[GDSOI2018 Simulation 4.19] Arrangement

topic

There are n numbers x1 ~xn . You need to find a permutation of them that satisfies m conditions, each of the form x_a must come before x_b. On this basis, you want to maximize the maximum subsegment sum of this permutation.

analyze

I'm not particularly sure of his rationale. . .

code

#include<iostream>
#include<cstring>
#include<cstdio>
#include<cmath>
#include<string>
#include<algorithm>
#include<queue>;

using namespace std;

const int INF=1e9;

struct arr{
    int x,y,w,next;
}edge[1000000];
int ls[100000];
int cur[100000];
int f[1000000];
int sign[100000]; 
int edge_m=1;
int ans;
int s,t;
int n,m;

int add(int x,int y,int w)//加边。
{
    edge_m++;
    edge[edge_m]=(arr){x,y,w,ls[x]},f[edge_m]=w,ls[x]=edge_m,cur[x]=edge_m; 
    edge_m++;
    edge[edge_m]=(arr){y,x,w,ls[y]},f[edge_m]=0,ls[y]=edge_m,cur[y]=edge_m;
} 

int dis[100000];

bool bfs()//bfs判断是否继续。 
{
    memset(dis,-1,sizeof(dis));
    queue<int> q;   
    dis[s]=0;
    q.push(s);
    do
    {
        int x=q.front();
        q.pop();
        for (int i=ls[x];i;i=edge[i].next)
        {
            if ((f[i])&&(dis[edge[i].y]==-1))
            {
                dis[edge[i].y]=dis[x]+1;
                q.push(edge[i].y);
                if (edge[i].y==t) return true;
            }
        }
    }while (!q.empty());
    return false;
}

int find(int x,int min_)//找增广路。
{
    if (x==t) return min_;
    int rec=min_;
    for (int i=cur[x];i;i=edge[i].next)
    {
        if ((f[i]!=0)&&(dis[x]+1==dis[edge[i].y]))
        {
            int k=find(edge[i].y,min(rec,f[i]));
            if (k!=0) sign[x]=edge[i].y;
            f[i]-=k;
            f[i^1]+=k;
            rec-=k;
            if (rec==0) return min_;
        }
    }
    if (rec==min_) dis[x]=-1;
    return min_-rec;
}

int dinic()
{
    while (bfs())
    {
        for (int i=0;i<=t;i++) cur[i]=ls[i];
        ans-=find(s,INF);
    }
    printf("%d",ans);
}

int a[100000];

int main()
{
    //freopen("permutation.in","r",stdin);
    //freopen("permutation.out","w",stdout);
    scanf("%d%d",&n,&m);
    s=n*2+1; t=s+1;
    ans=0;
    for (int i=1;i<=n;i++)
    {
        scanf("%d",&a[i]);
    }
    for (int i=1;i<=n;i++)
    {
        if (a[i]>0)
        {
            ans+=a[i];
            add(s,i,a[i]);
            add(n+i,t,a[i]);
        }
        else
            add(i,n+i,-a[i]);
    }
    for (int i=1;i<=m;i++)
    {
        int x,y;
        scanf("%d%d",&x,&y);
        add(x,y,INF);
        add(x+n,y+n,INF);
    }
    dinic();
}