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Posted 2020-10-05 冷暖知不知

int dp[1005][1005] = {0},len1,len2;
char a[1005],b[1005];

void lcs()
{
    for(int i = 1; i <= len1;i++)
    {
        for(int j = 1;j <= len2;j++)
        {
            if(a[i-1] == b[j-1])    dp[i][j] = dp[i-1][j-1]+1;
            else    dp[i][j] = max(dp[i-1][j],dp[i][j-1]);
        }
    }
}

void printflcs(int x,int y)
{
    if(x == 0 || y == 0)    return;
    if(a[x-1] == b[y-1])
    {
        printflcs(x-1,y-1);
        printf("%c",a[x-1]);
    }
    else if(dp[x][y-1] > dp[x-1][y])    printflcs(x,y-1);
    else    printflcs(x-1,y);
}

int main()
{
    scanf("%s%s",a,b);
    len1 = strlen(a);
    len2 = strlen(b);
    lcs();
    printflcs(len1,len2);
    printf("\n");
    return 0;
}

最长公共子序列LCS

---

int n,a[50005],b[50005] = {0};

int main()
{
    scanf("%d",&n);
    for(int i = 1;i <= n;i++)   scanf("%d",&a[i]);
    b[1] = a[1];
    int len = 1;
    for(int i = 2;i <= n;i++)
    {
        int t = lower_bound(b+1,b+len+1,a[i])-b;
        b[t] = a[i];
        if(t == len+1)  len++;
    }
    printf("%d\n",len);
    return 0;
}

最长上升子序列LIS

---

int n,m,w[3500],v[3500],dp[13000] = {0};
//数量n，背包m，重量w，价值v 

int main()
{
    scanf("%d%d",&n,&m);
    for(int i = 1;i <= n;i++)    scanf("%d%d",&w[i],&v[i]);
    for(int i = 1;i <= n;i++)
    {
        for(int j = m;j >= w[i];j--)    dp[j] = max(dp[j],dp[j-w[i]]+v[i]);    
    }
    printf("%d\n",dp[m]);
    return 0;
}

01背包

int dp[50005],v[1005],w[1005],n,W;
//数量n，背包W，重量w，价值v
//W很大，考虑每个价值的最小重量
int main()
{
    ios::sync_with_stdio(0);
    while(cin >> n >> W)
    {
        memset(dp, 0x3f, sizeof(dp));
        dp[0] = 0;
        int sum = 0;
        for(int i = 1;i <= n;i++)
        {
            cin >> w[i] >> v[i];
            sum += v[i];
        }
        for(int i = 1;i <= n;i++)
        {
            for(int j = sum;j >= v[i];j--)  dp[j] = min(dp[j-v[i]]+w[i],dp[j]);
        }
        for(int i = sum;i >= 0;i--)
        {
            if(dp[i] <= W)
            {
                cout << i << endl;
                break;
            }
        }
    }
    return 0;
}

01背包(容量很大)

int n,m,dp[105];

int main()
{
    ios::sync_with_stdio(false);
    int T;
    cin >> T;
    while(T--)
    {
        memset(dp,0,sizeof(dp));
        cin >> n >> m;
        for(int i = 1;i <= m;i++)
        {
            int p,h,c;    //花费，价值，数量
            cin >> p >> h >> c;
            int now = 1;
            while(1)
            {
                int t = min(now,c);
                c -= t;
                int pp = p*t,hh = h*t;
                for(int j = n;j >= pp;j--)  dp[j] = max(dp[j],dp[j-pp]+hh);
                if(c == 0)  break;
            }
        }
        cout << dp[n] << endl;
    }
    return 0;
}

多重背包二进制优化

完全背包初始化分两种情况：

1、如果背包要求正好装满则初始化 dp[0] = 0, dp[1~m] = -INF。

2、如果不需要正好装满 dp[0~m] = 0。

int n,m,e,f,w[505],v[505],dp[10005];
//数量n，背包m，重量w，价值v

int main()
{
    int T;
    scanf("%d",&T);
    while(T--)
    {
        scanf("%d%d%d",&e,&f,&n);
        m = f-e;
        memset(dp,0x3f,sizeof(dp));
        dp[0] = 0;
        for(int i = 1;i <= n;i++)    scanf("%d%d",&v[i],&w[i]);
        for(int i = 1;i <= n;i++)
        {
            for(int j = w[i];j <= m;j++)    dp[j] = min(dp[j],dp[j-w[i]]+v[i]);
        }
        if(dp[m] == INF)    printf("This is impossible.\n");
        else    printf("The minimum amount of money in the piggy-bank is %d.\n",dp[m]);
    }
    return 0;
}

//HDU1114

完全背包

完全背包

---

//不含62和4
int n,m,dp[15][5];
int a[15];

long long dfs(int now1,int now2,int limit)
{
    if(now2 == 2)   return 0;
    if(now1 == 0)   return 1;
    if(!limit && dp[now1][now2] != -1)  return dp[now1][now2];
    int endd = limit?a[now1]:9,ans = 0;
    for(int i = 0;i <= endd;i++)
    {
        if(i == 4)  continue;
        if(now2 == 1 && i == 2)    ans += dfs(now1-1,2,limit && i == endd);
        else if(i == 6) ans += dfs(now1-1,1,limit && i == endd);
        else    ans += dfs(now1-1,0,limit && i == endd);
    }
    if(!limit)  dp[now1][now2] = ans;
    return ans;
}

int main()
{
    memset(dp,-1,sizeof(dp));
    while(cin >> n >> m && (n || m))
    {
        n--;
        int cnt = 0;
        while(n)
        {
            a[++cnt] = n%10;
            n /= 10;
        }
        int ans = dfs(cnt,0,1);
        cnt = 0;
        while(m)
        {
            a[++cnt] = m%10;
            m /= 10;
        }
        cout << dfs(cnt,0,1)-ans << endl;
    }
    return 0;
}

数位dp

---

int a[15][15],sta[100],dp[15][100],sum[15][100],n,m;
int main()
{
    ios::sync_with_stdio(false);
    int T;
    cin >> T;
    while(T--)
    {
        memset(dp,0,sizeof(dp));
        memset(sum,0,sizeof(sum));
        cin >> n >> m;
        for(int i = 1;i <= n;i++)
        {
            for(int j = 1;j <= m;j++)   cin >> a[i][j];
        }
        int cnt = 0,endd = 1<<(m-1);
        for(int i = 0;i < endd;i++)
        {
            if(i & (i<<1))  continue;
            sta[++cnt] = i;
        }
        for(int i = 1;i < n;i++)
        {
            for(int j = 1;j <= cnt;j++)
            {
                for(int k = 1;k <= m-1;k++)
                {
                    if(1<<(k-1) & sta[j])
                    {
                        if(a[i][k] && a[i][k+1] && a[i+1][k] && a[i+1][k+1])    sum[i][j]++;
                    }
                }
            }
        }
        for(int i = 1;i <= cnt;i++) dp[1][i] = sum[1][i];
        for(int i = 1;i < n;i++)
        {
            for(int j = 1;j <= cnt;j++)
            {
                for(int k = 1;k <= cnt;k++)
                {
                    dp[i+2][k] = max(dp[i+2][k],dp[i][j]+sum[i+2][k]);
                    if(sta[j] & sta[k])         continue;
                    if(sta[j] & (sta[k]<<1))    continue;
                    if(sta[j] & (sta[k]>>1))    continue;
                    dp[i+1][k] = max(dp[i+1][k],dp[i][j]+sum[i+1][k]);
                }
            }
        }
        int ans = 0;
        for(int i = 1;i <= cnt;i++) ans = max(ans,dp[n-1][i]);
        cout << ans << endl;
    }
    return 0;
}

状压dp

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