L - Father Christmas flymouse

来源poj3160

After retirement as contestant from WHU ACM Team, flymouse volunteered to do the odds and ends such as cleaning out the computer lab for training as extension of his contribution to the team. When Christmas came, flymouse played Father Christmas to give gifts to the team members. The team members lived in distinct rooms in different buildings on the campus. To save vigor, flymouse decided to choose only one of those rooms as the place to start his journey and follow directed paths to visit one room after another and give out gifts en passant until he could reach no more unvisited rooms.

During the days on the team, flymouse left different impressions on his teammates at the time. Some of them, like LiZhiXu, with whom flymouse shared a lot of candies, would surely sing flymouse’s deeds of generosity, while the others, like snoopy, would never let flymouse off for his idleness. flymouse was able to use some kind of comfort index to quantitize whether better or worse he would feel after hearing the words from the gift recipients (positive for better and negative for worse). When arriving at a room, he chould choose to enter and give out a gift and hear the words from the recipient, or bypass the room in silence. He could arrive at a room more than once but never enter it a second time. He wanted to maximize the the sum of comfort indices accumulated along his journey.

Input

The input contains several test cases. Each test cases start with two integers N and M not exceeding 30 000 and 150 000 respectively on the first line, meaning that there were N team members living in N distinct rooms and M direct paths. On the next N lines there are N integers, one on each line, the i-th of which gives the comfort index of the words of the team member in the i-th room. Then follow M lines, each containing two integers i and j indicating a directed path from the i-th room to the j-th one. Process to end of file.

Output

For each test case, output one line with only the maximized sum of accumulated comfort indices.

Sample Input

2 2
14
21
0 1
1 0

Sample Output

35

Hint

32-bit signed integer type is capable of doing all arithmetic.

要走过节点最大，用tarjan算强连通，然后算出强连通里面的值是多少，把他看成一个点，建一个新的树，dfs

#include<iostream>
#include<stdio.h>
#include<stdlib.h>
#include <iomanip>
#include<cmath>
#include<float.h> 
#include<string.h>
#include<algorithm>
#define sf scanf
#define pf printf
#define mm(x,b) memset((x),(b),sizeof(x))
#include<vector>
#include<queue>
#include<map>
#define rep(i,a,n) for (int i=a;i<n;i++)
#define per(i,a,n) for (int i=a;i>=n;i--)
typedef long long ll;
const ll mod=1e9+100;
const double eps=1e-8;
using namespace std;
const double pi=acos(-1.0);
const int inf=0xfffffff;
const int N=30005;
struct Edge {
     int v,next;
}edge[5*N];
int dfn[N],low[N],sum[N];
int stack[N],node[N],visit[N],cnt,tot,index,value[N];
int belong[N],bcnt;
vector<int>tree[N];
vector<int>v;
void add_edge(int x,int y)
{
     edge[cnt].next=node[x];
     edge[cnt].v = y;
     node[x]=cnt++;
    return ;
 }
int dfs(int x,int y)
{
	int maxi=y;
	rep(i,0,tree[x].size())
	{
		maxi=max(maxi,dfs(tree[x][i],y+sum[tree[x][i]]));
	}
	return maxi;
}
void tarjan(int x)//代表第几个点在处理。递归的是点。
{
     dfn[x]=low[x]=++tot;// 新进点的初始化。
     stack[++index]=x;//进站
     visit[x]=1;//表示在栈里
    for(int i=node[x];i!=-1;i=edge[i].next)
     {
         if(!dfn[edge[i].v]) {//如果没访问过
            tarjan(edge[i].v);//往下进行延伸，开始递归
             low[x]=min(low[x],low[edge[i].v]);//递归出来，比较谁是谁的儿子／父亲，就是树的对应关系，涉及到强连通分量子树最小根的事情。
        }
        else if(visit[edge[i].v ]){  //如果访问过，并且还在栈里。
             low[x]=min(low[x],dfn[edge[i].v]);//比较谁是谁的儿子／父亲。就是链接对应关系
         }
     }
     if(low[x]==dfn[x]) //发现是整个强连通分量子树里的最小根。
    {
    	bcnt++;
         do{
         	belong[stack[index]]=bcnt;
             visit[stack[index]]=0;
             index--;
         }while(x!=stack[index+1]);//出栈 
     }
}
int in[N];
void solve(int n)
{
	tot=index=bcnt=0;
	v.clear();
	rep(i,0,n+1)
	tree[i].clear();
	mm(dfn,0);
	mm(in,0);
	mm(sum,0);
	mm(low,0);
	mm(belong,0);
	rep(i,1,n+1)
		if(!dfn[i])
		tarjan(i);
	rep(i,1,n+1)//计算入度 
	{
		if(value[i]>=0)
		sum[belong[i]]+=value[i];
		for(int j=node[i];j!=-1;j=edge[j].next)
		{
			if(belong[i]!=belong[edge[j].v])
			{
				in[belong[edge[j].v]]++;
				tree[belong[i]].push_back(belong[edge[j].v]);//连边
			}
		}
	}
	int ans=0;
	rep(i,1,bcnt+1)
	{
		if(!in[i])
		{
			ans=max(ans,dfs(i,sum[i]));
		}
	}
	pf("%d\n",ans);
}
int main()
{
	int n,m;
	while(~sf("%d%d",&n,&m))
	{
		cnt=0;
		mm(node,-1);
		rep(i,1,n+1)
		sf("%d",&value[i]);
		while(m--)
		{
			int x,y;
			sf("%d%d",&x,&y);
			add_edge(x+1,y+1);
		}
		solve(n);
	}
	return 0;
}