3

note：满足的情况只有[1,2],[2,1],[3,3]

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<vector>
#include<string>
#include<set>
#include<cmath>
#include<list>
#include<deque>
#include<cstdlib>
#include<bitset>
#include<stack>
#include<map>
#include<queue>
using namespace std;
typedef long long ll;
#define lson l,mid,rt<<1
#define rson mid+1,r,rt<<1|1
#define pushup() tree[rt]=tree[rt<<1]+tree[rt<<1|1]
const int INF=0x3f3f3f3f;
const double PI=acos(-1.0);
const double eps=1e-6;
const ll mod=1e9+7;
const int maxn=100005;
ll gcd(ll a,ll b){return b?gcd(b,a%b):a;}
ll lcm(ll a,ll b){return a/gcd(a,b)*b;}
ll dp[2000005][5];
ll n,l,r;
int main()
{ios::sync_with_stdio(false); cin.tie(0);cin>>n>>l>>r;ll num0=r/3-(l-1)/3;ll num1=num0;ll num2=num0;int i,j;for(i=l;i<=r;i++){if(i%3==0)break;else if(i%3==1)num1++;else num2++;}for(j=r;j>=i;j--){if(j%3==2)break;else if(j%3==1){num2--; break;}else if(j%3==0){num1--; num2--;break;}}dp[1][0]=num0; dp[1][1]=num1; dp[1][2]=num2;for(int i=2;i<=n;i++){dp[i][0]=((dp[i-1][0]*num0%mod+dp[i-1][1]*num2%mod)%mod+dp[i-1][2]*num1%mod)%mod;dp[i][1]=((dp[i-1][0]*num1%mod+dp[i-1][1]*num0%mod)%mod+dp[i-1][2]*num2%mod)%mod;dp[i][2]=((dp[i-1][0]*num2%mod+dp[i-1][1]*num1%mod)%mod+dp[i-1][2]*num0%mod)%mod;}cout<<dp[n][0]<<endl;return 0;
}

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<vector>
#include<string>
#include<set>
#include<cmath>
#include<list>
#include<deque>
#include<cstdlib>
#include<bitset>
#include<stack>
#include<map>
#include<queue>
using namespace std;
typedef long long ll;
#define lson l,mid,rt<<1
#define rson mid+1,r,rt<<1|1
#define pushup() tree[rt]=tree[rt<<1]+tree[rt<<1|1]
const int INF=0x3f3f3f3f;
const double PI=acos(-1.0);
const double eps=1e-6;
const ll mod=1e9+7;
const int maxn=100005;
ll gcd(ll a,ll b){return b?gcd(b,a%b):a;}
ll lcm(ll a,ll b){return a/gcd(a,b)*b;}
const int dir[4][2]={{1,0},{-1,0},{0,1},{0,-1}};
struct node{int x,y,id;node(int a,int b,int c):x(a),y(b),id(c){}
};
struct node1{int x,y,id,steps;node1(int a,int b,int c,int d):x(a),y(b),id(c),steps(d){}
};
int n,m,sump,s[10];
vector<node> p[10];
char mp[1050][1050];
queue<node> que;
queue<node1> que1;
void BFS()
{while(que.size()){node tmp=que.front();int id=tmp.id;que1.push(node1(tmp.x,tmp.y,tmp.id,0));while(que.size()&&que.front().id==id)  //判断第一个队列元素是否为当前压入队列是同一个玩家
        {que1.push(node1(que.front().x,que.front().y,que.front().id,0));que.pop();}while(que1.size()){node1 now=que1.front();que1.pop();if(now.steps==s[now.id]){que.push(node(now.x,now.y,now.id)); //走到最后一步继续压入第一个队列continue;}for(int i=0;i<4;i++){int dx=now.x+dir[i][0];int dy=now.y+dir[i][1];if(dx>=0&&dx<n&&dy>=0&&dy<m&&mp[dx][dy]=='.'){mp[dx][dy]='0'+now.id;que1.push(node1(dx,dy,now.id,now.steps+1));}}}}
}
int main()
{ios::sync_with_stdio(false); cin.tie(0);cin>>n>>m>>sump;for(int i=1;i<=sump;i++) cin>>s[i];for(int i=0;i<n;i++){for(int j=0;j<m;j++){cin>>mp[i][j];if(mp[i][j]>='0'&&mp[i][j]<='9')p[mp[i][j]-'0'].push_back(node(i,j,mp[i][j]-'0'));  //同一个玩家的城堡压入同一个向量里
        }}for(int i=1;i<=sump;i++)for(int j=0;j<p[i].size();j++)que.push(p[i][j]);BFS();int ans[10];memset(ans,0,sizeof(ans));for(int i=0;i<n;i++)for(int j=0;j<m;j++)for(int k=1;k<=sump;k++)if(mp[i][j]==(k+'0'))ans[k]++;cout<<ans[1];for(int i=2;i<=sump;i++)cout<<" "<<ans[i];cout<<endl;return 0;
}