Dynamic Programming (2) - Common Dynamic Programming Model

\ (1 \) digital triangle

Every time the lower left or lower right can go a grid, seeking the path of maximum weight.

\ (d (i, j) = max (d (i + 1, j), d (i + 1, j + 1)) + a (i, j). \) boundary is \ (d (n + 1 , J) = 0 \) , pushed from the bottom up (because to ensure \ (i + 1 \) line section \ (I \) before the update OK)

for(int i=1;i<=n+1;++i) d[n+1][i]=0;
for(int i=n;i>=1;--i)
{
    for(int j=1;j<=i;++j)
    {
        d[i][j]=max(d[i+1][j],d[i+1][j+1])+a[i][j];
    }
}

\ (2 \) nested rectangles

There \ (n-\) rectangles, each rectangle with a tuple \ ((a, b) \ ) FIG. We predetermined rectangular \ (X (a, b) \) can be nested in the rectangle \ (Y (c, d) \) if and only if \ (A <C, B <D \) , or \ (b < C, a <D \) (rotated \ (90 \) degrees). Selecting as many rectangle aligned so that except the last one, can be nested within each rectangle a lower rectangle. If multiple solutions, as small as possible to ensure the lexicographical

\ (DAG \) longest road problem

//dp[i]表示的是从i点出发的最长路
int dp(int x)
{
    int &ans=d[x];
    if(ans) return ans;
    for(int i=1;i<=n;++i)
    {
        if(G[x][i])
        {
            ans=max(ans,dp(i)+1);//注意记录的是从终点到起点的距离，这是为了方便字典序最小方案的输出
        }
    }
    d[x]=ans;
    return ans;
}
void print(int i)
{
    printf("%d ",i);
    for(int j=1;j<=n;++j)
    {
        if(G[i][j]&&d[j]+1==d[i])
        {
            print(j);
        }
    }
}


for(int i=1;i<=n;++i)
{
    scanf("%d%d",&a[i],&b[i]);
    if(a[i]>b[i]) swap(a[i],b[i]);
}
for(int i=1;i<=n;++i)
{
    for(int j=1;j<=n;++j)
    {
        if(a[i]>a[j]&b[i]>b[j])
        {
            G[j][i]=1;
        }
    }
}
int Max=0;
int endpos;
for(int i=1;i<=n;++i)
{
    if(dp(i)>Max)
    {
        Max=dp[i];
        endpos=i;
    }
}
print(endpos);

\ (3 \) coin issue

\ [F (i) = min (inf, f [i-Vi] +1 | Vi <= i), g (i) = max (-inf, g [i-Vi] +1 | Vi <= i) (Vi coin denominations) \]

\ (4.01 \) backpack has a special topic explain in detail

\ (5 \) point set matching problems (like pressure \ (DP \) )

\ (d (S) \) represents a set of \ (S \) shortest distance after pairing

double dis(int i,int j)
{
    return sqrt((a[i].x-a[j].x)*(a[i].x-a[j].x)+(a[i].y-a[j].y)*(a[i].y-a[j].y));
}

int main()
{
    int n;
    scanf("%d",&n);
    for(int i=0;i<n;++i) scanf("%d%d",&a[i].x,&a[i].y);
    memset(d,0x7f,sizeof(d));
    d[0]=0;
    for(int i=1;i<(1<<n);++i)//由于第一个点无论如何都是要配对的，所以无需枚举（否则时间复杂度会
    {                        //乘上一个n）
        int k=0;
        while(!(i&(1<<k))) ++k;
        for(int j=k+1;j<n;++j)
        {
            if(i&(1<<j)) d[i]=min(d[i],d[i^(1<<k)^(1<<j)]+dis(k,j));
        }
    }
}

\ (6 \) longest subsequence problem rises \ ((the LIS) \)

Primary: \ (O (the n-^ 2) \) (but not too good, and I learned the second, they never used it)

Advanced: \ (O (nlogn) \) \ (D [I] \) represented by \ (a [i] \) length of the longest sub-sequences of the rising end

for(int i=1;i<=n;+i) g[i]=inf;
for(int i=0;i<n;++i)
{
    int k=lower_bound(g+1,g+n+1,a[i])-g;
    d[i]=k;
    g[k]=a[i];
}

\ (7 \) the longest common subsequence problem \ ((the LCS) \) [Note: \ (the LCS \) sometimes referred to as the longest common suffix, and \ (the LCP \) longest common prefix correspond]

for(int i=0;i<la;++i)
{
    for(int j=0;j<lb;++j)
    {
        if(a[i]==b[j])
        {
            d[i][j]=max(d[i][j],d[i-1][j-1]+1);
        }
        else if(a[i]!=b[j])
        {
            d[i][j]=max(d[i-1][j],d[i][j-1]);
        }
    }
}

You can also scroll through an array optimization

int f=0;
for(int i=0;i<la;++i)
{
    f^=1;
    for(int j=0;j<lb;++j)
    {
        if(a[i]==b[j])
        {
            d[f][j]=max(d[f][j],[f^1][j-1]+1);
        }
        else if(a[i]!=b[j])
        {
            d[f][j]=max(d[f^1][j],d[f][j-1]);
        }
    }
}

\ (8 \) maximum continuous and

Prefix and practices:

for(int i=1;i<=n;++i)
{
    scanf("%d",&a[i]);
    sum[i]=sum[i-1]+a[i];
}
int Maxn=sum[n],Max=0;
for(int i=n-1;i>=1;--i)
{
    Max=max(Max,Maxn-sum[i]);
    Maxn=max(Maxn,sum[i]);
}

Dynamic programming practices: \ (D [i] \) represents the maximum continuous ending and i

for(int i=1;i<=n;++i) scanf("%d",&a[i]);
for(int i=1;i<=n;++i)
{
    d[i]=max(0,d[i-1])+a[i];
}