A BZOJVIP 1642

Discription
Bessie is a very hard-working cow. She always concentrates on improving her production. In order to produce more milk, she predicted the next N (1 < N < 1,000,000) hours, marked 0. N-1. Farmer John planned M (1 < M < 1,000) milking periods. Each time period has a start time (0 < start time < N), an end time (start time < end time < N), and a yield (1 < yield < 1,000,000) indicating the amount of milk that can be milked from Bessie. Farmer John milks from the start time to the end time, respectively. The whole time period must be used for each milking. But even Bessie has her production limit. After each milking, she had to rest R (1 < R < N) for an hour before the next milking. Given the time period of Farmer John's plan, please figure out the maximum milking amount in N hours.

Input
The first line has three integers N, M, R. The next line M has three integers Si, Ei, Pi.

Output
Maximum milk yield.

Sample Input12 4 2
1 2 8
10 12 19
3 6 24
7 10 31

Sample Output
43
Hint

Note: No milking at the end of the day

// dp after sorting

#include<stdio.h>
#include<string.h>
#include<algorithm>
using namespace std;
struct node{
    int x,y,z;
}a[1005];
int dp[1005];
int m,n,r,ans=-1;
int ma(int a,int b){return a>b?a:b;}
bool cmp(node a,node b){
    if(b.x==a.x)  return a.y<b.y;
    return a.x<b.x;
}
int main(){
    while(scanf("%d%d%d",&n,&m,&r)!=EOF){
        for(int i=1;i<=m;i++){
            scanf("%d%d%d",&a[i].x,&a[i].y,&a[i].z);
            a[i].y+=r;
        }
        sort(a+1,a+m+1,cmp);
        memset(dp,0,sizeof(dp));
        int tem,i,j,k;
        for(i=1;i<=m;i++){
            dp[i]=a[i].z;
            for(int j=1;j<i;j++){
                if(a[i].x>=a[j].y)
                dp[i]=ma(dp[j]+a[i].z,dp[i]);
            }
            //printf("%d %d %d\n",i,a[i].y,dp[a[i].y]);
            ans=ma(dp[i],ans);
        }
        printf("%d\n",ans);
    }
    return 0;
}

B CodeForces 366C

Discription
Dima, Inna and Seryozha have gathered in a room. That's right, someone's got to go. To cheer Seryozha up and inspire him to have a walk, Inna decided to cook something.
Dima and Seryozha have n fruits in the fridge. Each fruit has two parameters: the taste and the number of calories. Inna decided to make a fruit salad, so she wants to take some fruits from the fridge for it. Inna follows a certain principle as she chooses the fruits: the total taste to the total calories ratio of the chosen fruits must equal k. In other words, , where aj is the taste of the j-th chosen fruit and bj is its calories.
Inna hasn't chosen the fruits yet, she is thinking: what is the maximum taste of the chosen fruits if she strictly follows her principle? Help Inna solve this culinary problem — now the happiness of a young couple is in your hands!
Inna loves Dima very much so she wants to make the salad from at least one fruit.
Input
The first line of the input contains two integers n, k(1 ≤ n ≤ 100, 1 ≤ k ≤ 10). The second line of the input contains n integers a1, a2, ..., an(1 ≤ ai ≤ 100) — the fruits' tastes. The third line of the input contains n integers b1, b2, ..., bn(1 ≤ bi ≤ 100) — the fruits' calories. Fruit number i has taste ai and calories bi.

Output
If there is no way Inna can choose the fruits for the salad, print in the single line number -1. Otherwise, print a single integer — the maximum possible sum of the taste values of the chosen fruits.

Sample Input Input
3 2
10 8 1
2 7 1
Output
18
Input
5 3
4 4 4 4 4
2 2 2 2 2
Output
-1
The 01 knapsack problem after the formula is deformed has only negative volume. It should be noted that there are two ways to avoid it. One is to divide the positive and negative dp into two times, and the other is to enlarge the whole into positive numbers.

//enlarge
#include<stdio.h>
#include<string.h>
#include<algorithm>
using namespace std;
#define inf 0xcf
int dp[101][201000],n,k;
int a[111],b[111],c[111];
int ma(int a,int b){return a>b?a:b;}
int main(){
    while(scanf("%d%d",&n,&k)!=EOF){
        for(int i=1;i<=n;i++)    scanf("%d",&a[i]);
        for(int i=1;i<=n;i++)    scanf("%d",&b[i]);
        for(int i=1;i<=n;i++)    c[i]=a[i]-k*b[i];
        memset(dp,inf,sizeof(dp));
        dp[0][1000*n]=0;
        for(int i=1;i<=n;i++)
            for(int j=2000*n;j>=c[i]&&j>=0;j--){
                dp[i][j]=ma(dp[i-1][j],dp[i-1][j-c[i]]+a[i]);
                //printf("%d %d %d\n",i,j,dp[j]);
            }
        if(dp[n][n*1000]<=0)
            printf("-1\n");
        else
            printf("%d\n",dp[n][n*1000]);
    }
    return 0;
}

C -CF544C

** Full Backpack**

#include<stdio.h>
#include<string.h>
#include<algorithm>
using namespace std;
#define N 555
int a[N],m,n,mod,b;
int dp[N][N];
int main(){
    scanf("%d%d%d%d",&n,&m,&b,&mod);
    for(int i=0;i<n;i++)    scanf("%d",&a[i]);
    dp[0][0]=1;
    for(int i=0;i<n;i++){
        for(int j=1;j<=m;j++){
            for(int k=a[i];k<=b;k++){
                dp[j][k]=(dp[j][k]+dp[j-1][k-a[i]])%mod;
                //printf("%d %d %d\n",j,k,dp[j][k]);
            }
        }
    }
    int ans=0;
    for(int i=0;i<=b;i++)
        ans=(ans+dp[m][i])%mod;
    printf("%d\n",ans);
    return 0;
}

D-BZOJ1616

** Digital triangle, traversing tmn directly

#include<stdio.h>
#include<string.h>
#include<algorithm>
using namespace std;
#define N 111
char ma[N][N];
int m,n,t,dp[20][N][N],a,b,c,d;
int x[4]={0,0,1,-1},y[4]={1,-1,0,0};
int main(){
    while(scanf("%d%d%d",&m,&n,&t)!=EOF){
        for(int i=0;i<m;i++)
                scanf(" %s",&ma[i]);
        scanf("%d%d%d%d",&a,&b,&c,&d);
        memset(dp,0,sizeof(dp));
        dp[0][a-1][b-1]=1;
        for(int s=1;s<=t;s++)
            for(int i=0;i<m;i++)
                for(int j=0;j<n;j++)
                    if(ma[i][j]=='.')
                        for(int k=0;k<4;k++){
                            int xx=i+x[k],yy=j+y[k];
                            if(xx>=0&&yy>=0&&xx<m&&yy<n)
                            dp[s][i][j]+=dp[s-1][xx][yy];
                            //printf("%d %d %d %d\n",s,i,j,dp[s][i][j]);
                        }
        printf("%d\n",dp[t][c-1][d-1]);
    }
    return 0;
}

E-CF1197D

dp doesn't understand very well

/*For each element, there are m states, and the influence of the current element on the latter is different i n different states. So we can open an n* m size dp array dp[i][j], which represents the maximum value of the first element when the length of the j position is modeled on M.
The state transition equation is
(j == 1)dp[i][j] = max(a[i] - k, dp[i - 1][m] + a[i] - k);
(j > 1)dp[i][j] = dp[i - 1][j - 1] + a[i];
*/
#include <stdio.h>
#include <string.h>
using namespace std;
typedef long long ll;
const int N = 3e5 + 5;
ll a[N];
ll dp[N][20];
ll ma(ll a,ll b){return a>b?a:b;}
int main()
{
    int n, m, k;
    scanf("%d%d%d",&n,&m,&k);
    a[0] = 0;
    for (int i = 1; i <= m; i ++)dp[0][i] = -1e10;
    ll ans = 0;
    for (int i = 1; i <= n; i ++){
        scanf("%lld", &a[i]);
        for (int j = 1; j <= m; j ++){
            if (j == 1)dp[i][j] = ma(a[i] - k, dp[i - 1][m] + a[i] - k);
            else dp[i][j] = dp[i - 1][j - 1] + a[i];
            ans = ma(ans, dp[i][j]);
        }
    }
    printf("%lld\n", ans);
    return 0;
}

Deformation of F-CF583D LIS

Divide t <= N and T > n into two cases. Look at the code specifically.

#include<bits/stdc++.h>
#define inf 0x3f3f3f3f
using namespace std;
typedef long long ll;
int a[33333],dp[33333],num,n,t,cnt[333];
int ma(int a,int b){return a>b?a:b;}
int mi(int a,int b){return a<b?a:b;}
int main(){
    scanf("%d %d",&n,&t);
    for(int i=0;i<n;i++){
        scanf("%d",&a[i]);
        num=ma(num,++cnt[a[i]]);
    }
    int l=mi(n,t)*n;
    int ans=0;
    for(int i=n;i<l;i++)
        a[i]=a[i-n];
    for(int i=0;i<l;i++){
        dp[i]=1;
        for(int j=i-1;j>=ma(i-n,0);j--){
            if(a[i]>=a[j])
                dp[i]=ma(dp[i],dp[j]+1);
        }
        ans=ma(dp[i],ans);
    }
    if(t<=n)    printf("%d\n",ans);
    else        printf("%d\n",ans+num*(t-n));
    return 0;
}

G-CF10D

LCIS

//(to be completed)
/*dp[i][j]Express. The first i of A sequence is processed, and the ascending sequence ends with the longest subsequence of B[j] of B sequence. It's a good idea to feel how the state ends. Then the shift is obvious.
if(A[i]==B[j])
dp[i][j]=dp[i-1][k];//k It is less than j and B [k] < B [j]
else
dp[i][j]=dp[i-1][j];
We can loop I and j so that we can save the time to find k. Complexity O(n*m)
*/
#include<bits/stdc++.h>
using namespace std;
const int INF=0x3f3f3f3f;
const int maxn=100010;
typedef long long ll;
int dp[550][550],A[550],B[550],path[550][550],n,m;
bool print(int x)
{
    if(!x)
        return false;
    if(print(path[n][x]))
        printf(" %d",B[x]);
    else
        printf("%d",B[x]);
    return true;
}
int main()
{
    int i,j,tp,ans,pos;
    while(~scanf("%d",&n))
    {
        for(i=1;i<=n;i++)
            scanf("%d",&A[i]);
        scanf("%d",&m);
        for(i=1;i<=m;i++)
            scanf("%d",&B[i]);
        memset(dp,0,sizeof dp);
        for(i=1;i<=n;i++)
        {
            tp=pos=0;
            for(j=1;j<=m;j++)
            {
                dp[i][j]=dp[i-1][j];
                path[i][j]=path[i-1][j];
                if(A[i]==B[j]&&tp+1>dp[i][j])
                    dp[i][j]=tp+1,path[i][j]=pos;
                if(B[j]<A[i]&&dp[i-1][j]>tp)
                    tp=dp[i-1][j],pos=j;
            }
        }
        ans=1;
        for(i=1;i<=m;i++)
            if(dp[n][i]>dp[n][ans])
                ans=i;
        printf("%d\n",dp[n][ans]);
        if(dp[n][ans])
            print(ans);
        printf("\n");
    }
    return 0;
}