2016 湖南省省赛B题《有向无环图》

题目链接【https://vjudge.net/problem/CSU-1804】

题意： 给出一个有向无环图，然后让你算下面的结果，count（i，j）表示i->j之间的路径条数。

题解： 根据公式，可以把SUMa[i]提出来，然后对于没给我i点求SUMcount(i,j)*bj;

#include<bits/stdc++.h>
using namespace std;
typedef long long LL;
const int mod = 1e9 + 7;
const int maxn = 1e5 + 15;
int N, M;
LL A[maxn], B[maxn], dp[maxn];
int  vis[maxn];
struct Edge
{
    int to, next;
    Edge (int to = 0, int next = 0): to(to), next(next) {}
} E[maxn];
int head[maxn], tot;
void initedge()
{
    for(int i = 0; i <= N; i++) head[i] = dp[i] = -1, vis[i] = 0;
    tot = 0;
}
void addedge(int  u, int v)
{
    E[tot] = Edge(v, head[u]);
    head[u] = tot++;
}
LL DFS(int u)
{
    if(dp[u] != -1) return dp[u];
    vis[u] = 1;
    dp[u] = 0;
    for(int k = head[u]; ~k; k = E[k].next)
    {
        int v = E[k].to;
        dp[u] = (dp[u] + (B[v] + DFS(v)) % mod) % mod;
    }
    return dp[u];
}
int main ()
{
    while(~scanf("%d %d", &N, &M))
    {
        for(int i = 1; i <= N; i++)
            scanf("%lld %lld", &A[i], &B[i]);
        initedge();
        for(int i = 1; i <= M; i++)
        {
            int u, v;
            scanf("%d %d", &u, &v);
            addedge(u, v);
        }
        for(int i = 1; i <= N; i++)
            if(!vis[i]) DFS(i);
        LL ans = 0;
        for(int i = 1; i <= N; i++)
            ans = (ans + (dp[i] * A[i]) % mod) % mod;
        printf("%lld
", ans);
    }
    return 0;
}