Bzoj question record Part 4

[Bzoj1143] the question of ctsc... Use floyed to pass the closure, and then directly use the Hungary algorithm.

[Bzoj1452] a two-dimensional tree array that has never been written. It seems very simple ..

struct two_bit{  int f[305][305];  inline void add(int x,int z,int A)  {    for (;x<=n;x+=L(x))      for (int y=z;y<=m;y+=L(y))        f[x][y]+=A;  }  inline int ask(int x,int z)  {    int ans=0;    for (;x;x-=L(x))      for (int y=z;y;y-=L(y))        ans+=f[x][y];    return ans;  }}a[105];

[Bzoj1876] High-Precision GCD. But the implementation card is very tight. If it is a bare GCD, whether it is time-out or not, the code length alone is chilling. (For details, see my vijos1047 solution.) You can use the binary GCD for optimization.

If A and B are even numbers, gcd (a, B) = gcd (A/2, B/2)

If a is an odd number and B is an even number, gcd (a, B) = gcd (A, B/2)

If a is odd, B is odd, gcd (a, B) = gcd (A-B, B)

Until one of A and B is 0, or a = B.

[Bzoj3038 & 3211] initially thought it was a God question, and later found a nature: Even if a number is operated by a violent root number, it will become 1 after several times. (The major issue is 0 !!!) So we can use the line tree to modify it.

[Bzoj1695 & 1743] This permission question has left me weak for a long time. At first I went backwards, and there were child nodes in the letter tree to update the parent node-but there was more than crazy wa. So I started updating the subnode and finally got.

for (i=1;i<=n;i++)    for (j=m;j;j--)      for (k=1;k<=num[map[i][j]];k++)      {        x=record[map[i][j]][k];        if (flag[x]) for (p=1;p<=j;p++) c[x][p]++;        for (son=0;son<26;son++)          if ((t=tree[x][son])>-1)            for (p=1;p<=j;p++)              c[x][p]+=c[t][j];      }  for (son=0;son<26;son++)     if (tree[0][son]>-1)      ans+=c[tree[0][son]][1];

[Bzoj3170] Typical coordinate reconstruction.

for (i=1;i<=n;i++)   {    scanf("%lld%lld",&x,&y);a[i].x=x+y;a[i].y=x-y;    a[i].id=i;endx+=a[i].x;endy+=a[i].y;  }

[Bzoj3319] I honestly said: "violent rank1, can you believe it?

[Bzoj1765 & 1356] What is the largest area? Card time! First, locate all the diagonal lines and sort them in order. Enumeration from big to small is a set of equal diagonal lines. Once matched, the break is witty.

[Bzoj1879] looking at the question, it is a simple pressure DP.

f[0][0]=1;    for (i=0;i<len;i++)      for (status=0;status<(1<<n);status++)        if (f[i][status])          for (j=0;j<26;j++)            up(f[i+1][status|c[i+1][j]],f[i][status]);

[Bzoj1226] God-like pressure! Note a property: Bi <= 7, that is, for a certain point, only the first seven and the last seven points are affected.

PS: A or B-A and B is not a xor B?

    f[1][0][-1+A]=0;    for (i=1;i<=n;i++)      for (status=0;status<256;status++)        for (k=-8;k<=7;k++)          if (f[i][status][k+A]<INF)          {            if (status&1) {up(f[i+1][status>>1][k-1+A],f[i][status][k+A]);continue;}            r=INF;            for (j=0;j<=B[i];j++)              if (!(status&(1<<j)))              {                if (i+j>r) break;                up(r,B[i+j]+i+j);                up(f[i][status|(1<<j)][j+A],f[i][status][k+A]+cost(i+k,i+j));              }          }