题目链接：http://acm.csu.edu.cn/csuoj/problemset/problem?pid=1804

解法：先将每个点i对应的count(i,j)*bj算出来然后乘ai，累加就是答案，注意这里要类似拓扑排序那样，不过

要倒着做，避免后效性

#include <bits/stdc++.h>
using namespace std;
typedef long long LL;
const int maxn = 100010;
const int mod = 1e9+7;
struct edge{
    int v,next;
    edge(int v=0,int next=0):v(v),next(next){}
}E[maxn*2];
LL a[maxn], b[maxn], dp[maxn], du[maxn];
int head[maxn], edgecnt;
void init(){
    edgecnt=0;
    memset(head, -1, sizeof(head));
}
void add(int u, int v){
    E[edgecnt].v = v, E[edgecnt].next = head[u], head[u] = edgecnt++;
}
int n, m;
int main()
{
    while(scanf("%d%d",&n,&m)!=EOF)
    {
        init();
        memset(dp, 0, sizeof(dp));
        memset(du, 0, sizeof(du));
        for(int i=1; i<=n; i++) scanf("%lld%lld",&a[i],&b[i]);
        for(int i=1; i<=m; i++){
            int u,v;
            scanf("%d%d", &u,&v);
            add(v,u);
            du[u]++;
        }
        queue<int>q;
        for(int i=1; i<=n; i++) if(du[i]==0) q.push(i);
        while(!q.empty()){
            int u=q.front(); q.pop();
            for(int i=head[u]; ~i; i=E[i].next){
                int v = E[i].v;
                dp[v]=(dp[v]+(dp[u]+b[u])%mod)%mod;
                if(--du[v]==0) q.push(v);
            }
        }
        LL ans=0;
        for(int i=1; i<=n; i++){
            ans=(ans+1LL*dp[i]*a[i]%mod)%mod;
        }
        printf("%lld\n", ans);
    }
    return 0;
}

还可以直接记忆化搜索

#include<bits/stdc++.h>
using namespace std;
typedef long long LL;
const int maxn = 1e5 + 5;
LL dp[maxn];
LL a[maxn], b[maxn];
const int mod = 1e9 + 7;
struct EDGE
{
    int to, next;
    EDGE(int to = 0, int next = 0): to(to), next(next) {}
} edge[maxn];
int head[maxn], edgecnt;
void init()
{
    memset(dp, -1, sizeof(dp));
    memset(head, -1, sizeof(head));
    edgecnt = 0;
}
void add(int s, int t)
{
    edge[edgecnt] = EDGE(t, head[s]);
    head[s] = edgecnt++;
}
LL dfs(int u)
{
    if(dp[u] != -1)
        return dp[u];
    dp[u] = 0;
    for(int i = head[u]; ~i; i = edge[i].next)
    {
        int v = edge[i].to;
        dp[u] += b[v] + dfs(v);
        dp[u] %= mod;
    }
    return dp[u];
}
int main()
{
    int n, m;
    while(~scanf("%d %d", &n, &m))
    {
        init();
        for(int i = 1; i <= n; i++)
            scanf("%lld %lld", &a[i], &b[i]);
        while(m--)
        {
            int u, v;
            scanf("%d %d", &u, &v);
            add(u, v);
        }
        for(int i = 1; i <= n; i++)
            dfs(i);
        LL ans = 0;
        for(int i = 1; i <= n; i++)
        {
            ans += a[i] * dp[i];
            ans %= mod;
        }
        printf("%lld\n", ans);
    }
    return 0;
}