Spiderman poj-1925

DP

D [I] indicates that it takes at least a few steps from apartment to x-axis I .. P [I] the abscissa of building I, the ordinate of building I in H [I]

Note that the ordinate of Spiderman is always the height of the apartment !! So you don't need to worry about the ordinate !!

Traverse every building ..

The following requirements must be met when building I can be skipped from X coordinate J: (P [I]-j) ^ 2 <= H [I] ^ 2-(H [I]-H [1]) ^ 2

After the x-axis J passes through the building I, it reaches the x-axis 2 * P [I]-J ..

In summary, d [2 * P [I]-J] = min (d [2 * P [I]-J], d [J] + 1 )...

Finally, pay attention to the number of steps to reach the West tower.

/*
* 1925.cpp
*
* Created on: 2011-7-8
* Author:
*/
# Include <cstdio>
# Include <cmath>
# Include <cstring>
Using namespace STD;

Const int maxn = 5000 + 5;
Const int max_x = 2000000 + 5;
Constint INF = 1000000;

Int N;
Int P [maxn], H [maxn];
Double scale [maxn];
Int d [max_x];

Int main (){
Int tot;
Scanf ("% d", & ToT );
While (TOT --){
Scanf ("% d", & N );

For (INT I = 1; I <= N; I ++ ){
Scanf ("% d", & P [I], & H [I]);
// Scale [I] = SQRT (double) H [I] * H [I]-(H [I]-H [1]) * (H [I]-H [1]); // do not use SQRT; otherwise, TLE !! (Square is used for comparison later)
Scale [I] = H [I] * H [I]-(H [I]-H [1]) * (H [I]-H [1]);
}

Memset (D,-1, sizeof (INT) * max_x); // it can also be a for loop .. D [I] = inf to save time, set it to-1 and then judge ..
D [p [1] = 0;

//......
For (INT I = 2; I <= N; I ++ ){
For (Int J = P [I]-1; j> = P [1]; j --){
If (d [J] =-1) continue ;//!!!! Skip directly ~

Int Gap = P [I]-J;
If (GAP * gap> scale [I]) break; // compare the sum of squares of gap and Scale

Int aim = 2 * P [I]-J;
If (d [aim] =-1 | D [aim]> d [J] + 1 )){
D [aim] = d [J] + 1;
}

If (aim> = P [N] & (d [p [N] =-1 | D [p [N]> d [aim]) {// final processing
D [p [N] = d [aim];
}
}
}

If (d [p [N] <= 0)
Printf ("-1 \ n ");
Else
Printf ("% d \ n", d [p [N]);

}

Return 0;
}