APIO 2009 抢掠计划

题目问题自行解决。

总的来说这题还是比较简单的，缩完点之后就是一张有向无环图，然后spfa一遍最长路(是这么叫吧)就可以了。不过题目的数据坑爹，有链状的...所以缩点要手工栈非递归才能AC不然86分...囧。我是写不出来了...

 

代码很丑但还是贴一下防丢：

#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<queue>
#include<iostream>
using namespace std;
#define MaxN 700010
struct atp
{
    int x,y;
    }P[MaxN];
struct bian
{
    int y,next;
    }e[MaxN];int first[MaxN];

int d[MaxN],dfn[MaxN],low[MaxN],belong[MaxN],cost[MaxN];
bool b[MaxN];
bool bar[MaxN];
int stap[MaxN];
int tot=0,n,m,Bcnt,Index,stop,center,ans=0;
void add(int x,int y)
{
    tot++;
    e[tot].y=y;
    e[tot].next=first[x];
    first[x]=tot;
    }
    
void init()
{
    int x,t;
    scanf("%d%d",&n,&m);
    for (int i=1;i<=m;i++)
        scanf("%d%d",&P[i].x,&P[i].y),add(P[i].x,P[i].y);
    for (int i=1;i<=n;i++)
        scanf("%d",&d[i]);
    scanf("%d%d",&center,&x);
    for (int i=1;i<=x;i++)
    {
        scanf("%d",&t);
        bar[t]=true;
        }
    }

void tarjan(int i)
{
    int j;
    dfn[i]=low[i]=++Index;
    b[i]=true;
    stap[++stop]=i;
    for (int p=first[i];p;p=e[p].next)
    {
        j=e[p].y;
        if (!dfn[j])
        {
            tarjan(j);
            if(low[j]<low[i]) low[i]=low[j];
            } else
        if (b[j] && dfn[j]<low[i]) low[i]=dfn[j];
        }
    if (dfn[i]==low[i])
    {
        Bcnt++;
        do
        {
            j=stap[stop--];
            b[j]=false;
            belong[j]=Bcnt;
            }
        while (j!=i);
        }
    }
    



void spfa()
{
    int u;
    bool v[MaxN];
    memset(v,false,sizeof(false));
    queue<int>q;while (!q.empty()) q.pop();
    memset(d,0,sizeof(d));
    d[belong[center]]=cost[belong[center]];
    q.push(belong[center]);
    v[belong[center]]=true;
    while (!q.empty())
    {
        u=q.front();q.pop();v[u]=false;
        for (int p=first[u];p;p=e[p].next)
        {
            if (d[u]+cost[e[p].y]>d[e[p].y]) d[e[p].y]=d[u]+cost[e[p].y];
            if (!v[e[p].y])
            {
                           v[e[p].y]=true;
                           q.push(e[p].y);                           
                           }

            }
        }
    for (int i=1;i<=Bcnt;i++)
        if (b[i] &&d[i]>ans)ans=d[i];
    }
    
void work()
{
    stop=Bcnt=Index=0;
    memset(dfn,0,sizeof(dfn));
    for (int i=1;i<=n;i++)
        if (!dfn[i]) tarjan(i);
    memset(first,0,sizeof(first));tot=0;
    memset(b,false,sizeof(false));
    for (int i=1;i<=n;i++)
    {
        cost[belong[i]]+=d[i];
        if (bar[i]) b[belong[i]]=true;
        }
    for (int i=1;i<=m;i++)
        if (belong[P[i].x]!=belong[P[i].y])
        add(belong[P[i].x],belong[P[i].y]);
    spfa();
    printf("%d\n",ans);
    }
int main()
{
    freopen("atm.in","r",stdin);
    freopen("atm.out","w",stdout);
    init();
    work();
    return 0;
    }