Problem E - Steps

时间： 1ms 内存：128M

描述：

## Problem E - Steps

One steps through integer points of the straight line. The length of a step must be nonnegative and can be by one bigger than, equal to, or by one smaller than the length of the previous step.

What is the minimum number of steps in order to get from x to y? The length of the first and the last step must be 1.

Input consists of a line containing n, the number of test cases. For each test case, a line follows with two integers: 0 <= x <= y < 2^31. For each test case, print a line giving the minimum number of steps to get from x to y.

输入：

输出：

示例输入：

```
3
45 48
45 49
45 50
```

示例输出：

```
3
3
4
```

提示：

参考答案（内存最优[784]）：

```
/* Solution for "bridge" problem, a generalization of a problem
generally believed to be a Microsoft intelligence test.
*/
/* Algorithm
Observation 1: somebody has to get the flashlight back after every
forward trip. This should obviously be the fastest
person on the far shore.
Observation 2: in the general case, we have 4 or more people, the
flashlight is on the starting side.
Consider the two slowest people.
there are 2 sensible strategies; we pick the fastest:
1. we can send each with the fastest person, who returns
with the flashlight. this takes 2a+x+y time where a is
the fastest and x and y are the two slowest
2. we can send the fastest & 2nd fastest person, have
the fastest return with the flashlight, send the
slowpokes together, have the 2nd fastest return with
the flashlight. This takes a+2b+y where a is fastest,
b second fastest, and y slowest
BUT WAIT, you say. What if strategy 1 is better than
2 for x and y but perhaps x or y could have been paired
with somebody else. We already know that
2a+x+y <= a+2b+y
x <= 2b - a
Now if we had a different x' and y', it would be
the case that x' <= x and y' <= y so
x' <= 2b - a
so for any other pair 1 will be the best strategy.
BUT WAIT, you say again. What if strategy 2 is better
than strategy 1, but prevents strategy 2 from being used
later, which would have been even better.
Let a,b be the 2 fastest and w,x,y be the slowest.
If we do x,y together we (may) force w to be done alone.
x,y together: a+2b+y
w alone: a+w
total: 2a+2b+w+y
If we do y alone, then w,x together.
y alone: a+y
w,x together: a+2b+x
total: 2a+2b+x+y
Since x >= w, this is worse so there's no point in
deferring the pairing.
Observation 3: The end game. If there are 3 people left, the fastest
shuttles them across. If there are 2 people left,
they go together.
Observation 4: One person alone is a special case. Happens only if
this person started alone.
*/
#include <stdio.h>
#include <stdlib.h>
int i,j,k,n;
int t[1000];
char cross[2000][16];
int nc,tot;
compar(int **a, int **b){
return *a - *b;
}
main(){
scanf("%d",&n);
for (i=0;i<n;i++) scanf("%d",&t[i]);
qsort(t,n,sizeof(int),compar);
for (i=n-1;i>=3 && 2*t[1] < t[0]+t[i-1];i-=2){
sprintf(cross[nc++],"%d %d\n",t[0],t[1]);
sprintf(cross[nc++],"%d\n",t[0]);
sprintf(cross[nc++],"%d %d\n",t[i-1],t[i]);
sprintf(cross[nc++],"%d\n",t[1]);
tot = tot + t[0]+2*t[1]+t[i];
}
for (;i>=2;i--){
sprintf(cross[nc++],"%d %d\n",t[0],t[i]);
sprintf(cross[nc++],"%d\n",t[0]);
tot = tot + t[0] + t[i];
}
if (i==1) {
sprintf(cross[nc++],"%d %d\n",t[0],t[1]);
tot = tot + t[1];
}else{
sprintf(cross[nc++],"%d\n",t[0]);
tot = tot + t[0];
}
printf("%d\n",tot);
for (i=0;i<nc;i++) printf("%s",cross[i]);
}
```

参考答案（时间最优[16]）：

```
/* Solution for "bridge" problem, a generalization of a problem
generally believed to be a Microsoft intelligence test.
*/
/* Algorithm
Observation 1: somebody has to get the flashlight back after every
forward trip. This should obviously be the fastest
person on the far shore.
Observation 2: in the general case, we have 4 or more people, the
flashlight is on the starting side.
Consider the two slowest people.
there are 2 sensible strategies; we pick the fastest:
1. we can send each with the fastest person, who returns
with the flashlight. this takes 2a+x+y time where a is
the fastest and x and y are the two slowest
2. we can send the fastest & 2nd fastest person, have
the fastest return with the flashlight, send the
slowpokes together, have the 2nd fastest return with
the flashlight. This takes a+2b+y where a is fastest,
b second fastest, and y slowest
BUT WAIT, you say. What if strategy 1 is better than
2 for x and y but perhaps x or y could have been paired
with somebody else. We already know that
2a+x+y <= a+2b+y
x <= 2b - a
Now if we had a different x' and y', it would be
the case that x' <= x and y' <= y so
x' <= 2b - a
so for any other pair 1 will be the best strategy.
BUT WAIT, you say again. What if strategy 2 is better
than strategy 1, but prevents strategy 2 from being used
later, which would have been even better.
Let a,b be the 2 fastest and w,x,y be the slowest.
If we do x,y together we (may) force w to be done alone.
x,y together: a+2b+y
w alone: a+w
total: 2a+2b+w+y
If we do y alone, then w,x together.
y alone: a+y
w,x together: a+2b+x
total: 2a+2b+x+y
Since x >= w, this is worse so there's no point in
deferring the pairing.
Observation 3: The end game. If there are 3 people left, the fastest
shuttles them across. If there are 2 people left,
they go together.
Observation 4: One person alone is a special case. Happens only if
this person started alone.
*/
#include <stdio.h>
#include <stdlib.h>
int i,j,k,n;
int t[1000];
char cross[2000][16];
int nc,tot;
compar(int **a, int **b){
return *a - *b;
}
main(){
scanf("%d",&n);
for (i=0;i<n;i++) scanf("%d",&t[i]);
qsort(t,n,sizeof(int),compar);
for (i=n-1;i>=3 && 2*t[1] < t[0]+t[i-1];i-=2){
sprintf(cross[nc++],"%d %d\n",t[0],t[1]);
sprintf(cross[nc++],"%d\n",t[0]);
sprintf(cross[nc++],"%d %d\n",t[i-1],t[i]);
sprintf(cross[nc++],"%d\n",t[1]);
tot = tot + t[0]+2*t[1]+t[i];
}
for (;i>=2;i--){
sprintf(cross[nc++],"%d %d\n",t[0],t[i]);
sprintf(cross[nc++],"%d\n",t[0]);
tot = tot + t[0] + t[i];
}
if (i==1) {
sprintf(cross[nc++],"%d %d\n",t[0],t[1]);
tot = tot + t[1];
}else{
sprintf(cross[nc++],"%d\n",t[0]);
tot = tot + t[0];
}
printf("%d\n",tot);
for (i=0;i<nc;i++) printf("%s",cross[i]);
}
```

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