题意:需要你从从原点(0,0)出发走到平面第一象限中的终点(n,n),你只能向右或向上行走整数步,且过程中最多变换n-1次方向。现在告诉你n段路程分别的平均成本,试找到走到终点的最小消耗。
思路:
既然最多变换n-1次方向,即我们可以利用完n段线段的成本。
于是我们枚举最后一段时在第i次结束(走到终点),注意:这以为这之前 i-1 段路程都是交替变换方向的。
对于奇数位置,走到总距离和必须刚好等于n,于是我们找到该奇数位置 i 以及之前的所有奇数位置的最小值,让其走最多的步数,其余的位置都只走一步。偶数位置同理。不断更新答案取得最小值即可。
代码实现:
#include<bits/stdc++.h>
#define endl '\n'
#define null NULL
#define ll long long
#define int long long
#define pii pair<int, int>
#define lowbit(x) (x &(-x))
#define ls(x) x<<1
#define rs(x) (x<<1+1)
#define me(ar) memset(ar, 0, sizeof ar)
#define mem(ar,num) memset(ar, num, sizeof ar)
#define rp(i, n) for(int i = 0, i < n; i ++)
#define rep(i, a, n) for(int i = a; i <= n; i ++)
#define pre(i, n, a) for(int i = n; i >= a; i --)
#define IOS ios::sync_with_stdio(0); cin.tie(0);cout.tie(0);
const int way[4][2] = {
{1, 0}, {-1, 0}, {0, 1}, {0, -1}};
using namespace std;
const int inf = 0x3f3f3f3f3f3f3f3f;
const double PI = acos(-1.0);
const double eps = 1e-6;
const ll mod = 1e9+7;
const int N = 2e5 + 5;
inline void read(int &x){
char t=getchar();
while(!isdigit(t)) t=getchar();
for(x=t^48,t=getchar();isdigit(t);t=getchar()) x=x*10+(t^48);
}
int t, n, c[N];
int ma[N], mb[N], pra[N], prb[N];
signed main()
{
IOS;
cin >> t;
while(t --){
cin >> n;
me(ma); me(mb);
me(pra); me(prb);
for(int i = 1; i <= n; i ++){
cin >> c[i];
if(i&1){
if(i==1) ma[i] = pra[i] = c[i];
else ma[i] = min(ma[i-2], c[i]), pra[i] = pra[i-2]+c[i];
}
else{
if(i==2) mb[i] = prb[i] = c[i];
else mb[i] = min(mb[i-2], c[i]), prb[i] = prb[i-2]+c[i];
}
}
int ans = inf;
for(int i = 2; i <= n; i ++){
int tmp = 0, k = i/2;
if(i&1){
tmp += prb[i-1]+(n-k)*mb[i-1];
tmp += pra[i]+(n-k-1)*ma[i];
}
else{
tmp += prb[i]+(n-k)*mb[i];
tmp += pra[i-1]+(n-k)*ma[i-1];
}
ans = min(ans, tmp);
// cout << "i: " << i << " tmp: " << tmp << endl;
}
cout << ans << endl;
}
return 0;
}
Java练习:期间忘记tmp定义成long型 wa4,再又就是所有数组定义成long型的话如果都是1e5+10大小的话就会 t3,呜呜呜绝了
import java.util.Scanner;
import java.util.*;
public class Main {
public static void main(String[] args) {
Scanner in = new Scanner(System.in);
int t = in.nextInt();
while(t>0) {
t --;
int n = in.nextInt();
long [] c = new long[n+10];
long [] ma = new long[n+10];
long [] mb = new long[n+10];
long [] pra = new long[n+10];
long [] prb = new long[n+10];
for(int i = 1; i <= n; i ++){
c[i] = in.nextLong();
if(i%2!=0){
if(i==1) ma[i] = pra[i] = c[i];
else {
ma[i] = Math.min(ma[i-2], c[i]);
pra[i] = pra[i-2]+c[i];
}
}
else{
if(i==2) mb[i] = prb[i] = c[i];
else{
mb[i] = Math.min(mb[i-2], c[i]);
prb[i] = prb[i-2]+c[i];
}
}
}
long ans = 1L<<60;
for(int i = 2; i <= n; i ++){
long tmp = 0, k = i/2;
if(i%2!=0){
tmp += prb[i-1]+(n-k)*mb[i-1];
tmp += pra[i]+(n-k-1)*ma[i];
}
else{
tmp += prb[i]+(n-k)*mb[i];
tmp += pra[i-1]+(n-k)*ma[i-1];
}
ans = Math.min(ans, tmp);
}
System.out.println(ans);
}
in.close();
}
}