Xiao Ming is playing a very interesting game recently. The rules of the game are as follows:
There is one n ∗ m n*m n * m map. Every position on the map is either an open space, or a turret, or some enemies. Xiao Ming needs to manipulate the turret to attack the enemy.
The attack method is: for each turret, the game system has given the direction it can aim at (one of the top, bottom, left and right), Xiao Ming needs to choose its attack position. Each turret can only attack one position. The turret can only attack a certain position in its aiming direction. Of course, the turret can also not attack. The turret is powerful. It can and can only destroy all enemies at the target position.
For security reasons, the game system has ensured that there is no turret that can aim at another turret, that is, for any turret, there is no other turret at all possible attack positions. Moreover, if the starting point and end point of the turret are called the running track of the shell, the system does not allow the two tracks to intersect (including the starting point and end point).
Now, after selecting the target location, each turret fires at the same time. You have to tell Xiao Ming how many enemies he can kill at most.
analysisThe key to this problem is to build a map
First, find the maximum value that each turret can hit, and add it together to set it as tot
Then consider the conflict situation
Since the turrets cannot reach each other, the conflict must be horizontal and vertical
Cut each point into two points: horizontal point and vertical point. The horizontal point connects an inf edge to the vertical point
Then, for each transverse turret, connect an s to the inf side of the turret
Then connect the edges from the turret to the shell direction, and the edge weight is (the maximum value reached by the turret) - (the weight of the starting grid of the edge)
In the same way, the vertical turret is connected to the t-point, and the direction of the side is reversed, balabala
This illegal situation will connect s with t
Consider cutting off one ( i , j ) (i,j) (i,j) to ( i , j + 1 ) (i,j+1) The edge of (i,j+1) means that the corresponding turret is changed to attack ( i , j ) (i,j) (i,j) and lose the corresponding price
The final answer is tot minimum cut
#include<bits/stdc++.h> #define ll long long using namespace std; const int N=5050; const int M=1e9; ll read(){ ll x=0,f=1;char c=getchar(); while(!isdigit(c)); while(isdigit(c)) return x*f; } int n,m,s,t; struct node{ int to,nxt;ll cap; }p[3000500]; int fi[N],cnt; void addline(int x,int y,ll cap){ p[++cnt]=(node);fi[x]=cnt; p[++cnt]=(node);fi[y]=cnt; } int col[N],cur[N]; queue<int>q; int bfs(){ memset(col,0,sizeof(col));col[s]=1; q.push(s); while(!q.empty()){ int now=q.front();q.pop(); for(int i=cur[now]=fi[now];~i;i=p[i].nxt){ int to=p[i].to; if(col[to]||!p[i].cap) continue; col[to]=col[now]+1; q.push(to); } } return col[t]; } ll dfs(int x,ll lim){ if(x==t||!lim) return lim; ll res=0; for(int &i=cur[x];~i;i=p[i].nxt){ int to=p[i].to; if(!p[i].cap||col[to]!=col[x]+1) continue; ll add=dfs(to,min(lim,p[i].cap)); lim-=add;res+=add; p[i].cap-=add;p[i^1].cap+=add; if(!lim) break; } if(lim) col[x]=-1; return res; } ll dinic(){ ll res=0; while(bfs()){ while(ll tmp=dfs(s,2e15)) res+=tmp; } return res; } int vis[N]; void find1(int x){ if(vis[x]) return; vis[x]=1; for(int i=fi[x];~i;i=p[i].nxt){ if(!p[i].cap) continue; find1(p[i].to); } return; } void find2(int x){ if(vis[x]) return; vis[x]=2; for(int i=fi[x];~i;i=p[i].nxt){ if(!p[i^1].cap) continue; find2(p[i].to); } return; } #define id(x,y) ((x-1)*m+y) int num,x,tot; int dx[5]=,dy[5]=; inline bool exi(int x,int y){ return x>=1&&x<=n&&y>=1&&y<=m; } int a[55][55],mx[2505]; struct tower{ int x,y,op; int mx,xx,yy; }o[2505]; #define heng(x,y) ((x-1)*m+y) #define shu(x,y) ((x-1)*m+y+n*m) int main(){ memset(fi,-1,sizeof(fi));cnt=-1; n=read();m=read(); for(int i=1;i<=n;i++){ for(int j=1;j<=m;j++){ x=read(); if(x>=0) a[i][j]=x; else{ x=-x; o[++num]=(tower); } } }//printf("ok"); for(int i=1;i<=num;i++){ int x=o[i].x,y=o[i].y,op=o[i].op; while(x>=1&&x<=n&&y>=1&&y<=m){ x+=dx[op];y+=dy[op]; if(a[x][y]>o[i].mx){ o[i].mx=a[x][y]; o[i].xx=x;o[i].yy=y; } } tot+=o[i].mx; } for(int i=1;i<=n;i++){ for(int j=1;j<=m;j++) addline(heng(i,j),shu(i,j),2e15); } s=2*n*m+1;t=s+1; for(int i=1;i<=num;i++){ int x=o[i].x,y=o[i].y,op=o[i].op; int xx=o[i].xx,yy=o[i].yy,mx=o[i].mx; //a[x][y]=2e9; if(op<=2){ addline(shu(x,y),t,2e15); while(xx!=x||yy!=y){ // printf("(%d %d)->(%d %d) len=%d\n",xx,yy,xx-dx[op],yy-dy[op],a[xx-dx[op]][yy-dy[op]]); addline(shu(xx,yy),shu(xx-dx[op],yy-dy[op]),mx-a[xx-dx[op]][yy-dy[op]]); xx-=dx[op];yy-=dy[op]; } } else{ addline(s,heng(x,y),2e15); while(x!=xx||y!=yy){ // printf("(%d %d)->(%d %d) len=%d\n",x,y,x+dx[op],y+dy[op],a[x][y]); addline(heng(x,y),heng(x+dx[op],y+dy[op]),mx-a[x][y]); x+=dx[op];y+=dy[op]; } } } printf("%d\n",tot-dinic()); return 0; } /* 3 2 1 3 1 2 10 2 3 10 */