【51nod2484 小b和排序】dp

最新推荐文章于 2019-11-24 23:57:32 发布

ljq真的不能再吃了

最新推荐文章于 2019-11-24 23:57:32 发布

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分类专栏： ACM DP

本文链接：https://blog.youkuaiyun.com/heucodesong/article/details/89764460

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该博客探讨了51nod中的第2484道题目，涉及如何通过最小数量的交换操作，使得两个数组a和b都变得严格递增。博主分享了代码实现过程，特别强调了dp状态的设计，并提到了在解决问题时的一个关键错误——不能在某个if语句前加else子句。

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51nod2484
题意给你两个数组 a 和 b
问你交换最少多少次可以使得 a b 都严格递增
保证有解
想一下这个代码里的dp状态吧
一开始第二个if前面加个else 死活过不去其实不能加

/*
    if you can't see the repay
    Why not just work step by step
    rubbish is relaxed
    to ljq
*/
#include <cstdio>
#include <cstring>
#include <iostream>
#include <queue>
#include <cmath>
#include <map>
#include <stack>
#include <set>
#include <sstream>
#include <vector>
#include <stdlib.h>
#include <algorithm>
using namespace std;

#define dbg(x) cout<<#x<<" = "<< (x)<< endl
#define dbg2(x1,x2) cout<<#x1<<" = "<<x1<<" "<<#x2<<" = "<<x2<<endl
#define dbg3(x1,x2,x3) cout<<#x1<<" = "<<x1<<" "<<#x2<<" = "<<x2<<" "<<#x3<<" = "<<x3<<endl
#define max3(a,b,c) max(a,max(b,c))
#define min3(a,b,c) min(a,min(b,c))

typedef pair<int,int> pll;
typedef long long ll;
const int inf = 0x3f3f3f3f;
const int _inf = 0xc0c0c0c0;
const ll INF = 0x3f3f3f3f3f3f3f3f;
const ll _INF = 0xc0c0c0c0c0c0c0c0;
const ll mod =  (int)1e9+7;

ll gcd(ll a,ll b){return b?gcd(b,a%b):a;}
ll ksm(ll a,ll b,ll mod){int ans=1;while(b){if(b&1) ans=(ans*a)%mod;a=(a*a)%mod;b>>=1;}return ans;}
ll inv2(ll a,ll mod){return ksm(a,mod-2,mod);}
void exgcd(ll a,ll b,ll &x,ll &y,ll &d){if(!b) {d = a;x = 1;y=0;}else{exgcd(b,a%b,y,x,d);y-=x*(a/b);}}//printf("%lld*a + %lld*b = %lld\n", x, y, d);

/*namespace sgt
{
    #define mid ((l+r)>>1)

    #undef mid
}*/
const int MAX_N = 1025;
int a[MAX_N],b[MAX_N],dp[MAX_N][2];
int main()
{
    //ios::sync_with_stdio(false);
    //freopen("a.txt","r",stdin);
    //freopen("b.txt","w",stdout);
    int n,ans = 0,ans_ = 0;
    scanf("%d",&n);
    for(int i = 1;i<=n;++i) scanf("%d",&a[i]);
    for(int i = 1;i<=n;++i) scanf("%d",&b[i]);
    memset(dp,0x3f,sizeof(dp));
    dp[1][0] = 0;
    dp[1][1] = 1;
    for(int i = 2;i<=n;++i)
    {
        if(a[i-1]<a[i]&&b[i-1]<b[i])
        {
            dp[i][0] = min(dp[i][0],dp[i-1][0]);
            dp[i][1] = min(dp[i][1],dp[i-1][1]+1);
        }
        if(b[i-1]<a[i]&&a[i-1]<b[i])
        {
            dp[i][1] = min(dp[i-1][0]+1,dp[i][1]);
            dp[i][0] = min(dp[i][0],dp[i-1][1]);
        }
    }
    printf("%d\n",min(dp[n][0],dp[n][1]));
    //fclose(stdin);
    //fclose(stdout);
    //cout << "time: " << (long long)clock() * 1000 / CLOCKS_PER_SEC << " ms" << endl;
    return 0;
}