34 lines
1.1 KiB
C++
34 lines
1.1 KiB
C++
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#include <bits/stdc++.h>
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using namespace std;
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const int INF = 0x3f3f3f3f;
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const int N = 200010;
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int f[N];
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int w[N], n, m;
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// 通过了 11/12个数据
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int main() {
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/**
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普通DP大法
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状态表示：
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①集合:f[i]表示前i个灯塔满足条件，并且i点亮。
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②属性：最小代价
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状态计算：f[i]=min(f[j]+w[i]) (i−m≤j≤i−1)
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答案：(n−m+1)~n必须有灯塔亮，所以枚举一下求出最小值即可
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*/
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cin >> n >> m;
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for (int i = 1; i <= n; i++) cin >> w[i];
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//初始化
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memset(f, 0x3f, sizeof f); //预求最小，先设最大
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f[0] = 0; //需要手动初始化一下递推的base case
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//依题意，f[1]代表第1个灯塔点亮需要付出的代价是w[1],也就是f[0]=0,想想也正常，虚拟第0个点亮成本为0~
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for (int i = 1; i <= n; i++)
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for (int j = max(0, i - m); j <= i - 1; j++) //向前查看它之前的m个
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f[i] = min(f[i], f[j] + w[i]);
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int res = INF;
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for (int i = n + 1 - m; i <= n; i++) res = min(res, f[i]);
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printf("%d\n", res);
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return 0;
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}
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