JZOJ 3490. 旅游(travel)

题目大意

两个图，单调的跑

 

分析

 

• 暴力DP

代码

 

 

#pragma GCC optimize(2)
#pragma GCC optimize(3)
#include<bits/stdc++.h>
using namespace std;
struct sb
{
    long long x,y,t,val;
}a[1000010];
long long cnt=0;
long long read() {
    long long res = 0;
    char c = getchar();
    while(!isdigit(c)) c = getchar();
    while(isdigit(c)) res = (res << 1) + (res << 3) + c - 48, c = getchar();
    return res;
}
bool cmp(sb a,sb b)
{
    return a.val>b.val?true:false;
}
long long f[1000010];
int mapp[1001][1001];
int main ()
{
    long long n,m,x;
    n=read(); m=read(); 
    for (int i=1;i<=n;i++)
      for (int j=1;j<=m;j++)
           mapp[i][j]=read();
    cnt=0;
    for (int i=1;i<=n;i++)
      for (int j=1;j<=m;j++)
      {
           x=read();
           if (mapp[i][j]==0&&x==0) continue;
           a[++cnt].x=i; a[cnt].y=j; a[cnt].val=mapp[i][j]; a[cnt].t=x;
      }
    sort(a+1,a+1+cnt,cmp);
    long long tot=1,j=1;
    f[1]=a[1].t;
    while (a[j].val==a[j+1].val) tot++,j++,f[j+1]=a[j+1].t;
    long long ans=0;
    long long l=1,r=tot,noww=a[tot+1].val;
    for (int i=tot+1;i<=cnt;i++)
    {
        if (a[i].val!=a[r+1].val)
          l=r+1,r=i-1;
        for (int j=l;j<=r;j++)
          f[i]=max(f[i],f[j]+abs(a[i].x-a[j].x)+abs(a[i].y-a[j].y)+a[i].t);
        ans=max(ans,f[i]);
    }
    cout<<ans;
}