标签:链接 搜索 练手 memset name print php else ref
题目链接 http://acm.hdu.edu.cn/showproblem.php?pid=1176
题意不解释了
简单的记忆化搜索可以拿来练练手,注意要从pos = 5 开始搜索
#include <iostream> #include <cstring> #include <algorithm> #include <cstdio> #include <cmath> using namespace std; const int M = 1e5 + 10; int dp[15][M] , t2[M][12]; int dfs(int pos , int count) { if(count == 0) { if(pos == 6) { return 0; } else { return -1; } } if(dp[pos][count] != -1) return dp[pos][count]; int sum = 0; if(pos == 1) { sum += max(dfs(pos + 1 , count - 1) , dfs(pos , count - 1)); } else if(pos == 11) { sum += max(dfs(pos - 1 , count - 1) , dfs(pos , count - 1)); } else { sum += max(max(dfs(pos - 1 , count - 1) , dfs(pos + 1 , count - 1)) , dfs(pos , count - 1)); } if(sum != -1) { sum += t2[count][pos]; dp[pos][count] = sum; } return sum; } int main() { int n; while(scanf("%d" , &n) != EOF) { if(n == 0) break; memset(t2 , 0 , sizeof(t2)); int T = 0; for(int i = 0 ; i < n ; i++) { int x , y; scanf("%d%d" , &x , &y); t2[y][x + 1]++; T = max(T , y); } memset(dp , -1 , sizeof(dp)); int MAX = 0; for(int i = 1 ; i <= 11 ; i++) { MAX = max(MAX , dfs(i , T)); } printf("%d\n" , MAX); } return 0; }
标签:链接 搜索 练手 memset name print php else ref
原文地址:http://www.cnblogs.com/TnT2333333/p/6106735.html