POJ：3160-Father Christmas flymouse

Father Christmas flymouse

Time Limit: 1000MS

Memory Limit: 131072K

Description

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.

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解题心得：

1. 题意很简单，就是给你一个有向图，要你选择任意一个起点开始走，每一个点有一个权值（有正有负），你在每一个点可以选择是否加该点的权值，每个点可以多次走过但是权值只能加一次，问你走这个图得到的最大值是多少。
2. 每个点枚举跑DFS就不想了，处理负权直接当0来处理就行了，因为可以选择不加上去。这个题可以选择缩点之后再跑DFS，将一个联通块缩成一个点，这个点的的权值就是连通图里面权值的总和。

---

#include<stdio.h>
#include<cstring>
#include<iostream>
#include<vector>
#include<stack>
using namespace std;
const int maxn = 3e4+200;
vector<int> ve[maxn],shrink[maxn],maps[maxn];
int pre[maxn],w[maxn],n,m,dfn[maxn],low[maxn],num,tot,w1[maxn],Max;
bool vis[maxn];
stack<int> st;

void init()
{
    while(!st.empty())
        st.pop();
    for(int i=0; i<=n; i++)
    {
        ve[i].clear();
        maps[i].clear();
        shrink[i].clear();
    }
    num = tot = 0;
    memset(w1,0,sizeof(w1));
    memset(vis,0,sizeof(vis));
    memset(pre,0,sizeof(pre));
    memset(dfn,0,sizeof(dfn));
    memset(low,0,sizeof(low));
    for(int i=0; i<n; i++)
    {
        scanf("%d",&w[i]);
        w[i] = w[i]<0?0:w[i];
    }
    for(int i=0; i<m; i++)
    {
        int a,b;
        scanf("%d%d",&a,&b);
        ve[a].push_back(b);
    }
}

void tarjan(int x)
{
    dfn[x] = low[x] = ++tot;
    vis[x] = true;
    st.push(x);
    for(int i=0; i<ve[x].size(); i++)
    {
        int v = ve[x][i];
        if(!dfn[v])
        {
            tarjan(v);
            low[x] = min(low[x],low[v]);
        }
        else if(vis[v])
        {
            low[x] = min(low[x],dfn[v]);
        }
    }
    if(low[x] == dfn[x])
    {
        while(1)
        {
            int now = st.top();
            st.pop();
            shrink[num].push_back(now);
            pre[now] = num;
            w1[num] += w[now];
            vis[now] = false;
            if(now == x)
                break;
        }
        num++;
    }
}

void get_new_maps()//缩点之后建立新图
{
    memset(vis,0,sizeof(vis));
    for(int i=0; i<num; i++)
    {
        for(int j=0; j<shrink[i].size(); j++)
        {
            for(int k=0; k<ve[shrink[i][j]].size(); k++)
            {
                int v = ve[shrink[i][j]][k];
                if(pre[v] != i)//两点不在同一个联通块内
                    maps[i].push_back(pre[v]);
            }
        }
    }
}

void dfs(int pos,int sum_w)
{
    if(sum_w > Max)
        Max = sum_w;
    for(int i=0;i<maps[pos].size();i++)
    {
        int v = maps[pos][i];
        dfs(v,sum_w+w1[v]);
    }
}

int get_ans()
{
    Max = 0;
    for(int i=0;i<num;i++)
    {
        dfs(i,w1[i]);
    }
}

int main()
{
    while(cin>>n>>m)
    {
        init();
        for(int i=0; i<n; i++)
        {
            if(!dfn[i])
                tarjan(i);
        }
        get_new_maps();
        get_ans();
        printf("%d\n",Max);
    }
    return 0;
}