[USACO08JAN]Running

嘟嘟嘟

这很显然是一道dp题。

令dp[i][j]表示第 i 分钟末，疲劳度为 j 是的最大跑步距离，则

　　　dp[i][0] = max(dp[i - 1][0], max(dp[i - j][j]))

　　　dp[i][j] = max(dp[i - 1][j - 1] + a[i])

因为题中说即使疲劳值为0了，仍可以休息，所以dp[i][0]也可以从dp[i - 1][0]转换过来。

 #include<cstdio>
 #include<iostream>
 #include<cmath>
 #include<algorithm>
 #include<cstring>
 #include<cstdlib>
 #include<cctype>
 #include<vector>
 #include<stack>
 #include<queue>
 using namespace std;
 #define enter puts("")
 #define space putchar(' ')
 #define Mem(a) memset(a, 0, sizeof(a))
 typedef long long ll;
 typedef double db;
 const int INF = 0x3f3f3f3f;
 const int eps = 1e-;
 const int maxn = 1e4 + ;
 inline ll read()
 {
     ll ans = ;
     char ch = getchar(), last = ' ';
     while(!isdigit(ch)) {last = ch; ch = getchar();}
     while(isdigit(ch)) {ans = ans *  + ch - ''; ch = getchar();}
     if(last == '-') ans = -ans;
     return ans;
 }
 inline void write(ll x)
 {
     if(x < ) x = -x, putchar('-');
     if(x >= ) write(x / );
     putchar(x %  + '');
 }

 int n, m, a[maxn];
 int dp[maxn][maxn];

 int main()
 {
     n = read(); m = read();
     for(int i = ; i <= n; ++i) a[i] = read();
     for(int i = ; i <= n; ++i)
     {
         dp[i][] = dp[i - ][];
         for(int j = ; j <= min(i, m); ++j) dp[i][] = max(dp[i][], dp[i - j][j]);
         for(int j = ; j <= min(i, m); ++j) dp[i][j] = max(dp[i][j], dp[i - ][j - ] + a[i]);
     }
     write(dp[n][]); enter;
     return ;
 }