标签:style blog http color io os ar for sp
首先,我们要tarjan。。。 然后我们要缩点。。。
注意,缩点的时候两个新建的点会有重边,需要判重
正常的判重方法是bfs一边,但是我YY的比较奇葩,方法下面将。。。
缩好点就变成了一个DAG,然后就类似树形DP的方法求最大权值链
我是用记忆化搜索,当dfs某个点p时用数组vis记录一些东西:
首先vis[x] > 0代表x已经访问过了, vis[x] == p表示vis[x]上次是由p来的,于是就可以去重了。。。
然后就没有然后了,搞定yeah!
1 /************************************************************** 2 Problem: 1093 3 User: rausen 4 Language: C++ 5 Result: Accepted 6 Time:1604 ms 7 Memory:31220 kb 8 ****************************************************************/ 9 10 #include <cstdio> 11 #include <algorithm> 12 #include <cstring> 13 14 using namespace std; 15 const int inf = (int) 1e8; 16 struct edges{ 17 int from, to, next; 18 }e[2000005]; 19 20 int first[100005],tot; 21 int dfn[100005], low[100005], w[100005], sum[100005]; 22 int vis[100005], s[100005], f[100005], g[100005], F[100005]; 23 int n, m, MOD, X, Y, ans, top, cnt, num, ANS; 24 25 void add_edge(int x, int y){ 26 e[++tot].next = first[x]; 27 first[x] = tot; 28 e[tot].from = x, e[tot].to = y; 29 } 30 31 void tarjan(int p){ 32 dfn[p] = low[p] = ++cnt; 33 s[++top] = p; 34 vis[p] = 1; 35 int x, y; 36 for (x = first[p]; x; x = e[x].next){ 37 y = e[x].to; 38 if (!vis[y]) tarjan(y); 39 if (vis[y] < 2) 40 low[p] = min(low[p], low[y]); 41 } 42 if (dfn[p] == low[p]){ 43 ++num; 44 while (s[top + 1] != p){ 45 int y = s[top--]; 46 vis[y] = 2, w[y] = num; 47 ++sum[num]; 48 } 49 } 50 } 51 52 void dfs(int p){ 53 if (vis[p]) return; 54 vis[p] = inf; 55 int x, y; 56 for (int x = first[p]; x; x = e[x].next){ 57 y = e[x].to; 58 if (vis[y] == p) continue; 59 vis[y] = p; 60 dfs(y); 61 if (f[p] < f[y]) 62 f[p] = f[y], g[p] = g[y]; else 63 if (f[p] == f[y]) g[p] += g[y]; 64 } 65 g[p] %= MOD; 66 f[p] += sum[p]; 67 if (ans < f[p]) 68 ans = f[p], ANS = g[p]; else 69 if (ans == f[p]) ANS += g[p], ANS %= MOD; 70 } 71 72 int main(){ 73 scanf("%d%d%d", &n, &m, &MOD); 74 for (int i = 1; i <= m; ++i) 75 scanf("%d%d", &X, &Y), add_edge(X, Y); 76 for (int i = 1; i <= n; ++i) 77 if (!vis[i]) tarjan(i); 78 79 memset(first, 0, sizeof(first)); 80 tot = 0; 81 for (int i = 1; i <= m; ++i){ 82 X = e[i].from; 83 Y = e[i].to; 84 if (w[X] != w[Y]) add_edge(w[X], w[Y]); 85 } 86 memset(vis, 0, sizeof(vis)); 87 for (int i = 1; i <= num; ++i) 88 if (!first[i]) g[i] = 1; 89 for (int i = 1; i <= num; ++i) 90 if (!vis[i]) dfs(i); 91 printf("%d\n%d\n", ans, ANS % MOD); 92 return 0; 93 }
(p.s. 各种压代码+卡内存此题status终于上了前两页。。。)
标签:style blog http color io os ar for sp
原文地址:http://www.cnblogs.com/rausen/p/4008795.html