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  • BZOJ 1911 特别行动队(斜率优化DP)

    应该可以看出这是个很normal的斜率优化式子。推出公式搞一搞即可。

    # include <cstdio>
    # include <cstring>
    # include <cstdlib>
    # include <iostream>
    # include <vector>
    # include <queue>
    # include <stack>
    # include <map>
    # include <set>
    # include <cmath>
    # include <algorithm>
    using namespace std;
    # define lowbit(x) ((x)&(-x))
    # define pi 3.1415926535
    # define eps 1e-9
    # define MOD 10000
    # define INF 1000000000
    # define mem(a,b) memset(a,b,sizeof(a))
    # define FOR(i,a,n) for(int i=a; i<=n; ++i)
    # define FO(i,a,n) for(int i=a; i<n; ++i)
    # define bug puts("H");
    # define lch p<<1,l,mid
    # define rch p<<1|1,mid+1,r
    # define mp make_pair
    # define pb push_back
    typedef pair<int,int> PII;
    typedef vector<int> VI;
    # pragma comment(linker, "/STACK:1024000000,1024000000")
    typedef long long LL;
    int Scan() {
        int res=0, flag=0;
        char ch;
        if((ch=getchar())=='-') flag=1;
        else if(ch>='0'&&ch<='9') res=ch-'0';
        while((ch=getchar())>='0'&&ch<='9')  res=res*10+(ch-'0');
        return flag?-res:res;
    }
    void Out(int a) {
        if(a<0) {putchar('-'); a=-a;}
        if(a>=10) Out(a/10);
        putchar(a%10+'0');
    }
    const int N=1000005;
    //Code begin...
    
    int a[N], que[N], head, tail, A, B, C;
    LL dp[N], sum[N];
    
    bool check(int x, int y, int z){
        return dp[x]-dp[y]+A*(sum[x]*sum[x]-sum[y]*sum[y])>=(2*A*sum[z]+B)*(sum[x]-sum[y]);
    }
    bool sol(int x, int y, int z){
        return (dp[x]-dp[y]+A*(sum[x]*sum[x]-sum[y]*sum[y]))*(sum[y]-sum[z])>=(dp[y]-dp[z]+A*(sum[y]*sum[y]-sum[z]*sum[z]))*(sum[x]-sum[y]);
    }
    int main ()
    {
        int n;
        scanf("%d%d%d%d",&n,&A,&B,&C);
        FOR(i,1,n) scanf("%d",a+i), sum[i]=sum[i-1]+a[i];
        head=-1; tail=0; que[++head]=0;
        FOR(i,1,n) {
            while (head>tail&&check(que[tail+1],que[tail],i)) ++tail;
            int v=que[tail];
            dp[i]=dp[v]+A*(sum[i]-sum[v])*(sum[i]-sum[v])+B*(sum[i]-sum[v])+C;
            while (head>tail&&sol(i,que[head],que[head-1])) --head;
            que[++head]=i;
        }
        printf("%lld
    ",dp[n]);
        return 0;
    }
    View Code
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  • 原文地址:https://www.cnblogs.com/lishiyao/p/6662178.html
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