CF 987C Three displays DP或暴力 第十一题

Three displays

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

It is the middle of 2018 and Maria Stepanovna, who lives outside Krasnokamensk (a town in Zabaikalsky region), wants to rent three displays to highlight an important problem.

There are $\mathrm{nn displays\; placed\; along\; a\; road,\; and\; the ii-th\; of\; them\; can\; display\; a\; text\; with\; font\; size sisi only.\; Maria\; Stepanovna\; wants\; to\; rent\; such\; three\; displays\; with\; indices i<j<ki<j<k that\; the\; font\; size\; increases\; if\; you\; move\; along\; the\; road\; in\; a\; particular\; direction.\; Namely,\; the\; condition si<sj<sksi<sj<sk should\; be\; held.}$

The rent cost is for the $\mathrm{ii-th\; display\; is cici.\; Please\; determine\; the\; smallest\; cost\; Maria\; Stepanovna\; should\; pay.}$

Input

The first line contains a single integer $\mathrm{nn (3\le n\le 30003\le n\le 3000) —\; the\; number\; of\; displays.}$

The second line contains $\mathrm{nn integers s1,s2,\dots ,sns1,s2,\dots ,sn (1\le si\le 1091\le si\le 109) —\; the\; font\; sizes\; on\; the\; displays\; in\; the\; order\; they\; stand\; along\; the\; road.}$

The third line contains $\mathrm{nn integers c1,c2,\dots ,cnc1,c2,\dots ,cn (1\le ci\le 1081\le ci\le 108) —\; the\; rent\; costs\; for\; each\; display.}$

Output

If there are no three displays that satisfy the criteria, print -1. Otherwise print a single integer — the minimum total rent cost of three displays with indices $\mathrm{i<j<ki<j<k such\; that si<sj<sksi<sj<sk.}$

Examples

input

Copy

5
2 4 5 4 10
40 30 20 10 40

output

Copy

90

input

Copy

3
100 101 100
2 4 5

output

Copy

-1

input

Copy

10
1 2 3 4 5 6 7 8 9 10
10 13 11 14 15 12 13 13 18 13

output

Copy

33

Note

In the first example you can, for example, choose displays $\mathrm{11, 44 and 55,\; because s1<s4<s5s1<s4<s5 (2<4<102<4<10),\; and\; the\; rent\; cost\; is 40+10+40=9040+10+40=90.}$

In the second example you can't select a valid triple of indices, so the answer is -1.

 

题意： 给你n个数，每个数有两个权值（a，b），问取三个数，要求这三个数的a值递增，满足要求的最小三个数的和，没有满足要求的条件输出NO

暴力解法：

遍历中间一个数，然后两个循环分别找比他大的和比他小的，然后记录最小值

dp解法：

dp[i][j]，i为第几个选择的数，j是选择的数的位置

这样的话，状态转移方程是：

dp[1][i] 每个位置的数

dp[2][j]这个位置与前面任意位置组合成的递增的两个的数

dp[3][i]这个位置与前面两个位置的组合成递增的三个的数

dp[2][j]=min(dp[2][j],dp[1][i]+dp[1][j]);

dp[3][j]=min(dp[3][j],dp[2][i]+dp[1][j]);

 

暴力代码：

#include <map>
#include <set>
#include <cmath>
#include <queue>
#include <cstdio>
#include <vector>
#include <string>
#include <cstring>
#include <iostream>
#include <algorithm>
#define debug(a) cout << #a << " " << a << endl
using namespace std;
const int maxn = 1e6 + 10;
const int mod = 1e9 + 7;
typedef long long ll;
ll a[maxn], b[maxn];
int main(){
    std::ios::sync_with_stdio(false);
    ll n;
    while( cin >> n ) {
        for( ll i = 0; i < n; i ++ ) {
            cin >> a[i];
        }
        for( ll i = 0; i < n; i ++ ) {
            cin >> b[i];
        }
        ll ans = 1e12;
        bool flag = false;
        for( ll i = 1; i < n; i ++ ) {
            ll min1 = 1e12, min2 = 1e12;
            for( ll j = i+1; j < n; j ++ ) {
                if( a[i] < a[j] ) {
                    min1 = min( min1, b[j] );
                }
            }
            for( ll j = 0; j < i; j ++ ) {
                if( a[i] > a[j] ) {
                    min2 = min( min2, b[j] );
                }
            }
            if( min1 != 1e12 && min2 != 1e12 ) {
                flag = true;
                ans = min( ans, min1 + min2 + b[i] );
            }
        }
        if( flag ) {
            cout << ans << endl;
        } else {
            cout << -1 << endl;
        }
    }
    return 0;
}

 

dp代码：

#include <map>
#include <set>
#include <cmath>
#include <queue>
#include <cstdio>
#include <vector>
#include <string>
#include <cstring>
#include <iostream>
#include <algorithm>
#define debug(a) cout << #a << " " << a << endl
using namespace std;
const int maxn = 3*1e3 + 10;
const int mod = 1e9 + 7;
typedef long long ll;
ll a[maxn], b[maxn], dp[maxn][maxn];
int main(){
    std::ios::sync_with_stdio(false);
    ll n;
    while( cin >> n ) {
        for( ll i = 0; i < n; i ++ ) {
            cin >> a[i];
            dp[2][i] = 1e12, dp[3][i] = 1e12;
        }
        for( ll i = 0; i < n; i ++ ) {
            cin >> b[i];
        }
        ll ans = 1e12;
        for( ll j = 1; j < n; j ++ ) {
            for( ll i = 0; i < j; i ++ ) {
                if( a[i] < a[j] ) {
                    dp[2][j] = min( dp[2][j] , b[i] + b[j] );
                }
            }
        }
        for( ll j = 2; j < n; j ++ ) {
            for( ll i = 0; i < j; i ++ ) {
                if( a[i] < a[j] ) {
                    dp[3][j] = min( dp[3][j], dp[2][i] + b[j] );
                }
            }
            ans = min( ans, dp[3][j] );
        }
        if( ans != 1e12 ) {
            cout << ans << endl;
        } else {
            cout << -1 << endl;
        }
    }
    return 0;
}

 

转载于:https://www.cnblogs.com/l609929321/p/9221563.html