Ralph is going to collect mushrooms in the Mushroom Forest.
There are m directed paths connecting n trees in the Mushroom Forest. On each path grow some mushrooms. When Ralph passes a path, he collects all the mushrooms on the path. The Mushroom Forest has a magical fertile ground where mushrooms grow at a fantastic speed. New mushrooms regrow as soon as Ralph finishes mushroom collection on a path. More specifically, after Ralph passes a path the i-th time, there regrow i mushrooms less than there was before this pass. That is, if there is initially x mushrooms on a path, then Ralph will collect x mushrooms for the first time, x - 1 mushrooms the second time, x - 1 - 2 mushrooms the third time, and so on. However, the number of mushrooms can never be less than 0.
For example, let there be 9 mushrooms on a path initially. The number of mushrooms that can be collected from the path is 9, 8, 6 and 3when Ralph passes by from first to fourth time. From the fifth time and later Ralph can't collect any mushrooms from the path (but still can pass it).
Ralph decided to start from the tree s. How many mushrooms can he collect using only described paths?
The first line contains two integers n and m (1 ≤ n ≤ 106, 0 ≤ m ≤ 106), representing the number of trees and the number of directed paths in the Mushroom Forest, respectively.
Each of the following m lines contains three integers x, y and w (1 ≤ x, y ≤ n, 0 ≤ w ≤ 108), denoting a path that leads from tree x to tree ywith w mushrooms initially. There can be paths that lead from a tree to itself, and multiple paths between the same pair of trees.
The last line contains a single integer s (1 ≤ s ≤ n) — the starting position of Ralph.
Print an integer denoting the maximum number of the mushrooms Ralph can collect during his route.
2 2
1 2 4
2 1 4
1
16
3 3
1 2 4
2 3 3
1 3 8
1
8
In the first sample Ralph can pass three times on the circle and collect 4 + 4 + 3 + 3 + 1 + 1 = 16 mushrooms. After that there will be no mushrooms for Ralph to collect.
In the second sample, Ralph can go to tree 3 and collect 8 mushrooms on the path from tree 1 to tree 3.
大致题意：n个点m条有向边，每条边有一个权值，每条边可以经过多次，但是权值都会减少i,i是已经经过当前这条边的次数，不能减到0以下，现在给定起点，问能得到的最大的权值是多少？
分析：这道题算法比较好想，数学式子不大好推.
一个强连通分量里的所有边都是可以将权值取完的，将原图缩点后，图就变成了DAG,每条边只能经过1次，那么做一次dp,在加上经过的点内部可以得到的权值和就是答案.
关键就是给定一个n,如何计算这个n能够有多少贡献.最后的答案为n + (n - 1) + (n - 1 - 2) + (n - 1 - 2 - 3) +......因为是求和，从整体上来看，先把每一项变成n,项数为t,那么贡献为nt,剩下的t-1个数都要-1,t-2个数都要-2,根据这个来得到问题的解.首先考虑如何求出t来，因为式子减到0以下就不减了，后面的是一个等差数列，可以利用等差数列求和公式，设t' = t - 1,那么n = (t' + 1)*t' / 2,t'可以直接解一元二次方程得到，也可以二分得到，这里选用解方程的方法.之所以设t' = t-1是因为第一项-0，会多出一项.求出t之后，后面要减去的贡献可以表示为Σi*(t-i+1) (1 ≤ i ≤ t),利用乘法分配律展开可以得到t*(t+1)*(t+2)/6.那么贡献就是nt - t*(t+1)*(t+2)/6.
#include <cstdio> #include <cmath> #include <stack> #include <vector> #include <cstring> #include <iostream> #include <algorithm> using namespace std; typedef long long ll; const int maxn = 2000010; ll n, m, head[maxn], to[maxn], nextt[maxn], w[maxn], scc[maxn], pre[maxn], low[maxn], dfs_clock; ll tot = 1, f[maxn], cnt, head2[maxn], tot2 = 1, nextt2[maxn], to2[maxn], w2[maxn], tott; ll bg, ans, dp[maxn]; bool vis[maxn]; struct node { ll x, y, z; }e[maxn]; void add(ll x, ll y, ll z) { to[tot] = y; w[tot] = z; nextt[tot] = head[x]; head[x] = tot++; } void Add(ll x, ll y, ll z) { w2[tot2] = z; to2[tot2] = y; nextt2[tot2] = head2[x]; head2[x] = tot2++; } stack <ll> s; void tarjan(int u) { pre[u] = low[u] = ++dfs_clock; s.push(u); for (int i = head[u]; i; i = nextt[i]) { int v = to[i]; if (!pre[v]) { tarjan(v); low[u] = min(low[u], low[v]); } else if (!scc[v]) low[u] = min(low[u], pre[v]); } if (pre[u] == low[u]) { cnt++; while (1) { int t = s.top(); s.pop(); scc[t] = cnt; if (t == u) break; } } } ll solve(ll x) { ll t = (ll)floor((-1.0 + sqrt(1.0 + 8 * x)) / 2); ll s = (t + 1) * x - t * (t + 1) * (t + 2) / 6; return s; } ll dfss(int u) { if (dp[u] != -1) return dp[u]; ll res = f[u]; for (ll i = head2[u]; i; i = nextt2[i]) { ll v = to2[i]; res = max(res, f[u] + w2[i] + dfss(v)); } dp[u] = res; return res; } int main() { scanf("%I64d%I64d", &n, &m); for (int i = 1; i <= m; i++) { ll a, b, c; scanf("%I64d%I64d%I64d", &a, &b, &c); e[++tott].x = a; e[tott].y = b; e[tott].z = c; add(a, b, c); } scanf("%I64d", &bg); for (int i = 1; i <= n; i++) if (!pre[i]) tarjan(i); for (ll i = 1; i <= tott; i++) { ll x = e[i].x, y = e[i].y, z = e[i].z; if (scc[x] == scc[y]) f[scc[x]] += solve(z); } for (ll i = 1; i <= n; i++) for (ll j = head[i]; j; j = nextt[j]) { ll v = to[j]; if (scc[i] != scc[v]) Add(scc[i], scc[v], w[j]); } for (ll i = 1; i <= cnt; i++) dp[i] = -1; ans = dfss(scc[bg]); printf("%I64d ", ans); return 0; }