给定n个点m条边有向图及边权w,第i次经过一条边边权为w−1−2.−..−i,w≥0给定起点s问从起点出发最多能够得到权和,某条边可重复经过
有向图能够重复经过的边当且仅当成环,所以tarjan缩点成DAG,缩点后每个点内的权值可以通过算出,假设最大的n使得w−n(n+1)2≥0,那么该点值为(n+1)w−n(n+1)(n+2)/6,通过对DAG进行dp算出最长路就是答案
#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cstring>
#include <string>
#include <cmath>
#include<stack>
using namespace std;
typedef long long LL;
const int N=1000005;
int n,m,x,y,w,s;
int head[N],to[N],nxt[N],val[N],cnt;
void init(){memset(head,-1,sizeof(head)),cnt=0;}
void add(int u,int v,int w)
{
to[cnt]=v;val[cnt]=w;nxt[cnt]=head[u];head[u]=cnt++;
}
int head2[N],to2[N],nxt2[N],cnt2;LL val2[N];
void init2(){memset(head2,-1,sizeof(head2)),cnt2=0;}
void add2(int u,int v,int w)
{
to2[cnt2]=v;val2[cnt2]=w;nxt2[cnt2]=head2[u];head2[u]=cnt2++;
}
int dfs_ind = 1, dfn[N], low[N], sccno[N], scc_cnt=0;
LL w_[N];
stack<int> st;
void tarjan(int u)
{
dfn[u]=low[u]=dfs_ind++;
st.push(u);
for(int i=head[u];~i;i=nxt[i])
{
int v=to[i];
if(!dfn[v]){tarjan(v);low[u]=min(low[u],low[v]);}
else if(!sccno[v]){low[u]=min(low[u],dfn[v]);}
}
if(low[u]==dfn[u])
{
scc_cnt++;
while(1)
{
int x=st.top();
st.pop();
sccno[x]=scc_cnt;
if(x==u)
break;
}
}
}
LL cal(LL x)
{
LL n=sqrt(2.0*x+0.25)-0.5;
return (n+1)*x-(n+1)*(n+2)*n/6;
}
void DAG()
{
memset(dfn,0,sizeof(dfn));
memset(low,0,sizeof(low));
memset(sccno,0,sizeof(sccno));
memset(w_,0,sizeof(w_));
init2();
for(int i=1;i<=n;i++)
{
if(!dfn[i])
tarjan(i);
}
for(int i=1;i<=n;i++)
{
for(int j=head[i];~j;j=nxt[j])
{
int v=to[j];
if(sccno[i]!=sccno[v])
add2(sccno[i],sccno[v],1LL*val[j]);
else
w_[sccno[i]]+=cal(val[j]);
}
}
}
LL dp[N];
void dfs(int u)
{
if(~dp[u])
return ;
dp[u]=w_[u];
for(int i=head2[u];~i;i=nxt2[i])
{
dfs(to2[i]);
dp[u]=max(dp[u],w_[u]+val2[i]+dp[to2[i]]);
}
}
int main()
{
init();
scanf("%d%d",&n,&m);
for(int i=1;i<=m;i++)
scanf("%d%d%d",&x,&y,&w),add(x,y,w);
scanf("%d",&s);
DAG();
memset(dp,-1,sizeof(dp));
dfs(sccno[s]);
cout<<dp[sccno[s]]<<endl;
}