Uva 11069 - A Graph Problem 递推
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题意:有n个数,问满足下列两个条件时可以分成几个子集
1.子集中两个元素的差s要2<=s<=3
2.子集尽可能的长,如1 3 5,不能1 5,因为3还可以插到1 5中间
思路:用dp[i]表示用前i个数能形成的子集,对于dp[i]来说,决定它的是dp[i-2]或者dp[i-3],有第i个元素的子集必定有第i-2个或者第i-3个元素的子集
代码:
#include <iostream>#include <stdio.h>#include <cmath>#include <cstring>#include <map>#include <vector>#include <queue>#include <algorithm>const int M=100;using namespace std;long long dp[M];void init(){ dp[0]=dp[1]=1; dp[2]=2; for(int i=3;i<=M;i++) dp[i]=dp[i-2]+dp[i-3];}int main(){ int n; init(); while(scanf("%d",&n)==1) { printf("%lld\n",dp[n]); } return 0;}
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