Hdoj 4925 Apple Tree "min cut"

Title: Hdoj 4925 Apple Tree

Source: multi-university Training Contest 6

Test instructions: Give a matrix, and then each lattice in the number is 2^ (the number of adjacent lattice), and then asked not to take the adjacent number, so that the number of the largest.

Analysis: There are two ways to solve this problem, which is general solution. Network flow, there is also a law, because the problem of data is regular, so can find the law. A lot of people do that.

The following is a solution to the network flow, in fact, is a lattice fetch number problem.

is hdoj 1569 Click to open the version number of the link, only the data range is increased. It's just data water. Down the same effect. The same code can be AC.

ans = sum-min cut

Specific ideas see above link: Click to open the link

AC Code:

#include <cstdio> #include <cstring> #include <iostream> #include <string> #include < algorithm> #include <vector> #include <queue>using namespace std; #define Del (b) memset (A,b,sizeof (a)) const int N = 10200;const int inf = 0x3f3f3f3f;int n,m;struct node{int from,to,cap,flow;};  Vector<int> v[n];vector<node> E;int Vis[n];    Build the hierarchy diagram int cur[n];void add_node (int from,int to,int cap) {E.push_back ((Node) {from,to,cap,0});    E.push_back (Node) {to,from,0,0});    int tmp=e.size ();    V[from].push_back (tmp-2); V[to].push_back (tmp-1);}    BOOL BFs (int s,int t) {Del (vis,-1);    Queue<int> Q;    Q.push (s);    Vis[s] = 0;        while (!q.empty ()) {int X=q.front ();        Q.pop ();            for (int i=0; i<v[x].size (); i++) {Node TMP = e[v[x][i]];                if (vis[tmp.to]<0 && tmp.cap>tmp.flow)//The second condition guarantees {vis[tmp.to]=vis[x]+1;Q.push (tmp.to);    }}} if (vis[t]>0) return true; return false;}    int dfs (int o,int f,int t) {if (o==t | | f==0)//optimize return F;    int a = 0,ans=0;        for (int &i=cur[o]; i<v[o].size (); i++)//Note front ' & ', very important optimization {Node &tmp = e[v[o][i]];            if (vis[tmp.to]== (vis[o]+1) && (a = DFS (Tmp.to,min (f,tmp.cap-tmp.flow), T)) >0) {tmp.flow+=a; E[v[o][i]^1].flow-=a;            Ans+=a of the mode of storage and mapping;            F-=a;        if (f==0)//attention optimization break;  }} return ans;    Optimized}int dinci (int s,int t) {int ans=0;        while (BFS (s,t)) {Del (cur,0);        int Tm=dfs (S,INF,T);    Ans+=tm; } return ans;    int solve (int i,int j) {int ans=1;    if (i>1) ans*=2;    if (j>1) ans*=2;    if (i<n) ans*=2;    if (j<m) ans*=2; return ans;} int ID (int i,int j) {return (i-1) *m+j;}    int main () {int T;    scanf ("%d", &t); While(t--)        {scanf ("%d%d", &n,&m);        int s=0,t=m*n+1,x,sum=0;                for (int i=1;i<=n;i++) {for (int j=1;j<=m;j++) {x=solve (i,j);                Sum+=x;                    if ((i+j)%2) {Add_node (S,id (I,J), x);                    if (i>1) Add_node (ID (I,J), id (i-1,j), INF);                    if (j>1) Add_node (ID (I,J), id (i,j-1), INF);                    if (i<n) Add_node (ID (I,J), id (i+1,j), INF);                if (j<m) Add_node (ID (I,J), id (i,j+1), INF);            } else Add_node (ID (i,j), t,x);        }} printf ("%d\n", Sum-dinci (s,t));        for (int i=0;i<=t;i++) v[i].clear ();    E.clear (); } return 0;}

Hdoj 4925 Apple Tree "min cut"