昨天搜水题做，一不小心进一牛人的博客，
看到几道二分题的报告，忍不住做了一下，顺便把人家对题目的描述偷懒过来，说我邪恶吧( ⊙ o ⊙ )！
POJ 3258 River Hopscotch http://poj.org/problem?id=3258 长为L的河道，起点在0位置，终点为L位置，期间分布着n块石头，石头不会重叠 求拆掉其中的m块石头，使相邻两石头间的距离的最小值最大，并输出这个值。。。 额。。。貌似描述不清。。。将就一下，(*^__^*) 嘻嘻……
View Code
#include
<
stdio.h
>
#include < string .h >
#include < algorithm >
using namespace std;
const int N = 50000 + 10 ;
long rocks[N];
long l,n,m;
bool ok( long v)
{
int i;
int remove = 0 ;
int pre = 0 ;
for (i = 1 ;i < n;i ++ )
{
if (rocks[i] - rocks[pre] < v)
{
remove ++ ;
if (remove > m) return false ;
}
else
{
pre = i;
}
}
return true ;
}
int main()
{
while (scanf( " %ld%ld%ld " , & l, & n, & m) != EOF)
{
int i;
for (i = 1 ;i <= n;i ++ )
scanf( " %ld " , & rocks[i]);
rocks[ 0 ] = 0 ;
rocks[n + 1 ] = l;
n += 2 ;
sort(rocks,rocks + n);
long left = 0 ;
long right = l;
long ans = l;
while (left <= right)
{
long mid = (left + right) >> 1 ;
if (ok(mid))
{
ans = mid;
left = mid + 1 ;
} else
right = mid - 1 ;
}
printf( " %ld\n " ,ans);
}
return 0 ;
}
#include < string .h >
#include < algorithm >
using namespace std;
const int N = 50000 + 10 ;
long rocks[N];
long l,n,m;
bool ok( long v)
{
int i;
int remove = 0 ;
int pre = 0 ;
for (i = 1 ;i < n;i ++ )
{
if (rocks[i] - rocks[pre] < v)
{
remove ++ ;
if (remove > m) return false ;
}
else
{
pre = i;
}
}
return true ;
}
int main()
{
while (scanf( " %ld%ld%ld " , & l, & n, & m) != EOF)
{
int i;
for (i = 1 ;i <= n;i ++ )
scanf( " %ld " , & rocks[i]);
rocks[ 0 ] = 0 ;
rocks[n + 1 ] = l;
n += 2 ;
sort(rocks,rocks + n);
long left = 0 ;
long right = l;
long ans = l;
while (left <= right)
{
long mid = (left + right) >> 1 ;
if (ok(mid))
{
ans = mid;
left = mid + 1 ;
} else
right = mid - 1 ;
}
printf( " %ld\n " ,ans);
}
return 0 ;
}
poj 3273 Monthly Expense 题意：给你天数n，和每天需要花的钱，让你把这些天分成m份（每份都是连续的天）， 要求每份的和尽量少，输出这个和。 分析：二分时，left取支出最多的那一天的花费，right取总花费即可 http://poj.org/problem?id=3273
#include
<
stdio.h
>
const int N = 100000 + 10 ;
int day[N];
int n,m;
bool ok(__int64 v)
{
int tm = m;
int i = 0 ;
while (tm > 0 && i < n)
{
__int64 curs = v - day[i];
if (curs < 0 ) return false ;
i ++ ;
while (i < n && curs >= day[i])
{
curs -= day[i];
i ++ ;
}
tm -- ;
}
if (i == n) return true ;
return false ;
}
int main()
{
while (scanf( " %d%d " , & n, & m) != EOF)
{
int i;
__int64 ans;
__int64 sum = 0 ;
__int64 left = 0 ;
for (i = 0 ;i < n;i ++ )
{
scanf( " %d " , & day[i]);
if (day[i] > left)left = day[i];
sum += day[i];
}
__int64 right = sum;
ans = 0 ;
while (left <= right)
{
__int64 mid = (left + right) >> 1 ;
if (ok(mid))
{
ans = mid;
right = mid - 1 ;
}
else
left = mid + 1 ;
}
printf( " %I64d\n " ,ans);
}
return 0 ;
}
const int N = 100000 + 10 ;
int day[N];
int n,m;
bool ok(__int64 v)
{
int tm = m;
int i = 0 ;
while (tm > 0 && i < n)
{
__int64 curs = v - day[i];
if (curs < 0 ) return false ;
i ++ ;
while (i < n && curs >= day[i])
{
curs -= day[i];
i ++ ;
}
tm -- ;
}
if (i == n) return true ;
return false ;
}
int main()
{
while (scanf( " %d%d " , & n, & m) != EOF)
{
int i;
__int64 ans;
__int64 sum = 0 ;
__int64 left = 0 ;
for (i = 0 ;i < n;i ++ )
{
scanf( " %d " , & day[i]);
if (day[i] > left)left = day[i];
sum += day[i];
}
__int64 right = sum;
ans = 0 ;
while (left <= right)
{
__int64 mid = (left + right) >> 1 ;
if (ok(mid))
{
ans = mid;
right = mid - 1 ;
}
else
left = mid + 1 ;
}
printf( " %I64d\n " ,ans);
}
return 0 ;
}
poj 3122 Pie 题意是：某人在生日的时候请朋友一起吃pie，但是他的朋友十分挑剔，
如果有人分到了一块比其他人大的pie，那么其他的朋友就会抱怨，因此每个人，
包括主人，都必须分到大小一样的pie，但是每个人的pie只能是从一块大的圆形pie中切出来的，
要求求出每人能分到的最大的pie，给定圆形pie的数量 n，朋友数量m，以及每个pie的半径r
#include
<
stdio.h
>
#include < algorithm >
using namespace std;
const int N = 10000 + 10 ;
double s[N];
int r[N];
const double pi = 3.141592653589793238462643383279502884197169399 ;
int n,f;
bool ok( double v)
{
int i = n - 1 ;
double tf = f;
while (i >= 0 && tf > 0 )
{
int cur = ( int )(s[i] / v);
tf -= cur;
i -- ;
}
if (tf > 0 ) return false ;
return true ;
}
int main()
{
int T,i;
while (scanf( " %d " , & T) != EOF)
{
while (T -- )
{
scanf( " %d%d " , & n, & f);
f ++ ;
for (i = 0 ;i < n;i ++ )
scanf( " %d " , & r[i]);
sort(r,r + n);
for (i = 0 ;i < n;i ++ )
{
s[i] = ( double (r[i] * r[i]));
}
double left,right;
right = s[n - 1 ];
left = 0 ;
double ans = 0 ;
while ((right - left) >= 0.0000001 ) // 注意精度处理
{
double mid = (left + right) / 2 ;
if (ok(mid))
{
ans = mid;
left = mid;
}
else
right = mid;
}
printf( " %.3lf\n " ,ans * pi);
}
}
return 0 ;
}
#include < algorithm >
using namespace std;
const int N = 10000 + 10 ;
double s[N];
int r[N];
const double pi = 3.141592653589793238462643383279502884197169399 ;
int n,f;
bool ok( double v)
{
int i = n - 1 ;
double tf = f;
while (i >= 0 && tf > 0 )
{
int cur = ( int )(s[i] / v);
tf -= cur;
i -- ;
}
if (tf > 0 ) return false ;
return true ;
}
int main()
{
int T,i;
while (scanf( " %d " , & T) != EOF)
{
while (T -- )
{
scanf( " %d%d " , & n, & f);
f ++ ;
for (i = 0 ;i < n;i ++ )
scanf( " %d " , & r[i]);
sort(r,r + n);
for (i = 0 ;i < n;i ++ )
{
s[i] = ( double (r[i] * r[i]));
}
double left,right;
right = s[n - 1 ];
left = 0 ;
double ans = 0 ;
while ((right - left) >= 0.0000001 ) // 注意精度处理
{
double mid = (left + right) / 2 ;
if (ok(mid))
{
ans = mid;
left = mid;
}
else
right = mid;
}
printf( " %.3lf\n " ,ans * pi);
}
}
return 0 ;
}