NOIP2018 道路铺设

analysys

DP

设$f[i]$为填满1~i需要的天数

则
$$\begin{gathered} 当d[i]<d[i-1]时,f[i]=f[i-1] \\ 当d[i]>d[i-1]时,f[i]=f[i-1]+d[i]-d[i-1] \end{gathered}$$

code

#include
using namespace std;
#define ll long long
#define loop(i,start,end) for(register int i=start;i<=end;++i)
template<typename T>void read(T &x){
    x=0;char r=getchar();T neg=1;
    while(r>'9'||r<'0'){if(r=='-')neg=-1;r=getchar();}
    while(r>='0'&&r<='9'){x=(x<<1)+(x<<3)+r-'0';r=getchar();}
    x*=neg;
}
const int maxn=100000+10;
int d[maxn],n;
ll dp[maxn];
int main(){
    read(n);
    loop(i,1,n)read(d[i]);
    loop(i,1,n){
        if(d[i]<=d[i-1])dp[i]=dp[i-1];
        else if(d[i]>d[i-1])dp[i]=dp[i-1]+d[i]-d[i-1];
    }printf("%lld\n",dp[n]);
    return 0;
}