CodeForces - 894E Ralph and Mushrooms (强连通缩点+dp)

题意：一张有向图，每条边上都有wi个蘑菇，第i次经过这条边能够采到w-(i-1)*i/2个蘑菇，直到它为0。问最多能在这张图上采多少个蘑菇。

分析：在一个强连通分量内，边可以无限次地走直到该连通块内蘑菇被采完为止，因此每个强连通分量内的结果是确定的。

设一条边权值为w，最大走过次数为t，解一元二次方程得 t = (int)(1+sqrt(1+8w))；则该边对所在连通块的贡献为w*t - (t-1)*t*(t+1)/6。

而不在任何一个强连通分量内的边，最多只能走一次。所以在缩点后的DAG上进行dp即可。

#include<bits/stdc++.h>
using namespace std;
typedef long long LL;
const int maxn =1e6+5;
struct Edge{int v,next;LL val;  
}edges[maxn],E[maxn];
int head[maxn],tot,H[maxn],tt;
stack<int> S;
int pre[maxn],low[maxn],sccno[maxn],dfn,scc_cnt;
LL W[maxn];
LL dp[maxn];
void init()
{tot = dfn = scc_cnt=tt=0;memset(H,-1,sizeof(H));memset(W,0,sizeof(W));memset(dp,0,sizeof(dp));memset(pre,0,sizeof(pre));memset(sccno,0,sizeof(sccno));memset(head,-1,sizeof(head));
}void AddEdge(int u,int v,LL val)   {edges[tot] = (Edge){v,head[u],val};head[u] = tot++;
}void Tarjan(int u)
{int v;pre[u]=low[u]=++dfn;S.push(u);for(int i=head[u];~i;i=edges[i].next){v= edges[i].v;if(!pre[v]){Tarjan(v);low[u]=min(low[u],low[v]);}else if(!sccno[v]){low[u]=min(low[u],pre[v]);}}if(pre[u]==low[u]){int x;++scc_cnt;for(;;){x = S.top();S.pop();sccno[x]=scc_cnt;if(x==u)break;}}    
}void nAddEdge(int u,int v,LL w)
{E[tt] = (Edge){v,H[u],w};H[u] = tt++;
}LL dfs(int u)
{if(dp[u]) return dp[u];for(int i=H[u];~i;i=E[i].next){int v = E[i].v;dp[u] = max(dp[u],dfs(v)+E[i].val);}dp[u]+=W[u];return dp[u];
}int main()
{#ifndef ONLINE_JUDGEfreopen("in.txt","r",stdin);freopen("out.txt","w",stdout);#endifint N,M; while(scanf("%d%d",&N,&M)==2){init();int st,u,v; LL w;while(M--){scanf("%d%d%lld",&u,&v,&w);AddEdge(u,v,w);}scanf("%d",&st);for(int i=1;i<=N;++i){if(!pre[i]){Tarjan(i);}}for(int u =1;u<=N;++u){for(int i =head[u];~i;i = edges[i].next){v = edges[i].v;LL w = edges[i].val;if(sccno[u]!=sccno[v]){nAddEdge(sccno[u],sccno[v],w);}else{int t = (int)(1+sqrt(1+8*w))/2;W[sccno[u]] += (LL)t*w - (LL)(t-1)*t*(t+1)/6;}}}for(int i=1;i<=scc_cnt;++i){if(!dp[i]){dfs(i);}}printf("%lld\n",dp[sccno[st]]);}return 0;
}