Codeforces Round#447 E Ralph and Mushrooms

题目描述：
在一个有向图中，Ralph从某一个起点出发，去采蘑菇，每条道路的权重代表该道路上蘑菇的初始值，每次经过这条道路时，Ralph会将这条道路上的所有蘑菇都采掉，然后蘑菇会又长出来，但是会比上一次的蘑菇少，第一次走过后蘑菇会少一个，第二次走过后蘑菇会少两个，直到蘑菇少完。比如某条路一开始有5个蘑菇，第一次走过后，还有5-1=4个蘑菇，第二次走过后，还有4-2= 2个蘑菇，第三次走过后，不剩蘑菇。现在给出这个有向图和图中的边的权值，求Ralph最多能拿到多少蘑菇。
算法思想：
想一下，如果要拿到最多的蘑菇，能到达的环一定要把每个边上的所有蘑菇都拿得不能再拿了，简单来看就是先缩点，然后DAG上的最长路就可以了。
具体实现：
首先先跑一遍Tarjan，把SCC跑出来，再把SCC构建成一个新的DAG，在这个DAG上图DP跑最长路就可以得到最后结果了。

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贴一下代码

#include <cstdio>#include <cstring>#include <algorithm>#include <queue>using namespace std;typedef long long ll;const int maxn = 1000010;int head[maxn],head1[maxn],DFN[maxn],LOW[maxn];int belong[maxn],Stack[maxn];bool vis[maxn];bool instack[maxn];int du[maxn],L[maxn];ll d[maxn];int n,m;ll V[maxn];int Cnt,Bcnt,Index,Top,Cnt1;void init(){    Cnt = 1;    Cnt1 = 1;    Top = 0;    Index = 0;    Bcnt = 0;    memset(vis,false,sizeof(vis));    memset(d,0,sizeof(d));    memset(head,-1,sizeof(head));    memset(head1,-1,sizeof(head1));    memset(V,0,sizeof(V));    memset(DFN,0,sizeof(DFN));    memset(belong,0,sizeof(belong));    memset(instack,false,sizeof(instack));}struct Edge{    int to;    int next;    ll val;    int from;}edge[maxn],edge1[maxn];// edge is the graph before ,edge1 is the DAG of SCCvoid addEdge(int a,int b,ll v){    edge[Cnt].to = b;    edge[Cnt].next = head[a];    edge[Cnt].val = v;    edge[Cnt].from = a;    head[a] = Cnt++;}void Tarjan(int i){    int j;    DFN[i] = LOW[i] = ++Index;    instack[i] = true;    Stack[++Top] = i;    for(j = head[i];j != -1;j = edge[j].next)    {        int v = edge[j].to;        if(!DFN[v])        {            Tarjan(v);            if(LOW[v] < LOW[i])                LOW[i] = LOW[v];        }        else if(instack[v] && DFN[v] < LOW[i])            LOW[i] = DFN[v];    }    if(DFN[i] == LOW[i])    {        Bcnt++;        do{            j = Stack[Top--];            instack[j] = false;            belong[j] = Bcnt;        }while(j!=i);    }}ll Value(ll j)      // for every value j we can calc how                //many mushrooms can we get from that edge{    ll left = 1;    ll Right = j;    ll midd;    while(left < Right - 1)    {        midd = (left + Right) / 2;        if((midd*(midd+1))/2 < j)        {            left = midd;        }        else if((midd*(midd+1))/2 > j)        {            Right = midd;        }        else         {            break;        }    }    midd = (left+Right)/2;    ll ans = 0;    ans = ans + j*(midd+1);    ans = ans - (midd)*(midd+1)*(midd+2)/6;    return ans;}void addEdge1(int a,int b,ll v){    edge1[Cnt1].to = b;    edge1[Cnt1].next = head1[a];    edge1[Cnt1].val = v;    edge1[Cnt1].from = a;    head1[a] = Cnt1++;}void Build_Value()  //Calc every value of SCC{    int i;    for(i = 1;i <= Cnt;i ++)    {        if(belong[edge[i].from] == belong[edge[i].to])        {            V[belong[edge[i].from]] += Value(edge[i].val);        }        else         {            addEdge1(belong[edge[i].from],belong[edge[i].to],edge[i].val);        }    }}void toposort()     // Run topological sort{    memset(du,0,sizeof(du));    for(int i = 1;i <= Bcnt;++ i)    {        for(int j = head1[i];j != -1;j = edge1[j].next)        {            int v = edge1[j].to;            du[v] ++;            //printf("%d %d\n",edge1[j].from,edge1[j].to);        }    }    int tot = 0;    queue<int>Q;    for(int i = 1;i <= Bcnt;++ i)    {        if(!du[i])Q.push(i);    }    while(!Q.empty())    {        int x = Q.front();        Q.pop();        L[++tot] = x;        for(int j = head1[x];j != -1;j = edge1[j].next)        {            int v = edge1[j].to;            du[v] --;            if(!du[v])Q.push(v);        }    }}void dp(int j)  // Run Dp on DAG to find the Longest road{    d[j] += V[j];    int s;    for(s = head1[j];s != -1;s = edge1[s].next)    {        int v = edge1[s].to;        vis[v] = true;        d[v] = max(d[v],d[j] + edge1[s].val);    //  printf("%d\n",edge1[s].val);    }}void debug()        //just for debug{    for(int i = 1;i <= n;++ i)    {        //printf("%d\n",belong[i]);    }    for(int i = 1;i <= Bcnt;++ i)    {        printf("%d\n",L[i]);    }}int main(){    init();    scanf("%d%d",&n,&m);    for(int i = 1;i <= m;++ i)    {        int x,y;        ll v;        scanf("%d%d%lld",&x,&y,&v);        addEdge(x,y,v);    }    int st;    scanf("%d",&st);    for(int i = 1;i <= n;++ i)        if(!DFN[i])            Tarjan(i);    Build_Value();    toposort();    vis[belong[st]] = true;    for(int i = 1;i <= Bcnt;++ i)    {        if(vis[L[i]])            dp(L[i]);    }//  debug();    ll MAX = 0;    for(int i = 1;i <= Bcnt;++ i)    {        MAX = max(MAX,d[i]);    }    printf("%lld\n",MAX);    return 0;}

感觉整个题目只要思路清晰，解决起来是没有问题的。毕竟我这个大水X都写出来了