多维DP

P1508 Likecloud-吃、吃、吃

一眼看出来就是DP

状态转移： f[i][j] = max(f[i-1][j],f[i-1][j-1],f[i-1][j+1])+v[i][j];

 #include<bits/stdc++.h>
 using namespace std;
 #define LOACL  freopen("in","r",stdin);\
             freopen("out","w",stdout); 
 
 const int  inf = 987654321;
 const int sz = 1e6 + 5;
 const int mod = 1e9 + 7;
 const int sqrtn = 300; 
 
 #define add(u,v,w) (e[++tot]=(edge){v,head[u],1},head[u]=tot;) 
 #define f(i,l,r) for(int i=(int)l;i<=(int)r;++i)
 #define g(i,l,r) for(int i=(int)l;i>=(int)r;--i)
 #define CLR(arr,val) memset(arr,val,sizeof(arr)) 
 typedef long long ll; 
 int n,m,a[300][300],f[300][300],k;
 int main()
 {
     LOACL
     cin>>n>>m;
     k=m/2+1;
     f(i,1,n)f(j,1,m)cin>>a[i][j];
     f(i,1,n)f(j,1,m)
     {
         f[i][j]=max(max(f[i-1][j],f[i-1][j-1]),f[i-1][j+1])+a[i][j];
     }
     cout<<max(max(f[n][k],f[n][k-1]),f[n][k+1])<<endl;
     return 0;
 }

 

P1006 传纸条

金字塔升级版

f[i][j][k] 第几步 x1坐标x2坐标 的好感度
f[i][j][k] = max(f[i-1][j-1][k-1],f[i-1][j][k-1],f[i-1][j-1][k],f[i-1][j][k])+a[j][i-j+1]+a[k][i-k+1];

#include<bits/stdc++.h>
using namespace std;
#define LOACL  freopen("in","r",stdin);\
            freopen("out","w",stdout); 

const int  inf = 987654321;
const int sz = 1e6 + 5;
const int mod = 1e9 + 7;
const int sqrtn = 300; 

#define add(u,v,w) (e[++tot]=(edge){v,head[u],1},head[u]=tot;) 
#define f(i,l,r) for(int i=(int)l;i<=(int)r;++i)
#define g(i,l,r) for(int i=(int)l;i>=(int)r;--i)
#define CLR(arr,val) memset(arr,val,sizeof(arr)) 
typedef long long ll; 
int n,m;
int a[300][300],f[600][300][300];
int mm( int a,int b,int c,int d)
{
     return max(a,max(b,max(c,d)));
 }
int main()
{
    LOACL
    cin>>n>>m;
    f(i,1,n)f(j,1,m)cin>>a[i][j];
     f(k,1,n+m-1)    
    f(i,1,n)f(j,1,n)
    {
        if(k-i+1<1||k-j+1<1) continue;
     //   f[k][i][j]=mm(f[k-1][i][j],f[k-1][i-1][j-1],f[k-1][i][j-1],f[k-1][i-1][j]) + a[i][k-i+1] + a[j][k-j+1];
        f[k][i][j]=mm(f[k-1][i][j],f[k-1][i-1][j-1],f[k-1][i-1][j],f[k-1][i][j-1])+a[i][k-i+1]+a[j][k-j+1];
        if(i==j)f[k][i][j]-=a[i][k-i+1];
    }
 


    cout<<f[n+m-1][n][n]<<endl;

    return 0;
}

 

P1387 最大正方形

f[i][j] 运行到i行j 列  判断左方、上方、左上方的最小值，再+1
f[i][j]=min( f[i-1][j-1],f[i-1][j],f[i][j-1])+1;

#include<bits/stdc++.h>
using namespace std;
#define LOACL  freopen("in","r",stdin);\
            freopen("out","w",stdout); 

const int  inf = 987654321;
const int sz = 1e6 + 5;
const int mod = 1e9 + 7;
const int sqrtn = 300; 

#define add(u,v,w) (e[++tot]=(edge){v,head[u],1},head[u]=tot;) 
#define f(i,l,r) for(int i=(int)l;i<=(int)r;++i)
#define g(i,l,r) for(int i=(int)l;i>=(int)r;--i)
#define CLR(arr,val) memset(arr,val,sizeof(arr)) 
typedef long long ll; 

int n,m;
int a[300][300],f[300][300];
int main()
{
    LOACL
    cin>>n>>m;
    f(i,1,n)f(j,1,m)cin>>a[i][j];
    f(i,1,n)f(j,1,m)
    {
        if(a[i][j]==0) continue;
        f[i][j]=min(min(f[i-1][j-1],f[i-1][j]),f[i][j-1])+1;
    }
    int ans = 0;
    f(i,1,n)f(j,1,m) ans=max(ans,f[i][j]);
    cout<<ans<<endl;
    return 0;
}

 

P1417 烹调方案

划重点
a[x]-(p+c[x])*b[x]+a[y]-(p+c[x]+c[y])*by
a[y]-(p+c[y])*b[y]+a[x]-(p+c[y]+c[x])*bx
对这两个式子化简，得到①＞②的条件是c[x]*b[y]<c[y]*b[x].
发现只要满足这个条件的物品对(x,y)，x在y前的代价永远更优。

#include<bits/stdc++.h>
using namespace std;
#define LOACL  freopen("in","r",stdin);\
            freopen("out","w",stdout); 

const int  inf = 987654321;
const int sz = 1e6 + 5;
const int mod = 1e9 + 7;
const int sqrtn = 300; 

#define add(u,v,w) (e[++tot]=(edge){v,head[u],1},head[u]=tot;) 
#define f(i,l,r) for(ll i=(ll)l;i<=(ll)r;++i)
#define g(i,l,r) for(ll i=(ll)l;i>=(ll)r;--i)
#define CLR(arr,val) memset(arr,val,sizeof(arr)) 
typedef long long ll; 
ll t,n;
struct node
{
    ll a,b,c;
}e[sz];
ll f[sz];
bool cmp(node a,node b)
{
    return a.b*b.c >a.c*b.b;
}
int main()
{
    LOACL
    cin>>t>>n;
    f(i,1,n)cin>>e[i].a;
    f(i,1,n)cin>>e[i].b;
    f(i,1,n)cin>>e[i].c;
    sort(e+1,e+n+1,cmp);
 
    f(i,1,n)
    g(j,t,e[i].c)
    //     for(ll j = t; j - e[i].c >= 0; j--)
             f[j] = max(f[j], f[j-e[i].c] + e[i].a - j * e[i].b);
    ll maxx =0;
    f(i,1,t)
    maxx=max(maxx,f[i]);
    cout<<maxx<<endl;
     
    return 0;
}