)
 65         f[root] = max(f[root], a[i] + min(i - 1, T - i + 1));
 66 }
 67  
 68 void dfs(int p){
 69     low[p] = dfn[p] = ++cnt;
 70     int x, y;
 71     for (x = first[p]; x; x = e[x].next)
 72         if ((y = e[x].to) != fa[p]){
 73             if (!dfn[y]){
 74                 fa[y] = p, dep[y] = dep[p] + 1;
 75                 dfs(y);
 76                 low[p] = min(low[p], low[y]);
 77             }else low[p] = min(low[p], dfn[y]);
 78             if (dfn[p] < low[y]){
 79                 ans = max(ans, f[p] + f[y] + 1);
 80                 f[p] = max(f[p], f[y] + 1);
 81             }
 82         }
 83     for (x = first[p]; x; x = e[x].next)
 84         if (fa[(y = e[x].to)] != p && dfn[p] < dfn[y])
 85             DP(p, y);
 86 }
 87  
 88 int main(){
 89     n = read(), m = read();
 90     int k, X, Y, i, j;
 91     for (i = 1; i <= m; ++i){
 92         k = read(), X = read();
 93         for (j = 2; j <= k; ++j){
 94             Y = read();
 95             add_Edges(X, Y);
 96             X = Y;
 97         }
 98     }
 99     dfs(1);
100     printf("%d\n", ans);
101     return 0;
102 }
View Code

 

posted on 2014-10-22 20:47  Xs酱~  阅读(701)  评论(0)  编辑  收藏  举报