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 ] − c ∗ i − c ∗ j + a [ x ] [ y ] + c ∗ x + c ∗ y
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 ] [ y ] + 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d p [ 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 ] [ y ] + 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d p [ 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d p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\n",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 '\n' #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\n",res); return 0; } /* -4999000000****1061109567****3****2 -4999000000****1061109567****2****3 -3937890433 2001000000 3001000000 4000000001 3001000000 4001000000 5001000000 4000000001 5001000000 6001000000 */
