nyoj 20 吝啬的国度
2015-08-01 19:17
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图论 2012-07-06 18:42
1057人阅读 评论(0)收藏举报
还是《吝啬的国度》,研读了别人的较为高效的代码,无论是时间开销,还是空间开销都要小很多啊,截图为证:
我的所有运行:
52ms的是我参照别人的写的代码,时间开销比我的小将近一半,内存开销仅有我的五分之一。
不过,我自己又将他的改进了一下,虽然内存稍稍大了点,但时间开销为40ms,排名第一啊,哈哈!
WrongAnswer那次是因为忘了在递归中标记start,RuntimeError那次是为了调试时查看数组内容,就把数组开了100,结果调试完忘了改回来,直接提交了,数组越界,唉...
题目中要求从起始城市出发,输出经过每个城市时,之前的那个城市的编号。这样,相邻两个城市之间的关系实质上已经表示出来了,也就是说,存放之前城市编号的那个数组,存储了一个有向图,如下图所示:
输入的测试数据为:10 18 1010 33 710 41 91 88 61 29 5
红色箭头建立的的图,即为数组中存放之前城市编号所建立的有向图。
按照这种思路建立有向图,需要解决箭头方向的问题,即确定哪个城市离出发城市更近。这里有两种方法:
(一),建图的过程中,不管箭头方向,在整个图建立完成之后,从出发城市开始调整箭头方向。
(二),建图的过程中,根据出发城市的位置调整箭头方向,整个图建立完成时,也就是一个正确的有向图。
思路(一)也就是高人的思路,代码如下:
[cpp]
view plaincopyprint?
#include <stdio.h>
#include <memory.h>
int map[100005];
void Adjust(int currentCity)
{
int priorCity = map[currentCity];
if (priorCity != 0)
{
Adjust(priorCity);
map[priorCity] = currentCity;
}
}
int main()
{
int i, testNum, cityNum, startCity, cityA, cityB;
scanf("%d", &testNum);
while (testNum-- != 0)
{
scanf("%d%d", &cityNum, &startCity);
memset(map, 0, sizeof(int)*cityNum + 1);
for (i = 1; i < cityNum; i++)
{
scanf("%d%d", &cityA, &cityB);
if (map[cityB] == 0)
{
map[cityB] = cityA;
}
else
{
Adjust(cityA);
map[cityA] = cityB;
}
}
Adjust(startCity);
map[startCity] = - 1;
for (i = 1; i < cityNum; i++)
{
printf("%d ", map[i]);
}
printf("%d\n", map[i]);
}
return 0;
}
思路(二)是我自己的,因为我发现不考虑方向建立图时,会出现方向的多次调整,影响效率,所以我就想在建立图的过程中就考虑方向,代码如下:
[cpp]
view plaincopyprint?
#include <stdio.h>
#include <memory.h>
int map[100005];
bool flag[100005];
void AdjustIncludeStart(int currentCity)
{
int priorCity = map[currentCity];
if (priorCity != 0)
{
AdjustIncludeStart(priorCity);
map[priorCity] = currentCity;
flag[priorCity] = true;
}
}
void AdjustExcludeStart(int currentCity)
{
int priorCity = map[currentCity];
if (priorCity != 0)
{
AdjustExcludeStart(priorCity);
map[priorCity] = currentCity;
}
}
int main()
{
int i, testNum, cityNum, startCity, cityA, cityB;
scanf("%d", &testNum);
while (testNum-- != 0)
{
scanf("%d%d", &cityNum, &startCity);
memset(map, 0, sizeof(int)*cityNum + 1);
memset(flag, false, sizeof(bool)*cityNum + 1);
map[startCity] = - 1;
flag[startCity] = true;
for (i = 1; i < cityNum; i++)
{
scanf("%d%d", &cityA, &cityB);
if (flag[cityA])
{
if (map[cityB] != 0)
{
AdjustIncludeStart(cityB);
}
map[cityB] = cityA;
flag[cityB] = true;
}
else if (flag[cityB])
{
if (map[cityA] != 0)
{
AdjustIncludeStart(cityA);
}
map[cityA] = cityB;
flag[cityA] = true;
}
else
{
if (map[cityB] == 0)
{
map[cityB] = cityA;
}
else
{
AdjustExcludeStart(cityA);
map[cityA] = cityB;
}
}
}
for (i = 1; i < cityNum; i++)
{
printf("%d ", map[i]);
}
printf("%d\n", map[i]);
}
return 0;
}
NYOJ 20 吝啬的国度 续
分类:图论 2012-07-06 18:42
1057人阅读 评论(0)收藏举报
还是《吝啬的国度》,研读了别人的较为高效的代码,无论是时间开销,还是空间开销都要小很多啊,截图为证:
我的所有运行:
52ms的是我参照别人的写的代码,时间开销比我的小将近一半,内存开销仅有我的五分之一。
不过,我自己又将他的改进了一下,虽然内存稍稍大了点,但时间开销为40ms,排名第一啊,哈哈!
WrongAnswer那次是因为忘了在递归中标记start,RuntimeError那次是为了调试时查看数组内容,就把数组开了100,结果调试完忘了改回来,直接提交了,数组越界,唉...
题目中要求从起始城市出发,输出经过每个城市时,之前的那个城市的编号。这样,相邻两个城市之间的关系实质上已经表示出来了,也就是说,存放之前城市编号的那个数组,存储了一个有向图,如下图所示:
输入的测试数据为:10 18 1010 33 710 41 91 88 61 29 5
红色箭头建立的的图,即为数组中存放之前城市编号所建立的有向图。
按照这种思路建立有向图,需要解决箭头方向的问题,即确定哪个城市离出发城市更近。这里有两种方法:
(一),建图的过程中,不管箭头方向,在整个图建立完成之后,从出发城市开始调整箭头方向。
(二),建图的过程中,根据出发城市的位置调整箭头方向,整个图建立完成时,也就是一个正确的有向图。
思路(一)也就是高人的思路,代码如下:
[cpp]
view plaincopyprint?
#include <stdio.h>
#include <memory.h>
int map[100005];
void Adjust(int currentCity)
{
int priorCity = map[currentCity];
if (priorCity != 0)
{
Adjust(priorCity);
map[priorCity] = currentCity;
}
}
int main()
{
int i, testNum, cityNum, startCity, cityA, cityB;
scanf("%d", &testNum);
while (testNum-- != 0)
{
scanf("%d%d", &cityNum, &startCity);
memset(map, 0, sizeof(int)*cityNum + 1);
for (i = 1; i < cityNum; i++)
{
scanf("%d%d", &cityA, &cityB);
if (map[cityB] == 0)
{
map[cityB] = cityA;
}
else
{
Adjust(cityA);
map[cityA] = cityB;
}
}
Adjust(startCity);
map[startCity] = - 1;
for (i = 1; i < cityNum; i++)
{
printf("%d ", map[i]);
}
printf("%d\n", map[i]);
}
return 0;
}
#include <stdio.h> #include <memory.h> int map[100005]; void Adjust(int currentCity) { int priorCity = map[currentCity]; if (priorCity != 0) { Adjust(priorCity); map[priorCity] = currentCity; } } int main() { int i, testNum, cityNum, startCity, cityA, cityB; scanf("%d", &testNum); while (testNum-- != 0) { scanf("%d%d", &cityNum, &startCity); memset(map, 0, sizeof(int)*cityNum + 1); for (i = 1; i < cityNum; i++) { scanf("%d%d", &cityA, &cityB); if (map[cityB] == 0) { map[cityB] = cityA; } else { Adjust(cityA); map[cityA] = cityB; } } Adjust(startCity); map[startCity] = - 1; for (i = 1; i < cityNum; i++) { printf("%d ", map[i]); } printf("%d\n", map[i]); } return 0; }
思路(二)是我自己的,因为我发现不考虑方向建立图时,会出现方向的多次调整,影响效率,所以我就想在建立图的过程中就考虑方向,代码如下:
[cpp]
view plaincopyprint?
#include <stdio.h>
#include <memory.h>
int map[100005];
bool flag[100005];
void AdjustIncludeStart(int currentCity)
{
int priorCity = map[currentCity];
if (priorCity != 0)
{
AdjustIncludeStart(priorCity);
map[priorCity] = currentCity;
flag[priorCity] = true;
}
}
void AdjustExcludeStart(int currentCity)
{
int priorCity = map[currentCity];
if (priorCity != 0)
{
AdjustExcludeStart(priorCity);
map[priorCity] = currentCity;
}
}
int main()
{
int i, testNum, cityNum, startCity, cityA, cityB;
scanf("%d", &testNum);
while (testNum-- != 0)
{
scanf("%d%d", &cityNum, &startCity);
memset(map, 0, sizeof(int)*cityNum + 1);
memset(flag, false, sizeof(bool)*cityNum + 1);
map[startCity] = - 1;
flag[startCity] = true;
for (i = 1; i < cityNum; i++)
{
scanf("%d%d", &cityA, &cityB);
if (flag[cityA])
{
if (map[cityB] != 0)
{
AdjustIncludeStart(cityB);
}
map[cityB] = cityA;
flag[cityB] = true;
}
else if (flag[cityB])
{
if (map[cityA] != 0)
{
AdjustIncludeStart(cityA);
}
map[cityA] = cityB;
flag[cityA] = true;
}
else
{
if (map[cityB] == 0)
{
map[cityB] = cityA;
}
else
{
AdjustExcludeStart(cityA);
map[cityA] = cityB;
}
}
}
for (i = 1; i < cityNum; i++)
{
printf("%d ", map[i]);
}
printf("%d\n", map[i]);
}
return 0;
}
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