atcoder AtCoder Beginner Contest 210 D

传送门

思路：

首先考虑暴力的做法，枚举两个点计算距离后取最小值，复杂度(O(n^{4}))
考虑怎么优化。
首先可以确定的是，一定要枚举一个点，那么可以通过控制坐标的大小关系将绝对值去掉。
假设对于点((i,j))，只考虑点((x,y))满足(i<=x<=n,j<=y<=m)
那么代价就变成了$$a[i][j]+a[x][y]+c(x-i)+c(y-j)$$
拆开得到$$a[i][j]-ci-cj+a[x][j]+cx+cy$$
对于点((i,j))的贡献(a[i][j]-c*i-c*j)，可以通过枚举计算；
对于点((x,y))的贡献(a[x][j]+c*x+c*y)，由于题目要求代价最小，所以应当维护最小值。
也就是说(dp[i][j])表示的是对于满足(i<=x<=n,j<=y<=m)的所有的点((x,y))的贡献(a[x][j]+c*x+c*y)的最小值。
所以每次枚举到((i,j))时，只需要假定这个点必选，再加上(dp[i][j])表示上一个点的贡献，所有的答案取最小值即可，注意更新完答案后应当更新(dp)数组。
这样会产生的一个问题就是，只是考虑了两个点的位置大致是呈左上-右下的趋势的情况，如果一个点在右上，另一个点在左下，这种情况没有被枚举到。
所以还应该再跑一遍，基本流程同上文类似。

代码：

又臭又长的垃圾代码：（下一个是比较简洁的代码！）

ll n,m,c,a[1100][1100],dp[1100][1100],dp1[1100][1100];

int main(){
	n=read,m=read,c=read;
	rep(i,1,n) rep(j,1,m) a[i][j]=read;
	rep(i,0,n+1) rep(j,0,m+1) dp[i][j]=1e18,dp1[i][j]=1e18;
	ll res=1e18;
	for(int i=n;i;i--)
		for(int j=m;j;j--){
			ll now=a[i][j]-c*i-c*j;
			ll las=1e18;
			if(i+1<=n&&j+1<=m) las=min(las,dp[i+1][j+1]);
			if(i+1<=n) las=min(las,dp[i+1][j]);
			if(j+1<=m) las=min(las,dp[i][j+1]);
			res=min(res,now+las);
			dp[i][j]=min(dp[i][j],a[i][j]+c*i+c*j);
			if(i+1<=n&&j+1<=m) dp[i][j]=min(dp[i][j],dp[i+1][j+1]);
			if(i+1<=n) dp[i][j]=min(dp[i][j],dp[i+1][j]);
			if(j+1<=m) dp[i][j]=min(dp[i][j],dp[i][j+1]);
		}
	for(int i=1;i<=n;i++){
		for(int j=m;j;j--){
			ll now=a[i][j]+c*i-c*j;
			ll las=1e18;
			if(i-1>=1&&j+1<=m) las=min(las,dp1[i-1][j+1]);
			if(i-1>=1) las=min(las,dp1[i-1][j]);
			if(j+1<=m) las=min(las,dp1[i][j+1]);
			res=min(res,now+las);
			dp1[i][j]=min(dp1[i][j],a[i][j]-c*i+c*j);
			if(i-1>=1&&j+1<=m) dp1[i][j]=min(dp1[i][j],dp1[i-1][j+1]);
			if(i-1>=1) dp1[i][j]=min(dp1[i][j],dp1[i-1][j]);
			if(j+1<=m) dp1[i][j]=min(dp1[i][j],dp1[i][j+1]);
		}
	}
	printf("%lld
",res);
	
	return 0;
}

跟我的思路类似但是代码超级简洁的巨巨代码：

#include <iostream>
#include <algorithm>
#include <vector>
#include <cstring>
#include <cstdio>
#define endl '
'
#define int long long
#define pb push_back
#define mp make_pair
#define INF 0x3f3f3f3f
#define Inf 1000000000000000000LL
#define F first
#define S second
using namespace std;
typedef pair<int,int>pii;
int n,m,C;
int a[1010][1010];
int dp[1010][1010];
int ans=0x3f3f3f3f3f3f3f3f;
void DP(){
	memset(dp,0x3f,sizeof dp);
	for(int i=1;i<=n;i++){
		for(int j=1;j<=m;j++){
			dp[i][j]=min(dp[i][j-1],dp[i-1][j]);
			ans=min(ans,dp[i][j]+a[i][j]+C*(i+j));
			dp[i][j]=min(dp[i][j],a[i][j]-C*(i+j));
		}
	}
}
signed main(){
	cin>>n>>m>>C;
	for(int i=1;i<=n;i++)
		for(int j=1;j<=m;j++)
			cin>>a[i][j];
	DP();
	for(int i=1;i<=n;i++){
		int l=1,r=m;
		while(l<r){
			swap(a[i][l],a[i][r]);
			l++,r--;
		}
	}
	DP();
	cout<<ans<<endl;
	return 0;
}

赛时写假了边界，宇巨一波奇技淫巧强行AC的代码：

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
typedef pair<ll, ll>PLL;
typedef pair<int, int>PII;
typedef pair<double, double>PDD;
#define I_int ll
inline ll read()
{
    ll x = 0, f = 1;
    char ch = getchar();
    while(ch < '0' || ch > '9')
    {
        if(ch == '-')f = -1;
        ch = getchar();
    }
    while(ch >= '0' && ch <= '9')
    {
        x = x * 10 + ch - '0';
        ch = getchar();
    }
    return x * f;
}

inline void out(ll x){
	if (x < 0) x = ~x + 1, putchar('-');
	if (x > 9) out(x / 10);
	putchar(x % 10 + '0');
}

inline void write(ll x){
	if (x < 0) x = ~x + 1, putchar('-');
	if (x > 9) write(x / 10);
	putchar(x % 10 + '0');
	puts("");
}

#define read read()
#define closeSync ios::sync_with_stdio(0);cin.tie(0);cout.tie(0)
#define multiCase int T;cin>>T;for(int t=1;t<=T;t++)
#define rep(i,a,b) for(int i=(a);i<=(b);i++)
#define repp(i,a,b) for(int i=(a);i<(b);i++)
#define per(i,a,b) for(int i=(a);i>=(b);i--)
#define perr(i,a,b) for(int i=(a);i>(b);i--)
ll ksm(ll a, ll b, ll p)
{
    ll res = 1;
    while(b)
    {
        if(b & 1)res = res * a % p;
        a = a * a % p;
        b >>= 1;
    }
    return res;
}
const int inf = 0x3f3f3f3f;
#define PI acos(-1)
const int maxn=4e5+100;

ll n,m,c,a[1100][1100],dp[1100][1100],dp1[1100][1100];
struct Node{
	ll x,y,val;
}t[1000009];
bool cmp(Node x,Node y)
{
	return x.val<y.val;
}
int main(){
	n=read,m=read,c=read;
	ll cnt=0;
	rep(i,1,n) rep(j,1,m) {
		a[i][j]=read,dp[i][j]=1e18,dp1[i][j]=1e18;
		t[++cnt] = {i,j,a[i][j]};
		}
	if(n*m<=30000)
	{
		
		ll anss = 1e18;
		sort(t+1,t+1+cnt,cmp);
		rep(i,1,min(n*m,30000ll))		rep(j,1,min(n*m,30000ll))
		{
			if(i==j) continue;
			ll x1,y1,x2,y2,val1,val2;
			x1 = t[i].x,y1 = t[i].y,val1 = t[i].val;
			x2 = t[j].x,y2 = t[j].y,val2 = t[j].val;
			ll temp = c*(abs(x1-x2)+abs(y1-y2)) +val1+val2;
			anss = min(anss,temp);
		}
		
		printf("%lld",anss);
		return 0;
	}
	ll res=1e18;
	for(int i=n;i;i--)
		for(int j=m;j;j--){
			if(i==n&&j==m){
				dp[i][j]=a[i][j]+c*i+c*j;
				continue;
			}
			ll now=a[i][j]-c*i-c*j;
			ll las=1e18;
			if(i+1<=n&&j+1<=m) las=min(las,dp[i+1][j+1]);
			if(i+1<=n) las=min(las,dp[i+1][j]);
			if(j+1<=m) las=min(las,dp[i][j+1]);
			res=min(res,now+las);
			dp[i][j]=min(dp[i][j],a[i][j]+c*i+c*j);
			if(i+1<=n&&j+1<=m) dp[i][j]=min(dp[i][j],dp[i+1][j+1]);
			if(i+1<=n) dp[i][j]=min(dp[i][j],dp[i+1][j]);
			if(j+1<=m) dp[i][j]=min(dp[i][j],dp[i][j+1]);
		}
	for(int i=1;i<=n;i++){
		for(int j=m;j;j--){
			if(i==1&&j==m){
				dp1[i][j]=a[i][j]+c*i+c*j;
				continue;
			}
			ll now=a[i][j]+c*i-c*j;
			ll las=1e18;
			if(i-1>=1&&j+1<=m) las=min(las,dp1[i-1][j+1]);
			if(i-1>=1) las=min(las,dp1[i-1][j]);
			if(j+1<=m) las=min(las,dp1[i][j+1]);
			res=min(res,now+las);
			dp1[i][j]=min(dp1[i][j],a[i][j]-c*i+c*j);
			if(i-1>=1&&j+1<=m) dp1[i][j]=min(dp1[i][j],dp1[i-1][j+1]);
			if(i-1>=1) dp1[i][j]=min(dp1[i][j],dp1[i-1][j]);
			if(j+1<=m) dp1[i][j]=min(dp1[i][j],dp1[i][j+1]);
		}
	}
	printf("%lld
",res);
	
	return 0;
}
/*
-4999000000****1061109567****3****2
-4999000000****1061109567****2****3
-3937890433
2001000000 3001000000 4000000001 
3001000000 4001000000 5001000000 
4000000001 5001000000 6001000000 


*/