ABC 210D (动态规划)

题目链接在这里：D - National Railway (atcoder.jp)

首先需要明确的是在地图上选择两点，两点的位置关系有两种，一种是左上右下，一种是左下右上！！！

然后这是一种比较经典的操作吧，就是维护一个二维的前缀最小值，注意由于位置关系有两种，所以维护的最小值也是有两种，一种是左对角线，一种是右对角线

#include "bits/stdc++.h"
using namespace std;
const int MAX=1005;
typedef long long LL;
LL n,m,c,ans;
LL a[MAX][MAX],d[MAX][MAX],b[MAX][MAX];
int main(){
    freopen ("d.in","r",stdin);
    freopen ("d.out","w",stdout);
    int i,j,k,w;
    scanf("%lld%lld%lld",&n,&m,&c);
    for (i=1;i<=n;i++) for (j=1;j<=m;j++) scanf("%lld",&a[i][j]);
    b[1][1]=9000000000000000000ll;d[1][1]=a[1][1]-c*2;
    for (i=2;i<=n;i++) b[i][1]=min(b[i-1][1],a[i-1][1]-c*i),d[i][1]=a[i][1]-c*(i+1);
    for (i=2;i<=m;i++) b[1][i]=min(b[1][i-1],a[1][i-1]-c*i),d[1][i]=a[1][i]-c*(i+1);
    for (i=2;i<=n;i++){
        for (j=2;j<=m;j++){
            d[i][j]=a[i][j]-c*(i+j);
            b[i][j]=min(b[i-1][j],min(b[i][j-1],min(d[i-1][j],d[i][j-1])));
        }
    }
    ans=9000000000000000000ll;
    for (i=1;i<=n;i++){
        for (j=1;j<=m;j++){
            if (i==1 && j==1) continue;
            ans=min(ans,b[i][j]+a[i][j]+c*(i+j));
        }
    }
    for (i=1;i<=n;i++) for (j=1;j<=m/2;j++) swap(a[i][j],a[i][m-j+1]);
    memset(b,0,sizeof(b));
    b[1][1]=9000000000000000000ll;d[1][1]=a[1][1]-c*2;
    for (i=2;i<=n;i++) b[i][1]=min(b[i-1][1],a[i-1][1]-c*i),d[i][1]=a[i][1]-c*(i+1);
    for (i=2;i<=m;i++) b[1][i]=min(b[1][i-1],a[1][i-1]-c*i),d[1][i]=a[1][i]-c*(i+1);
    for (i=2;i<=n;i++){
        for (j=2;j<=m;j++){
            d[i][j]=a[i][j]-c*(i+j);
            b[i][j]=min(b[i-1][j],min(b[i][j-1],min(d[i-1][j],d[i][j-1])));
        }
    }
    for (i=1;i<=n;i++){
        for (j=1;j<=m;j++){
            if (i==1 && j==1) continue;
            ans=min(ans,b[i][j]+a[i][j]+c*(i+j));
        }
    }
    printf("%lld",ans);
    return 0;
}