题目链接:
https://cpc.csgrandeur.cn/csgoj/problemset/problem?pid=1135
这个题的代码真不懂,求助大家
#include<cstdio>
#include<queue>
#include<iostream>
#include<vector>
#include<map>
#include<cstring>
#include<string>
#include<set>
#include<stack>
#include<algorithm>
#define cle(a) memset(a,0,sizeof(a))
#define inf(a) memset(a,0x3f,sizeof(a))
#define ll long long
#define Rep(i,a,n) for(int i=a;i<=n;i++)
using namespace std;
#define INF2 9223372036854775807ll
const int INF = ( 2e9 ) + 2;
const ll maxn = 1e5+10;
int d[maxn],a[maxn],b[maxn],In[maxn];
vector<int> g[maxn];
const int mod = 1e9+7;
ll topsort(int n)
{
ll ans=0;
priority_queue<int> q;
for(int i=1;i<=n;i++)
{
if(In[i]==0)
{
q.push(i);
}
}
while(!q.empty())
{
int u=q.top();q.pop();
d[u]=(d[u]+a[u])%mod;
for(int i=0,L=g[u].size();i<L;i++)
{
int v=g[u][i];
In[v]--;
d[v] = (d[u]+d[v])%mod;
if(In[v]==0)
q.push(v);
ans = (ans+1LL*d[u]*b[v])%mod;
}
}
return ans%mod;
}
int main()
{
int n,m;
while(~scanf("%d%d",&n,&m))
{
for(int i=1;i<=n;i++)g[i].clear();
memset(In,0,sizeof(In));
memset(d,0,sizeof(d));
for(int i=1;i<=n;i++)
scanf("%d%d",&a[i],&b[i]);
for(int i=0;i<m;i++)
{
int u,v;
scanf("%d%d",&u,&v);
g[u].push_back(v);
In[v]++;
}
ll ans=topsort(n);
cout<<ans<<endl;
}
}
思路
拓扑排序,因为是有向无环图,所以应该要找到入度为0的点,沿着这点所到的点,入度为0的点都能到达,将入度为0的点的a[i] ”传递“到它能直接到达的点,累加过去。