csu oj 1804: 有向无环图 (dfs回溯)

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

中文题意就不说了。

dfs从底到根回溯即可，看代码应该能清楚。

 //#pragma comment(linker, "/STACK:102400000, 102400000")
 #include <algorithm>
 #include <iostream>
 #include <cstdlib>
 #include <cstring>
 #include <cstdio>
 #include <vector>
 #include <cmath>
 #include <ctime>
 #include <list>
 #include <set>
 #include <map>
 using namespace std;
 typedef long long LL;
 typedef pair <int, int> P;
 const int N = 1e5 + ;
 struct Edge {
     int next, to;
 }edge[N];
 int head[N], tot, in[N];
 LL a[N], b[N], ans, d[N], mod = 1e9 + ; //d[i]表示b[i.son]*count[i,j]+b[i]
 bool vis[N];
 
 void init(int n) {
     for(int i = ; i <= n; ++i) {
         head[i] = -;
         in[i] = ;
         vis[i] = false;
     }
     tot = ;
     ans = ;
 }
 
 inline void add_edge(int u, int v) {
     edge[tot].next = head[u];
     edge[tot].to = v;
     head[u] = tot++;
 }
 
 void dfs(int u) {
     d[u] = b[u] % mod;
     for(int i = head[u]; ~i; i = edge[i].next) {
         int v = edge[i].to;
         if(!vis[v]) { //说明这个点以及子树没访问
             dfs(v);
             vis[v] = true;
         }
         ans = (ans + a[u] * d[v] % mod) % mod;
         d[u] = (d[v] + d[u]) % mod;
     }
 }
 
 int main()
 {
     int n, m, u, v;
     while(scanf("%d %d", &n, &m) != EOF) {
         for(int i = ; i <= n; ++i) {
             scanf("%lld %lld", a + i, b + i);
         }
         init(n);
         for(int i = ; i <= m; ++i) {
             scanf("%d %d", &u, &v);
             add_edge(u, v);
             ++in[v];
         }
         for(int i = ; i <= n; ++i) {
             if(!in[i]) { //入度为0
                 dfs(i);
             }
         }
         printf("%lld\n", ans);
     }
     return ;
 }