Floyd 算法
/* THE PROGRAM IS MADE BY PYY */
/*----------------------------------------------------------------------------//
Copyright (c) 2011 panyanyany All rights reserved.
URL : http://acm.hdu.edu.cn/showproblem.php?pid=1690
Name : 1690 Bus System
Date : Saturday, January 21, 2012
Time Stage : 7 hours
Result:
5283668 2012-01-21 17:27:37 Accepted 1690
78MS 288K 2952 B
C++ pyy
Test Data :
Review :
一共WA25次，不仅题意坑爹，数据也很坑爹，总结，这是一个坑爹的世界……
本题的路径之和貌似是无限接近 0x7fff,ffff,ffff,ffff 的，所以无穷大已经不保险了，
直接 -1 才是正道啊！
感谢华神的指点！
//----------------------------------------------------------------------------*/
#include <cstdio>
#include <cstring>
#include <queue>
#include <cstdlib>
using namespace std ;
#define INF (-1)
#define MAXN 102
typedef __int64 LL ;
#define min(x, y) ((x) < (y) ? (x) : (y))
#define max(x, y) ((x) > (y) ? (x) : (y))
#define MEM(a, v) memset (a, v, sizeof (a))
bool used[MAXN] ;
int n, m ;
int L[4], C[4], id[MAXN] ;
LL map[MAXN][MAXN] ;
void floyd ()
{
int i, j, k ;
for (k = 1 ; k <= n ; ++k)
for (i = 1 ; i <= n ; ++i)
for (j = 1 ; j <= n ; ++j)
if (map[i][k] != INF && map[k][j] != INF)
{
if (map[i][j] == INF || map[i][j] > map[i][k] + map[k][j])
map[i][j] = map[i][k] + map[k][j] ;
}
}
// 这个已经没用了
int getid (int x)
{
int i ;
for (i = 1 ; i <= n ; ++i)
if (id[i] == x)
return i ;
return 0 ;
}
int getdist (int i, int j)
{
int tmp = id[i] - id[j] ;
if (tmp < 0)
tmp = -tmp ;
if (0 < tmp && tmp <= L[0])
return C[0] ;
if (L[0] < tmp && tmp <= L[1])
return C[1] ;
if (L[1] < tmp && tmp <= L[2])
return C[2] ;
if (L[2] < tmp && tmp <= L[3])
return C[3] ;
return INF ;
}
void makemap ()
{
int i, j ;
MEM (map, INF) ;
for (i = 1 ; i <= n ; ++i)
{
for (j = i + 1 ; j <= n ; ++j)
{
map[i][j] = map[j][i] = getdist (i, j) ;
}
}
}
int main ()
{
int i, j, k ;
int x, y, tcase ;
LL ret ;
scanf ("%d", &tcase) ;
{
k = 0 ;
while (k++ < tcase)
{
for (i = 0 ; i < 4 ; ++i)
{
scanf ("%d", &L[i]) ;
}
for (i = 0 ; i < 4 ; ++i)
{
scanf ("%d", &C[i]) ;
}
scanf ("%d%d", &n, &m) ;
for (i = 1 ; i <= n ; ++i)
{
scanf ("%d", &id[i]) ;
}
makemap () ;
floyd () ;
printf ("Case %d:\n", k) ;
for (i = 0 ; i < m ; ++i)
{
// 这里输入的数字，直接就是下标了，不用 getid() 了
// 一直以为 x,y 有可能是 -10000…… 到 10000……
scanf ("%d%d", &x, &y) ;
ret = map[x][y] ;
if (ret == INF)
printf ("Station %d and station %d are not attainable.\n",
x, y) ;
else
printf (
"The minimum cost between station %d and station %d is %I64d.\n",
x, y, ret) ;
}
}
}
return 0 ;
}
Spfa 算法
/* THE PROGRAM IS MADE BY PYY */
/*----------------------------------------------------------------------------//
Copyright (c) 2011 panyanyany All rights reserved.
URL : http://acm.hdu.edu.cn/showproblem.php?pid=1690
Name : 1690 Bus System
Date : Saturday, January 21, 2012
Time Stage : 7 hours
Result:
5283757 2012-01-21 18:08:50 Accepted 1690
234MS 304K 3430 B
C++ pyy
Test Data :
Review :
本题的路径之和貌似是无限接近 0x7fff,ffff,ffff,ffff 的，所以无穷大已经不保险了，
直接 -1 才是正道啊！
感谢华神的指点！
//----------------------------------------------------------------------------*/
#include <cstdio>
#include <cstring>
#include <queue>
#include <cstdlib>
using namespace std ;
#define INF (-1)
#define MAXN 102
typedef __int64 LL ;
#define min(x, y) ((x) < (y) ? (x) : (y))
#define max(x, y) ((x) > (y) ? (x) : (y))
#define MEM(a, v) memset (a, v, sizeof (a))
bool used[MAXN] ;
int n, m ;
int L[4], C[4], id[MAXN] ;
LL dist[MAXN], map[MAXN][MAXN] ;
LL spfa (const int beg, const int end)
{
int i, t ;
queue<int> q ;
MEM (used, 0) ;
MEM (dist, INF) ;
q.push (beg) ;
used[beg] = 1 ;
dist[beg] = 0 ;
while (!q.empty ())
{
t = q.front () ;
q.pop () ;
for (i = 1 ; i <= n ; ++i)
{
if (dist[t] == INF || map[t][i] == INF)
continue ;
if (dist[i] == INF || dist[i] > dist[t] + map[t][i])
{
dist[i] = dist[t] + map[t][i] ;
if (!used[i])
{
used[i] = 1 ;
q.push (i) ;
}
}
}
used[t] = 0 ;
}
return dist[end] ;
}
// 这个已经没用了
int getid (int x)
{
int i ;
for (i = 1 ; i <= n ; ++i)
if (id[i] == x)
return i ;
return 0 ;
}
int getdist (int i, int j)
{
int tmp = id[i] - id[j] ;
if (tmp < 0)
tmp = -tmp ;
if (0 < tmp && tmp <= L[0])
return C[0] ;
if (L[0] < tmp && tmp <= L[1])
return C[1] ;
if (L[1] < tmp && tmp <= L[2])
return C[2] ;
if (L[2] < tmp && tmp <= L[3])
return C[3] ;
return INF ;
}
void makemap ()
{
int i, j ;
MEM (map, INF) ;
for (i = 1 ; i <= n ; ++i)
{
for (j = i + 1 ; j <= n ; ++j)
{
map[i][j] = map[j][i] = getdist (i, j) ;
}
}
}
int main ()
{
int i, k ;
int x, y, tcase ;
LL ret ;
scanf ("%d", &tcase) ;
{
k = 0 ;
while (k++ < tcase)
{
for (i = 0 ; i < 4 ; ++i)
{
scanf ("%d", &L[i]) ;
}
for (i = 0 ; i < 4 ; ++i)
{
scanf ("%d", &C[i]) ;
}
scanf ("%d%d", &n, &m) ;
for (i = 1 ; i <= n ; ++i)
{
scanf ("%d", &id[i]) ;
}
makemap () ;
printf ("Case %d:\n", k) ;
for (i = 0 ; i < m ; ++i)
{
// 这里输入的数字，直接就是下标了，不用 getid() 了
// 一直以为 x,y 有可能是 -10000…… 到 10000……
scanf ("%d%d", &x, &y) ;
ret = spfa (x, y) ;
if (ret == INF)
printf ("Station %d and station %d are not attainable.\n",
x, y) ;
else
printf (
"The minimum cost between station %d and station %d is %I64d.\n",
x, y, ret) ;
}
}
}
return 0 ;
}
Dijkstra 算法
/* THE PROGRAM IS MADE BY PYY */
/*----------------------------------------------------------------------------//
Copyright (c) 2011 panyanyany All rights reserved.
URL : http://acm.hdu.edu.cn/showproblem.php?pid=1690
Name : 1690 Bus System
Date : Saturday, January 21, 2012
Time Stage : 7 hours
Result:
5283728 2012-01-21 17:54:39 Accepted 1690
468MS 292K 3628 B
C++ pyy
Test Data :
Review :
因为一张图要反复利用，所以 Floyd 会比 Dijkstra 快
本题的路径之和貌似是无限接近 0x7fff,ffff,ffff,ffff 的，所以无穷大已经不保险了，
直接 -1 才是正道啊！
感谢华神的指点！
//----------------------------------------------------------------------------*/
#include <cstdio>
#include <cstring>
#include <queue>
#include <cstdlib>
using namespace std ;
#define INF (-1)
#define MAXN 102
typedef __int64 LL ;
#define min(x, y) ((x) < (y) ? (x) : (y))
#define max(x, y) ((x) > (y) ? (x) : (y))
#define MEM(a, v) memset (a, v, sizeof (a))
bool used[MAXN] ;
int n, m ;
int L[4], C[4], id[MAXN] ;
LL dist[MAXN], map[MAXN][MAXN] ;
LL dijkstra (const int beg, const int end)
{
int i, j ;
int iMinPath ;
LL MinPath ;
MEM (used, 0) ;
for (i = 1 ; i <= n ; ++i)
dist[i] = map[beg][i] ;
for (i = 1 ; i <= n ; ++i)
{
iMinPath = 0 ;
MinPath = INF ;
for (j = 1 ; j <= n ; ++j)
{
if (used[j] || dist[j] == INF)
continue ;
if (MinPath == INF || dist[j] < MinPath)
{
iMinPath = j ;
MinPath = dist[j] ;
}
}
used[iMinPath] = 1 ;
for (j = 1 ; j <= n ; ++j)
{
if (used[j])
continue ;
if (dist[iMinPath] == INF || map[iMinPath][j] == INF)
continue ;
if (dist[j] == INF || dist[iMinPath] + map[iMinPath][j] < dist[j])
dist[j] = dist[iMinPath] + map[iMinPath][j] ;
}
}
return dist[end] ;
}
// 这个已经没用了
int getid (int x)
{
int i ;
for (i = 1 ; i <= n ; ++i)
if (id[i] == x)
return i ;
return 0 ;
}
int getdist (int i, int j)
{
int tmp = id[i] - id[j] ;
if (tmp < 0)
tmp = -tmp ;
if (0 < tmp && tmp <= L[0])
return C[0] ;
if (L[0] < tmp && tmp <= L[1])
return C[1] ;
if (L[1] < tmp && tmp <= L[2])
return C[2] ;
if (L[2] < tmp && tmp <= L[3])
return C[3] ;
return INF ;
}
void makemap ()
{
int i, j ;
MEM (map, INF) ;
for (i = 1 ; i <= n ; ++i)
{
for (j = i + 1 ; j <= n ; ++j)
{
map[i][j] = map[j][i] = getdist (i, j) ;
}
}
}
int main ()
{
int i, k ;
int x, y, tcase ;
LL ret ;
scanf ("%d", &tcase) ;
{
k = 0 ;
while (k++ < tcase)
{
for (i = 0 ; i < 4 ; ++i)
{
scanf ("%d", &L[i]) ;
}
for (i = 0 ; i < 4 ; ++i)
{
scanf ("%d", &C[i]) ;
}
scanf ("%d%d", &n, &m) ;
for (i = 1 ; i <= n ; ++i)
{
scanf ("%d", &id[i]) ;
}
makemap () ;
printf ("Case %d:\n", k) ;
for (i = 0 ; i < m ; ++i)
{
// 这里输入的数字，直接就是下标了，不用 getid() 了
// 一直以为 x,y 有可能是 -10000…… 到 10000……
scanf ("%d%d", &x, &y) ;
ret = dijkstra (x, y) ;
if (ret == INF)
printf ("Station %d and station %d are not attainable.\n",
x, y) ;
else
printf (
"The minimum cost between station %d and station %d is %I64d.\n",
x, y, ret) ;
}
}
}
return 0 ;
}
