题目描述
Problem E: Hotter Colder
The children's game Hotter Colder is played as follows. Player A leaves the room while player B hides an object somewhere in the room. Player A re-enters at position (0,0) and then visits various other positions about the room. When player A visits a new position, player B announces "Hotter" if this position is closer to the object than the previous position; player B announces "Colder" if it is farther and "Same" if it is the same distance.
Input consists of up to 50 lines, each containing an x,y coordinate pair followed by "Hotter", "Colder", or "Same". Each pair represents a position within the room, which may be assumed to be a square with opposite corners at (0,0) and (10,10). For each line of input print a line giving the total area of the region in which the object may have been placed, to 2 decimal places. If there is no such region, output 0.00.
输入
暂无
输出
暂无
样例输入
10.0 10.0 Colder
10.0 0.0 Hotter
0.0 0.0 Colder
10.0 10.0 Hotter
样例输出
50.00
37.50
12.50
0.00
参考代码
#include <stdio.h>
#include <math.h>
char reach[1<<24];
double w[24], x[24], len, wt, sumw, sumx, cg;
int n;
int solve(int m, double sx, double sw)
{
int i;
double sumt1, sumt2;
if (reach[m]) return 0;
reach[m] = 1;
cg = sx / sw;
if (cg < -1.5 || cg > 1.5) return 0;
if (m == (1<<n)-1) return 1;
for (i=0;i<n;i++)
{
if (!(m & (1<<i)) && solve(m|(1<<i),sx+x[i]*w[i],sw+w[i]))
{
printf("%lg %lgn",x[i],w[i]);
return 1;
}
}
return 0;
}
main(){
int cc,i;
for (cc=1; 3 == scanf("%lf%lf%d",&len,&wt,&n) && n; cc++) {
memset(reach,0,1<<n);
for (i=0;i<n;i++) scanf("%lf%lf",&x[i],&w[i]);
printf("Case %d:n",cc);
if (!solve(0,0,wt)) printf("Impossiblen");
}
}
解析
暂无
