标签:tin fine 方向 head mod size cst return names
题意:给一个n节点的有向无环图,要找一个这样的点:该点到其它n-1要逆转的道路最少。
析:很明显的树形DP,两次dfs,对于边,进行处理,如果是正向就是1,反向是-1,先进行dfs,计算出向子结点的方向要反转几条边,然后再第二次考虑,从父结点和子结点考虑。
代码如下:
#pragma comment(linker, "/STACK:1024000000,1024000000") #include <cstdio> #include <string> #include <cstdlib> #include <cmath> #include <iostream> #include <cstring> #include <set> #include <queue> #include <algorithm> #include <vector> #include <map> #include <cctype> #include <cmath> #include <stack> #include <sstream> #define debug() puts("++++"); #define gcd(a, b) __gcd(a, b) #define lson l,m,rt<<1 #define rson m+1,r,rt<<1|1 #define freopenr freopen("in.txt", "r", stdin) #define freopenw freopen("out.txt", "w", stdout) using namespace std; typedef long long LL; typedef unsigned long long ULL; typedef pair<int, int> P; const int INF = 0x3f3f3f3f; const LL LNF = 1e16; const double inf = 0x3f3f3f3f3f3f; const double PI = acos(-1.0); const double eps = 1e-6; const int maxn = 200000 + 10; const int mod = 1e9 + 7; const int dr[] = {-1, 0, 1, 0}; const int dc[] = {0, 1, 0, -1}; const char *de[] = {"0000", "0001", "0010", "0011", "0100", "0101", "0110", "0111", "1000", "1001", "1010", "1011", "1100", "1101", "1110", "1111"}; int n, m; const int mon[] = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31}; const int monn[] = {0, 31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31}; inline bool is_in(int r, int c){ return r > 0 && r <= n && c > 0 && c <= m; } struct Edge{ int to, next, val; }; Edge edge[maxn<<1]; int head[maxn], cnt; int in[maxn]; int ans[maxn], res; int dp[maxn]; void add_edge(int u, int v, int val){ edge[cnt].to = v; edge[cnt].val = val; edge[cnt].next = head[u]; head[u] = cnt++; } void dfs1(int u, int fa){ dp[u] = 0; for(int i = head[u]; ~i; i = edge[i].next){ int v = edge[i].to; if(v == fa) continue; dfs1(v, u); if(edge[i].val == 1) dp[u] += dp[v]; else dp[u] += dp[v] + 1; } } void dfs2(int u, int fa, int fv){ ans[u] = dp[u] + fv; res = min(res, ans[u]); for(int i = head[u]; ~i; i = edge[i].next){ int v = edge[i].to; if(v == fa) continue; dfs2(v, u, fv+dp[u]-dp[v]+edge[i].val); } } int main(){ scanf("%d", &n); memset(head, -1, sizeof head); for(int i = 1; i < n; ++i){ int u, v; scanf("%d %d", &u, &v); add_edge(u, v, 1); add_edge(v, u, -1); } dfs1(1, -1); res = n - 1; dfs2(1, -1, 0); printf("%d\n", res); for(int i = 1; i <= n; ++i) if(res == ans[i]) printf("%d ", i); printf("\n"); return 0; }
CodeForces 219D Choosing Capital for Treeland (树形DP)
标签:tin fine 方向 head mod size cst return names
原文地址:http://www.cnblogs.com/dwtfukgv/p/7353304.html