It was a dark and stormy night that ripped the roof and gates off the stalls that hold Farmer John's cows. Happily, many of the cows were on vacation, so the barn was not completely full.
The cows spend the night in stalls that are arranged adjacent to each other in a long line. Some stalls have cows in them; some do not. All stalls are the same width.
Farmer John must quickly erect new boards in front of the stalls, since the doors were lost. His new lumber supplier will supply him boards of any length he wishes, but the supplier can only deliver a small number of total boards. Farmer John wishes to minimize the total length of the boards he must purchase.
Given M (1 <= M <= 50), the maximum number of boards that can be purchased; S (1 <= S <= 200), the total number of stalls; C (1 <= C <= S) the number of cows in the stalls, and the C occupied stall numbers (1 <= stall_number <= S), calculate the minimum number of stalls that must be blocked in order to block all the stalls that have cows in them.
Print your answer as the total number of stalls blocked.
PROGRAM NAME: barn1
INPUT FORMAT
| Line 1: | M, S, and C (space separated) |
| Lines 2-C+1: | Each line contains one integer, the number of an occupied stall. |
SAMPLE INPUT (file barn1.in)
4 50 18 3 4 6 8 14 15 16 17 21 25 26 27 30 31 40 41 42 43
OUTPUT FORMAT
A single line with one integer that represents the total number of stalls blocked.SAMPLE OUTPUT (file barn1.out)
25[One minimum arrangement is one board covering stalls 3-8, one covering 14-21, one covering 25-31, and one covering 40-43.]
解题思路：题意是将一个区间分为N个小区间使得小区间间隔的和最小。
1：将区间排序为有序区间
2：算出每个区间间隔并排序
3：将区间尾首相减在减去允许分的M个最大间隔
当允许划分的间隔大于区间个数直接输出区间个数即可
/*
ID:nealgav1
PROG:barn1
LANG:C++
*/
#include<cstdio>
#include<algorithm>
#define N 1234
using namespace std;
int line[N];
bool cmp(int a,int b)
{
return a<b;
}
int main()
{
freopen("barn1.in","r",stdin);
freopen("barn1.out","w",stdout);
int n,m,cas;
while(scanf("%d%d%d",&n,&m,&cas)!=EOF)
{
for(int i=0;i<cas;i++)
scanf("%d",&line[i]);
if(n>cas)
printf("%d\n",cas);
else
{
sort(line,line+cas,cmp);
int ans=line[cas-1]-line[0];
for(int i=0;i<cas-1;i++)
line[i]=line[i+1]-line[i];
sort(line,line+cas-1,cmp);
int j=cas-2;
for(int i=0;i<n-1;i++)//减少一个
ans-=line[j--];
printf("%d\n",ans+n);
}
}
}
If we can purchase M boards, then we can leave unblocked M-1 runs of stalls without cows in them, in addition to any stalls on the leftmost side that don't have cows and any stalls on the rightmost side that don't have cows.
We input the list of cows in stalls, storing into an array whether or not there is a cow in a particular stall. Then we walk the array counting sizes of runs of cowless stalls. We sort the list of sizes and pick the M-1 largest ones as the stalls that will remain uncovered.
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <assert.h>
#define MAXSTALL 200
int hascow[MAXSTALL];
int
intcmp(const void *va, const void *vb)
{
return *(int*)vb - *(int*)va;
}
void
main(void)
{
FILE *fin, *fout;
int n, m, nstall, ncow, i, j, c, lo, hi, nrun;
int run[MAXSTALL];
fin = fopen("barn1.in", "r");
fout = fopen("barn1.out", "w");
assert(fin != NULL && fout != NULL);
fscanf(fin, "%d %d %d", &m, &nstall, &ncow);
for(i=0; i<ncow; i++) {
fscanf(fin, "%d", &c);
hascow[c-1] = 1;
}
n = 0; /* answer: no. of uncovered stalls */
/* count empty stalls on left */
for(i=0; i<nstall && !hascow[i]; i++)
n++;
lo = i;
/* count empty stalls on right */
for(i=nstall-1; i>=0 && !hascow[i]; i--)
n++;
hi = i+1;
/* count runs of empty stalls */
nrun = 0;
i = lo;
while(i < hi) {
while(hascow[i] && i<hi)
i++;
for(j=i; j<hi && !hascow[j]; j++)
;
run[nrun++] = j-i;
i = j;
}
/* sort list of runs */
qsort(run, nrun, sizeof(run[0]), intcmp);
/* uncover best m-1 runs */
for(i=0; i<nrun && i<m-1; i++)
n += run[i];
fprintf(fout, "%d\n", nstall-n);
exit(0);
}
Alexandru Tudorica's solution might be simpler: var f:text;
a,b:array[1..1000] of longint;
i,m,s,c,k:longint;
procedure qsort(l,r:longint);
var i,j,x,y:longint;
begin
i:=l; j:=r; x:=a[(l+r) div 2];
repeat
while a[i]<x do i:=i+1;
while x<a[j] do j:=j-1;
if i<=j then begin
y:=a[i]; a[i]:=a[j]; a[j]:=y;
i:=i+1;
j:=j-1;
end;
until i>j;
if l<j then qsort(l,j);
if i<r then qsort(i,r);
end;
procedure qsortb(l,r:longint);
var i,j,x,y:longint;
begin
i:=l; j:=r; x:=b[(l+r) div 2];
repeat
while b[i]<x do i:=i+1;
while x<b[j] do j:=j-1;
if i<=j then
begin
y:=b[i]; b[i]:=b[j]; b[j]:=y;
i:=i+1;
j:=j-1;
end;
until i>j;
if l<j then qsortb(l,j);
if i<r then qsortb(i,r);
end;
begin
assign(f,'barn1.in');
reset(f);
readln(f,m,k,c);
for i:=1 to c do readln(f,a[i]);
qsort(1,c);
for i:=1 to c-1 do b[i]:=a[i+1]-a[i]-1;
qsortb(1,c-1);
for i:=c-1 downto (c-m+1) do s:=s+b[i];
close(f);
assign(f,'barn1.out');
rewrite(f);
writeln(f,a[c]-a[1]-s+1);
close(f);
end.
USER: Neal Gavin Gavin [nealgav1] TASK: barn1 LANG: C++ Compiling... Compile: OK Executing... Test 1: TEST OK [0.000 secs, 3352 KB] Test 2: TEST OK [0.000 secs, 3352 KB] Test 3: TEST OK [0.000 secs, 3352 KB] Test 4: TEST OK [0.000 secs, 3352 KB] Test 5: TEST OK [0.000 secs, 3352 KB] Test 6: TEST OK [0.000 secs, 3352 KB] Test 7: TEST OK [0.000 secs, 3352 KB] Test 8: TEST OK [0.000 secs, 3352 KB] Test 9: TEST OK [0.000 secs, 3352 KB] Test 10: TEST OK [0.000 secs, 3352 KB] All tests OK.
Your program ('barn1') produced all correct answers! This is your submission #2 for this problem. Congratulations!
Here are the test data inputs:
------- test 1 ---- 4 50 17 3 4 6 8 14 15 16 17 25 26 27 30 31 40 41 42 43 ------- test 2 ---- 2 10 4 2 4 6 8 ------- test 3 ---- 3 27 16 2 3 5 6 8 9 10 13 14 15 16 19 20 21 22 27 ------- test 4 ---- 1 200 8 101 105 102 106 103 107 104 99 ------- test 5 ---- 50 200 10 18 69 195 38 73 28 6 172 53 99 ------- test 6 ---- 50 30 6 30 25 20 15 10 5 ------- test 7 ---- 20 200 80 65 178 64 70 18 32 88 90 98 20 152 31 118 117 127 81 175 73 136 161 165 63 130 133 190 10 4 138 200 43 189 37 86 182 145 110 67 126 114 153 99 25 155 119 176 55 48 197 62 147 125 60 12 23 112 96 27 122 35 50 36 49 149 108 100 188 77 191 6 121 166 132 82 95 150 89 22 40 128 56 ------- test 8 ---- 4 200 100 72 180 46 198 196 131 165 112 52 133 187 93 57 35 128 65 127 130 12 49 88 155 122 193 101 164 98 143 54 149 38 84 45 139 79 16 102 20 14 150 188 33 176 135 29 80 19 74 11 114 95 185 137 59 32 189 66 67 191 91 77 134 18 10 7 200 8 13 55 24 142 184 17 6 109 105 43 181 85 94 151 160 115 25 116 111 37 104 144 97 90 141 120 119 152 182 123 172 40 23 ------- test 9 ---- 20 195 100 1 2 3 4 5 11 12 13 14 15 21 22 23 24 25 31 32 33 34 35 41 42 43 44 45 51 52 53 54 55 61 62 63 64 65 71 72 73 74 75 81 82 83 84 85 91 92 93 94 95 101 102 103 104 105 111 112 113 114 115 121 122 123 124 125 131 132 133 134 135 141 142 143 144 145 151 152 153 154 155 161 162 163 164 165 171 172 173 174 175 181 182 183 184 185 191 192 193 194 195 ------- test 10 ---- 1 200 2 1 200
