【APIO2010】【斜率优化DP】特别行动队

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看到这道题会很容易想到是动态规划。

然后朴素的方程也很容易写出:

用f[i]表示将前i个士兵分组得到的最大战斗力，sum[i]表示前i个士兵战斗力总和

f[i] = max{f[i], f[j]+ a * (sum[i] - sum[j]) ^ 2 + b * (sum[i] - sum[j]) + c}

但是这样做是O(n^2)的，对于n的范围来数肯定超时。

于是我们考虑f[i]的决策j,k < i且j < k

当决策j比决策k更优时:

f[j] + a * (sum[i] - sum[j]) ^2 + b * (sum[i] - sum[j]) + c > f[k] + a * (sum[i] - sum[j]) ^ 2 + b * (sum[i] - sum[j]) + c

-> f[i] + a *sum[i] ^ 2 - 2 * a * sum[i] * sum[j] +a * sum[j] ^ 2 + b * (sum[i] - sum[j])

> f[k] + a *sum[k] ^ 2 - 2 * a * sum[i] * sum[k] + a * sum[k] ^ 2 + b * (sum[i] - sum[k])

-> f[j] - f[k] + a * (sum[j] ^ 2 - sum[k] ^ 2) - b * (sum[j] - sum[k]) > 2 * a * sum[i] * (sum[j] - sum[k])

->[(f[j] + a * sum[j] ^ 2 - b *sum[j]) - (f[k] + a * sum[k] ^ 2- b * sum[j])] / (sum[j] - sum[k]) > 2 * a * sum[i]

令x = sum[],y = f[] + a * sum[] ^ 2 - b * sum[]

经过转化后，我们可以发现上式是单调的，于是我们可以维护一个上凸的凸包，每次用平摊O(1)的时间决策，这样总的时间复杂度就优化至O(n)了

代码:

#include<cstdio>#include<cstring>using namespace std;const int maxn = 1000000 + 10;int n,A,B,C;long long sum[maxn],f[maxn];int deq[maxn];void init(){freopen("commando.in","r",stdin);freopen("commando.out","w",stdout);}void readdata(){scanf("%d",&n);scanf("%d%d%d",&A,&B,&C);for(int i = 1;i <= n;i++){scanf("%lld",&sum[i]);sum[i] += sum[i-1];}}double slope(int j,int k){long long ret = (f[j] + A * (sum[j] * sum[j]) - B * sum[j]) - (f[k] + A * (sum[k] * sum[k]) - B * sum[k]);return (double)ret / (double)(sum[j] - sum[k]);}void solve(){memset(deq,0,sizeof(deq));int s = 0,e = 0;for(int i = 1;i <= n;i++){while(s < e && slope(deq[s+1],deq[s]) >= (double)(2 * A * sum[i]))s++;long long tmp = sum[i] - sum[deq[s]];f[i] = f[deq[s]] + (long long)A * tmp * tmp + (long long)B * tmp + C;    while(s < e && slope(deq[e-1],deq[e]) < slope(deq[e],i))e--;deq[++e] = i;}printf("%lld\n",f[n]);}int main(){init();readdata();solve();return 0;}