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[CSUOJ1804]有向无环图(树dp)

题目链接:http://acm.csu.edu.cn/OnlineJudge/problem.php?id=1804

题意:中文题面,那个式子也很清楚。

1e5这个点数量很大了,由于是个DAG,可以认为整张图是一个森林,关注点应该放在每一棵树上的每一个点对整题的贡献。

记每一个起点u,v是u的后继。v有多个,记作vi,u到vi的贡献是count(u,vi)*a(u)*b(vi),提出a(u),剩下count(u,vi)*b(vi),count(u,vi)可以后序遍历得到,整个结果存在dp(u)中,后序更新答案时dp(u)*a(u)即可。随后可以更新b(vi)部分,每次累加到dp(u)上,因为下一次的访问(u的父亲)必定包括了dp(u)的结果。

  1 /*  2 ━━━━━┒ギリギリ♂ eye!  3 ┓┏┓┏┓┃キリキリ♂ mind!  4 ┛┗┛┗┛┃\○/  5 ┓┏┓┏┓┃ /  6 ┛┗┛┗┛┃ノ)  7 ┓┏┓┏┓┃  8 ┛┗┛┗┛┃  9 ┓┏┓┏┓┃ 10 ┛┗┛┗┛┃ 11 ┓┏┓┏┓┃ 12 ┛┗┛┗┛┃ 13 ┓┏┓┏┓┃ 14 ┃┃┃┃┃┃ 15 ┻┻┻┻┻┻ 16 */ 17 #include <algorithm> 18 #include <iostream> 19 #include <iomanip> 20 #include <cstring> 21 #include <climits> 22 #include <complex> 23 #include <fstream> 24 #include <cassert> 25 #include <cstdio> 26 #include <bitset> 27 #include <vector> 28 #include <deque> 29 #include <queue> 30 #include <stack> 31 #include <ctime> 32 #include <set> 33 #include <map> 34 #include <cmath> 35 using namespace std; 36 #define fr first 37 #define sc second 38 #define cl clear 39 #define BUG puts("here!!!") 40 #define W(a) while(a--) 41 #define pb(a) push_back(a) 42 #define Rint(a) scanf("%d", &a) 43 #define Rll(a) scanf("%I64d", &a) 44 #define Rs(a) scanf("%s", a) 45 #define Cin(a) cin >> a 46 #define FRead() freopen("in", "r", stdin) 47 #define FWrite() freopen("out", "w", stdout) 48 #define Rep(i, len) for(int i = 0; i < (len); i++) 49 #define For(i, a, len) for(int i = (a); i < (len); i++) 50 #define Cls(a) memset((a), 0, sizeof(a)) 51 #define Clr(a, x) memset((a), (x), sizeof(a)) 52 #define Full(a) memset((a), 0x7f7f7f, sizeof(a)) 53 #define lrt rt << 1 54 #define rrt rt << 1 | 1 55 #define pi 3.14159265359 56 #define RT return 57 #define lowbit(x) x & (-x) 58 #define onecnt(x) __builtin_popcount(x) 59 typedef long long LL; 60 typedef long double LD; 61 typedef unsigned long long ULL; 62 typedef pair<int, int> pii; 63 typedef pair<string, int> psi; 64 typedef pair<LL, LL> pll; 65 typedef map<string, int> msi; 66 typedef vector<int> vi; 67 typedef vector<LL> vl; 68 typedef vector<vl> vvl; 69 typedef vector<bool> vb; 70  71 const int maxn = 100100; 72 const LL mod = 1e9+7; 73 int n, m; 74 int a[maxn], b[maxn]; 75 vi G[maxn]; 76 bool vis[maxn]; 77 LL dp[maxn]; 78 LL ret; 79  80 void dfs(int u) { 81     Rep(i, G[u].size()) { 82         int v = G[u][i]; 83         if(!vis[v]) { 84             vis[v] = 1; 85             dfs(v); 86         } 87         dp[u] = (dp[u] + dp[v]) % mod; 88     } 89     ret = (ret + (dp[u] * a[u] % mod)) % mod; 90     dp[u] = (dp[u] + b[u]) % mod; 91 } 92  93 int main() { 94     // FRead(); 95     int u, v; 96     while(~Rint(n)&&~Rint(m)) { 97         ret = 0; Cls(vis); Cls(dp); 98         For(i, 1, n+1) { 99             Rint(a[i]), Rint(b[i]);100             G[i].clear();101         }102         Rep(i, m) {103             Rint(u); Rint(v);104             G[u].pb(v);105         }106         For(i, 1, n+1) {107             if(!vis[i]) dfs(i);108         }109         For(i, 1, n+1) cout << dp[i] << " ";110         cout << endl;111         cout << ret << endl;112     }113     RT 0;114 }

 

[CSUOJ1804]有向无环图(树dp)