题意:中文题
思路:简单递推
AC代码:
#include "iostream" #include "iomanip" #include "string.h" #include "stack" #include "queue" #include "string" #include "vector" #include "set" #include "map" #include "algorithm" #include "stdio.h" #include "math.h" #pragma comment(linker, "/STACK:102400000,102400000") #define bug(x) cout<<x<<" "<<"UUUUU"<<endl; #define mem(a,x) memset(a,x,sizeof(a)) #define step(x) fixed<< setprecision(x)<< #define mp(x,y) make_pair(x,y) #define pb(x) push_back(x) #define ll long long #define endl (" ") #define ft first #define sd second #define lrt (rt<<1) #define rrt (rt<<1|1) using namespace std; const long long INF = 1e18+1LL; const int inf = 1e9+1e8; const int N=1e5+100; const ll mod=1e9+7; int n,dp[3][11]; int M[N][11]; int re(int t, int x){ return M[t][x]; } int main(){ ios::sync_with_stdio(false),cin.tie(0),cout.tie(0); while(cin>>n&&n){ int tm=0; mem(dp,0), mem(M,0); for(int i=1; i<=n; ++i){ int t,x; cin>>x>>t; M[t][x]++; tm=max(tm,t); } dp[1][4]=re(1,4),dp[1][5]=re(1,5),dp[1][6]=re(1,6); int cur=1; for(int i=1; i<tm; ++i,cur^=1){ for(int j=0; j<=10; ++j){ dp[cur^1][j]=max(dp[cur^1][j],dp[cur][j]+re(i+1,j)); if(j-1>=0) dp[cur^1][j-1]=max(dp[cur^1][j-1],dp[cur][j]+re(i+1,j-1)); if(j+1<=10) dp[cur^1][j+1]=max(dp[cur^1][j+1],dp[cur][j]+re(i+1,j+1)); } } int ans=0; for(int i=0; i<=10; ++i) ans=max(ans,dp[cur][i]); cout<<ans<<endl; } return 0; }