codeforces 894e Ralph and Mushrooms

http://www.elijahqi.win/archives/1627
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 3 when 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?

Input
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 y with 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.

Output
Print an integer denoting the maximum number of the mushrooms Ralph can collect during his route.

Examples
Input
2 2
1 2 4
2 1 4
1
Output
16
Input
3 3
1 2 4
2 3 3
1 3 8
1
Output
8
Note
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.

题目 给定边权 每次可以取边上x x-1 x-1-2 x-1-2-3 直到减到0为止

那么首先可以tarjan缩点 就是所有强连通的点上可以都取到 那么 我们可以采用数学方法计算强连通

点的权值 这个假设我们 最大值是x 然后 求出1+2+3+..x<=n 这个最大的x的整数解出现在哪里

那么整个环的答案就是：首先求出n∗(n+1)<=w的最大的n，然后价值就是n∗w−∑i∗(i+1)/2+w(1<=i&i<=n)，也就是n∗w−n∗(n+1)∗(n+2)/6+w

然后根据拓扑序求出我的最长路即可 然后我就把点权直接记在边上 跑拓扑序dp

最后输出答案即可 注意有些地方需要使用Long long

#include<cstdio>
#include<stack>
#include<cmath>
#include<queue>
#include<algorithm>
#define ll long long
#define inf 1LL<<60
#define N 1100000
using namespace std;
inline char gc(){
    static char now[1<<16],*S,*T;
    if (T==S){T=(S=now)+fread(now,1,1<<16,stdin);if (T==S) return EOF;}
    return *S++;
}
inline int read(){
    int x=0;char ch=gc();
    while (ch<'0'||ch>'9') ch=gc();
    while (ch<='9'&&ch>='0'){x=x*10+ch-'0';ch=gc();}
    return x;
}
stack<int>q;
struct node{
    int x,y,next;ll z;
}data[N<<1],data1[N<<1];
ll ans=-1,f[N],w[N];
int b[N],stackf[N],dfn[N],low[N],h1[N],h[N],num,s,n,m,in[N];
void tarjan(int x){
    dfn[x]=low[x]=++num;stackf[x]=1;q.push(x);
    for (int i=h[x];i;i=data[i].next){
        int y=data[i].y;
        if (!dfn[y]) tarjan(y),low[x]=min(low[x],low[y]);else if (stackf[y]) low[x]=min(low[x],dfn[y]);
    }
    if (dfn[x]==low[x]){
        int y;++s;
        do{
            y=q.top();q.pop();
            b[y]=s;stackf[y]=0;
        }while (y!=x);
    }
}
inline ll calc(ll x){
    ll tt=sqrt(2*x+0.25)-0.5;
    return x+tt*x-(tt+1)*(tt+2)*tt/6;
}
inline void insert1(int x,int y,ll z){
    data1[++num].y=y;data1[num].z=z;data1[num].next=h1[x];h1[x]=num;data1[num].x=x;
}
int main(){
    freopen("cf.in","r",stdin);
    n=read();m=read();
    for (int i=1;i<=m;++i){
        int x=read(),y=read(),z=read();
        data[++num].y=y;data[num].z=z;data[num].next=h[x];h[x]=num;data[num].x=x;
    }
    int st=read();num=0;
    for (int i=1;i<=n;++i) if (!dfn[i]) tarjan(i);
    for (int i=1;i<=m;++i){
        int x=data[i].x,y=data[i].y;
        if (b[x]==b[y]) w[b[x]]+=calc(data[i].z);
    }num=0;
    for (int i=1;i<=n;++i){
        for (int j=h[i];j;j=data[j].next){
            int y=data[j].y;if (b[y]==b[i]) continue;
            insert1(b[i],b[y],data[j].z+w[b[y]]);
        }
    }queue<int>q;
    for (int i=1;i<=s;++i) f[i]=-inf;f[b[st]]=w[b[st]];
    for (int i=1;i<=num;++i) in[data1[i].y]++;
    for (int i=1;i<=s;++i)  if (!in[i]) q.push(i);
    while(!q.empty()){
        int x=q.front();q.pop();
        for (int i=h1[x];i;i=data1[i].next){
            int y=data1[i].y;ll z=data1[i].z;if (--in[y]==0) q.push(y);f[y]=max(f[y],f[x]+z);
        }
    }
    for (int i=1;i<=s;++i) ans=max(ans,f[i]);
    printf("%I64d",ans);
    return 0;
}