51Node 1499 Figure

1499 Figure
Base time limit: 2 seconds Space limit: 262144 KB score: 80 Difficulty: 5-level algorithm topic collection focus on
to a graph, you want to divide these points into a, b two sets, so that the export sub-graph satisfying A is a complete graph, and B's export sub-graph is a non-edge Figure.
But in fact you don't have to be able to do it, so you allow mistakes. We define a perfect value of:
1. If the two points in a are connected, the perfect value of |i-j| is increased.
2. If two points in B are unbounded, increase the perfect value of the |i-j|.
(I,j is the number of these two points)
so we can make the perfect value the best.
N <=, M <= 200000
Input
N, m for points and sides
m rows, and
U,v for an no-forward edge.
(there will be no heavy and self-loops)
Output
A number that represents the maximum perfect value.
Input Example
5 5
1 2
1 3
1 4
1 5
2 3
Output example
one

Normal solution #include <bits/stdc++.h> #define FIN freopen ("Input.txt", "R", stdin);
#define FOUT freopen ("Output.txt", "w+", stdout);
using namespace Std;

typedef long Long LL;
const int INF = 0X3F3F3F3F;
const int mod = 998244353;
const double eps=1e-8;
Const double Pi=acos (-1.0);

const int N = 400010; The int map[1005][1005];//adjacency matrix, which holds the point-to-struct node {int x,y,id of the connected Edge;}
Node[n];
struct Edge {int to,cap,rev;//end point, capacity, reverse side};
The adjacency matrix of the vector<edge>g[n];//graph represents the distance designator int iter[n];//the access token int level[n];//vertex to the source point used in the bool Used[n];//dfs, the current arc, which is useless on the previous side.
    BOOL CMP (Node A,node b) {if (a.x==b.x) return a.y>b.y;
else return a.x<b.x;
    }//To add an edge void Addedge (int u,int v,int W) {G[u].push_back (edge) {v,w,g[v].size ()}) from S to t capacity to cap for the graph; G[v].push_back (Edge) {u,w,g[u].size ()-1});//If it is a non-direction graph, just change the 0 to W can}//through the BFS calculate the distance from the source point label void BFs (int s) {memset (level,-1,
    sizeof (level));
    Queue<int> que;
    level[s]=0;
    Que.push (s); while (!que.empty ()) {int V=que. Front ();
        Que.pop ();
            for (int i=0;i<g[v].size (); i++) {Edge &e=G[v][i];
                if (e.cap>0&&level[e.to]<0) {level[e.to]=level[v]+1;
            Que.push (e.to);
    }}}//Through DFS find augmented path int dfs (int v,int t,int f) {if (v==t) return F;
        for (int &i=iter[v];i<g[v].size (); i++) {Edge &e=G[v][i];
            if (e.cap>0&&level[v]<level[e.to]) {int D=dfs (e.to,t,min (F,e.cap));
                if (d>0) {e.cap-=d;
                G[e.to][e.rev].cap+=d;
            return D;
}}} return 0;
    }//solves the maximum stream int max_flow (int s,int t) from S to t {int flow=0;
    for (;;)
        {BFS (s);
        if (level[t]<0) return flow;
        memset (Iter,0,sizeof (ITER));
        int F;
    while ((F=dfs (S,t,inf)) >0) flow+=f;
}} void Init (){memset (map,0,sizeof (MAP));
for (int i=0;i<n;i++) g[i].clear ();
    } int main () {int n,m;
        while (~SCANF ("%d%d", &n,&m)) {Init ();
        int s=0,t=n+1;
        int u,v;
            for (int i=0; i<m; i++) {scanf ("%d%d", &u,&v);
            Map[u][v]=1;
        Map[v][u]=1;
        } int sum=0; for (int i=1, i<=n; i++) for (int j=1; j<=n; j + +) Sum+=abs (i-j);//Find out the value of all point pairs//prin
        TF ("sum:%d\n", sum); for (int i=1, i<=n; i++) {for (int j=i+1; j<=n; J + +) {Addedge (i,j,ab S (i-j));//Any point pair between a |u-v| edge if (map[i][j]==1)//For each edge (u,v), a U-, t capacity of |u-v|, even a V-t capacity of |u-v|
                The Edge.
                    {//printf ("<--%d%d-->\n", i,j);
                    Addedge (I,t,abs (i-j));
                Addedge (J,t,abs (i-j)); } else//no edge for each point pair (u,v), with an S-u capacity of |u-v|, and an edge of S-v capacity of |u-v|.
                    {//printf ("!--%d%d--! \ n", i,j);
                    Addedge (S,i,abs (i-j));
                Addedge (S,j,abs (i-j));
        }}} int Mincut=max_flow (s,t);//For Minimum cut//printf ("mincut:%d\n", mincut);
 printf ("%d\n", (sum-mincut)/2);//Final result}}