「TJOI 2013」攻击装置

题目链接

戳我

(solution)

这道题和网络24题之骑士共存问题很相似
只是输入方式不一样而已
详细见:这儿

(Code)

#include<bits/stdc++.h>
#define file(a) freopen(a".in","r",stdin);freopen(a".out","w",stdout)
#define rg register
using namespace std;
typedef long long ll;
int n,m,s,t,z,y,x;
inline int read(){
    int x=0,f=1;char c=getchar();
    while(c<'0'||c>'9')  f= (c=='-')?-1:1,c=getchar();
    while(c>='0'&&c<='9')  x=x*10+c-48,c=getchar();
    return f*x;
}
struct node{
    int to,next,v;
}a[400001];
int cnt,head[200001],cur[200001];
void add(int x,int y,int c){
    a[++cnt].to=y;
    a[cnt].next=head[x];
    a[cnt].v=c;
    head[x]=cnt;
}
queue <int> q;
int dep[100001];
int bfs(){
    memset(dep,0,sizeof(dep));
    q.push(s),dep[s]=1;
    while(!q.empty()){
        int now=q.front();
        q.pop();
        for(int i=head[now];i;i=a[i].next){
            int v=a[i].to;
            if(!dep[v]&&a[i].v)
                q.push(v),dep[v]=dep[now]+1;
        }
    }
    if(dep[t]!=0)
        return 1;
    else
        return 0;
}
int dfs(int k,int list){
    if(k==t)
        return list;
    for(int &i=cur[k];i;i=a[i].next){
        int v=a[i].to;
        if(a[i].v&&dep[v]==dep[k]+1){
            int p=dfs(v,min(list,a[i].v));
            if(p){
                a[i].v-=p;
                if(i%2)
                    a[i+1].v+=p;
                else
                    a[i-1].v+=p;
                return p;
            }
        }
    }
    return 0;
}
int Dinic(){
    int ans=0;
    while(bfs()){
        for(int i=0;i<t;i++)
            cur[i]=head[i];
        int k=dfs(s,2147483);
        while(k)
            ans+=k,k=dfs(s,2147483);
    }
    return ans;
}
int f[201][201];
int fx[10]={0,1,1,-1,-1,2,2,-2,-2};
int fy[10]={0,2,-2,2,-2,1,-1,1,-1};
char S[1000];
int main(){
    n=read(),s=0,t=n*n+1;
    int sum=0;
    for(int i=1;i<=n;i++){
        cin>>S;
        for(int j=0;j<strlen(S);j++){
            f[i][j+1]=S[j]-'0';
            if(f[i][j+1])
                m++;
        }
    }
    for(int i=1;i<=n;i++)
        for(int j=1;j<=n;j++){
            if(f[i][j])
                continue;
            if(((i+j)%2)==0)
                add((i-1)*n+j,t,1),add(t,(i-1)*n+j,0);
            else
                add(s,(i-1)*n+j,1),add((i-1)*n+j,s,0);
        }
    for(int i=1;i<=n;i++)
        for(int j=1;j<=n;j++)
            for(int k=1;k<=8;k++){
                int x=i+fx[k],y=j+fy[k],p=(x-1)*n+y;
                if(x>0&&y>0&&x<=n&&y<=n&&((i+j)%2)==1&&!f[i][j]&&!f[x][y])
                    add((i-1)*n+j,p,10000000),add(p,(i-1)*n+j,0);
            }
    int ans=Dinic();
    printf("%d",n*n-m-ans);
}