Bitmask DP 2 problems added

Author: zarif98sjs

Dynamic Programming/27 Basic Bitmask DP.cpp

@@ -0,0 +1,111 @@
+
+/**
+
+Problem : LightOJ 1011
+
+**/
+
+/**Which of the favors of your Lord will you deny ?**/
+
+#include<bits/stdc++.h>
+using namespace std;
+
+#define LL long long
+#define PII pair<int,int>
+#define PLL pair<LL,LL>
+#define MP make_pair
+#define F first
+#define S second
+#define INF INT_MAX
+
+#define ALL(x) (x).begin(), (x).end()
+#define DBG(x) cerr << __LINE__ << " says: " << #x << " = " << (x) << endl
+
+#include <ext/pb_ds/assoc_container.hpp>
+#include <ext/pb_ds/tree_policy.hpp>
+using namespace __gnu_pbds;
+
+template<class TIn>
+using indexed_set = tree<
+                    TIn, null_type, less<TIn>,
+                    rb_tree_tag, tree_order_statistics_node_update>;
+
+/*
+PBDS
+-------------------------------------------------
+1) insert(value)
+2) erase(value)
+3) order_of_key(value) // 0 based indexing
+4) *find_by_order(position) // 0 based indexing
+
+*/
+
+inline void optimizeIO()
+{
+    ios_base::sync_with_stdio(false);
+    cin.tie(NULL);
+}
+
+const int nmax = 2e5+7;
+const LL LINF = 1e17;
+
+string to_str(LL x)
+{
+    stringstream ss;
+    ss<<x;
+    return ss.str();
+}
+
+//bool cmp(const PII &A,const PII &B)
+//{
+//
+//}
+
+
+int n;
+LL ara[16][16];
+LL dp[16][1<<16];
+
+LL solve(int r,int mask)
+{
+    if(r==n) return 0;
+
+    LL &ret = dp[r][mask];
+
+    if(ret!=-1) return ret;
+
+    for(int c = 0;c<n;c++)
+        if(!(mask&(1<<c)))
+            ret = max(ret, ara[r][c] + solve(r+1,mask | (1<<c)));
+
+    return ret;
+
+}
+
+int main()
+{
+    //freopen("out.txt","w",stdout);
+
+    optimizeIO();
+
+    int tc;
+    cin>>tc;
+
+    for(int qq=1; qq<=tc; qq++)
+    {
+        cin>>n;
+
+        for(int i=0;i<n;i++)
+            for(int j=0;j<n;j++)
+                cin>>ara[i][j];
+
+        memset(dp,-1,sizeof dp);
+        LL ans = solve(0,0);
+
+        cout<<"Case "<<qq<<": "<<ans<<endl;
+    }
+
+    return 0;
+}
+
+

Dynamic Programming/28 Brush IV.cpp

@@ -0,0 +1,136 @@
+
+/**
+
+Problem : LightOJ 1018
+
+**/
+
+/**Which of the favors of your Lord will you deny ?**/
+
+#include<bits/stdc++.h>
+using namespace std;
+
+#define LL long long
+#define PII pair<int,int>
+#define PLL pair<LL,LL>
+#define MP make_pair
+#define F first
+#define S second
+#define INF INT_MAX
+
+#define ALL(x) (x).begin(), (x).end()
+#define DBG(x) cerr << __LINE__ << " says: " << #x << " = " << (x) << endl
+
+#include <ext/pb_ds/assoc_container.hpp>
+#include <ext/pb_ds/tree_policy.hpp>
+using namespace __gnu_pbds;
+
+template<class TIn>
+using indexed_set = tree<
+                    TIn, null_type, less<TIn>,
+                    rb_tree_tag, tree_order_statistics_node_update>;
+
+/*
+PBDS
+-------------------------------------------------
+1) insert(value)
+2) erase(value)
+3) order_of_key(value) // 0 based indexing
+4) *find_by_order(position) // 0 based indexing
+
+*/
+
+inline void optimizeIO()
+{
+    ios_base::sync_with_stdio(false);
+    cin.tie(NULL);
+}
+
+const int nmax = 2e5+7;
+const LL LINF = 1e17;
+
+string to_str(LL x)
+{
+    stringstream ss;
+    ss<<x;
+    return ss.str();
+}
+
+//bool cmp(const PII &A,const PII &B)
+//{
+//
+//}
+
+
+int n;
+int X[16];
+int Y[16];
+int col[16][16];
+int dp[1<<16];
+
+bool collinear(int i, int j, int k) {
+  return ((X[i] - X[j]) * (Y[k] - Y[j]) - (Y[i] - Y[j]) * (X[k] - X[j])) == 0;
+}
+
+void precalc()
+{
+    memset(col,0,sizeof col);
+
+    for(int i=0;i<n;i++)
+        for(int j=0;j<n;j++)
+            for(int k=0;k<n;k++)
+                if(i!=j && collinear(i,j,k))
+                    col[i][j] |= (1<<k); /** Number of 1's in col[i][j] = How many Points are collinear with ith and jth point **/
+
+}
+
+int solve(int mask)
+{
+    if(mask==((1<<n)-1)) return 0;
+
+    if (__builtin_popcount(mask) == (n - 1)) /** n-1 other points are already collinear with this one **/
+        return 1;
+
+    int &ret = dp[mask];
+
+    if(ret!=-1) return ret;
+
+    int i;
+    for(i=0;mask&(1<<i);i++); /** Go to the lsb which is not collinear **/
+    /**Suppose you have satisfied bitmask M of the points. Let the lowest index dust point which hasn't been cleaned yet be x. Note that you are going to have to clean x eventually. Therefore you lose nothing by assuming you definitely clean x now. Then at each state you narrow down to N edges because you have one point which must be on the line **/
+
+    ret = INT_MAX;
+    for(int j=0;j<n;j++)
+        if(i!=j && !(mask&(1<<j)))
+            ret = min(ret,1 + solve(mask|col[i][j]));
+
+    return ret;
+}
+
+int main()
+{
+    //freopen("out.txt","w",stdout);
+
+    optimizeIO();
+
+    int tc;
+    cin>>tc;
+
+    for(int qq=1; qq<=tc; qq++)
+    {
+        cin>>n;
+
+        for(int i=0;i<n;i++)
+            cin>>X[i]>>Y[i];
+
+        precalc();
+        memset(dp,-1,sizeof dp);
+        int ans = solve(0);
+
+        cout<<"Case "<<qq<<": "<<ans<<endl;
+    }
+
+    return 0;
+}
+
+