题意

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$w-1-2.-..-i$，$w\ge 0$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二分算出，假设最大的$n$使得$w-\frac{n(n+1)}{2}\ge 0$，那么该点值为$(n+1)w-\frac{n(n+1)(n+2)}{6}$，通过对DAG进行dp算出最长路就是答案

代码

#include <bits/stdc++.h>
#define inf 0x7f7f7f7f
using namespace std;
typedef long long LL;
const int N = 1000005;
int n, m, x, y, w, s;
int head[N], nxt[N], to[N], val[N], cnt;
inline void init(){memset(head, -1, sizeof(head)); cnt = 0;}
inline void add(int u, int v, int w) {to[cnt] = v, val[cnt] = w, nxt[cnt] = head[u], head[u] = cnt++;}
int head2[N], nxt2[N], to2[N], cnt2; LL val2[N];
inline void init2() {memset(head2, -1, sizeof(head)); cnt2 = 0;}
inline void add2(int u, int v, LL 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(dfn[u] == low[u]) {scc_cnt++;while(1) {int x = st.top(); st.pop();sccno[x] = scc_cnt;if(x == u) break;}}
}
inline 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] + dp[to2[i]] + val2[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;return 0;
}