BZOJ 3891 USACO 2014 Dec Piggy Back 搜索 最短路
2016-01-31 17:01
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第一个人从1出发边权A,第二个人从2出发边权B,走到一起边权C,求最短方案使两人走到n。
分别从1,2,n跑一次最短路,然后枚举走到一起的点计算答案即可。
没开快速读入 104ms,开快速读入84ms,第二页和第一页的区别,然而很小。。
#include <cstdio>
#include <cstring>
#include <algorithm>
const int N = 40001, M = 80001;
using namespace std;
int read() {
int s = 0, f = 1; char ch = getchar();
for (; ch < '0' || ch > '9'; ch = getchar()) if (ch == '-') f = -1;
for (; '0' <= ch && ch <= '9'; ch = getchar()) s = s * 10 + ch - '0';
return s * f;
}
int h
, p[M], v[M], cnt = 0;
int f
, g
, t
;
void bfs(int *d, int s) {
static bool vis
;
static int q
;
int i, f = 0, r = 0;
memset(vis, 0, sizeof vis);
q[r++] = s; vis[s] = 1;
while (f < r) {
int u = q[f++];
for(i = h[u]; i; i = p[i])
if (!vis[v[i]]) {
vis[v[i]] = 1; d[v[i]] = d[u] + 1;
q[r++] = v[i];
}
}
}
int main() {
int i, x, y, a = read(), b = read(), c = read(), n = read(), m = read();
long long ans = 1ll<<60;
for (i = 1; i <= m; i++) {
x = read(); y = read();
p[++cnt] = h[x]; v[cnt] = y; h[x] = cnt;
p[++cnt] = h[y]; v[cnt] = x; h[y] = cnt;
}
bfs(f, 1); bfs(g, 2); bfs(t, n);
for (i = 1; i <= n; i++)
ans = min(ans, (long long) a * f[i] + b * g[i] + c * t[i]);
printf("%lld", ans);
return 0;
}
Submit: 164 Solved: 127
[Submit][Status][Discuss]
to rest. Being clever bovines, they come up with a plan to minimize the total amount of energy they both spend while walking. Bessie spends B units of energy when walking from a field to an adjacent field, and Elsie spends E units of energy when she walks
to an adjacent field. However, if Bessie and Elsie are together in the same field, Bessie can carry Elsie on her shoulders and both can move to an adjacent field while spending only P units of energy (where P might be considerably less than B+E, the amount
Bessie and Elsie would have spent individually walking to the adjacent field). If P is very small, the most energy-efficient solution may involve Bessie and Elsie traveling to a common meeting field, then traveling together piggyback for the rest of the journey
to the barn. Of course, if P is large, it may still make the most sense for Bessie and Elsie to travel separately. On a side note, Bessie and Elsie are both unhappy with the term "piggyback", as they don't see why the pigs on the farm should deserve all the
credit for this remarkable form of transportation. Given B, E, and P, as well as the layout of the farm, please compute the minimum amount of energy required for Bessie and Elsie to reach the barn.
给定一个N个点M条边的无向图,其中Bessie在1号点,Elsie在2号点,它们的目的地为N号点。Bessie每经过一条边需要消耗B点能量,Elsie每经过一条边需要消耗E点能量。当它们相遇时,它们可以一起行走,此时它们每经过一条边需要消耗P点能量。求它们两个到达N号点时最少消耗多少能量?
described above. N is the number of fields in the farm (numbered 1..N, where N >= 3), and M is the number of connections between fields. Bessie and Elsie start in fields 1 and 2, respectively. The barn resides in field N. The next M lines in the input each
describe a connection between a pair of different fields, specified by the integer indices of the two fields. Connections are bi-directional. It is always possible to travel from field 1 to field N, and field 2 to field N, along a series of such connections.
Elsie collectively need to spend to reach the barn. In the example
shown here, Bessie travels from 1 to 4 and Elsie travels from 2 to 3
to 4. Then, they travel together from 4 to 7 to 8.
1 4
2 3
3 4
4 7
2 5
5 6
6 8
7 8
分别从1,2,n跑一次最短路,然后枚举走到一起的点计算答案即可。
没开快速读入 104ms,开快速读入84ms,第二页和第一页的区别,然而很小。。
#include <cstdio>
#include <cstring>
#include <algorithm>
const int N = 40001, M = 80001;
using namespace std;
int read() {
int s = 0, f = 1; char ch = getchar();
for (; ch < '0' || ch > '9'; ch = getchar()) if (ch == '-') f = -1;
for (; '0' <= ch && ch <= '9'; ch = getchar()) s = s * 10 + ch - '0';
return s * f;
}
int h
, p[M], v[M], cnt = 0;
int f
, g
, t
;
void bfs(int *d, int s) {
static bool vis
;
static int q
;
int i, f = 0, r = 0;
memset(vis, 0, sizeof vis);
q[r++] = s; vis[s] = 1;
while (f < r) {
int u = q[f++];
for(i = h[u]; i; i = p[i])
if (!vis[v[i]]) {
vis[v[i]] = 1; d[v[i]] = d[u] + 1;
q[r++] = v[i];
}
}
}
int main() {
int i, x, y, a = read(), b = read(), c = read(), n = read(), m = read();
long long ans = 1ll<<60;
for (i = 1; i <= m; i++) {
x = read(); y = read();
p[++cnt] = h[x]; v[cnt] = y; h[x] = cnt;
p[++cnt] = h[y]; v[cnt] = x; h[y] = cnt;
}
bfs(f, 1); bfs(g, 2); bfs(t, n);
for (i = 1; i <= n; i++)
ans = min(ans, (long long) a * f[i] + b * g[i] + c * t[i]);
printf("%lld", ans);
return 0;
}
3891: [Usaco2014 Dec]Piggy Back
Time Limit: 10 Sec Memory Limit: 128 MBSubmit: 164 Solved: 127
[Submit][Status][Discuss]
Description
Bessie and her sister Elsie graze in different fields during the day, and in the evening they both want to walk back to the barnto rest. Being clever bovines, they come up with a plan to minimize the total amount of energy they both spend while walking. Bessie spends B units of energy when walking from a field to an adjacent field, and Elsie spends E units of energy when she walks
to an adjacent field. However, if Bessie and Elsie are together in the same field, Bessie can carry Elsie on her shoulders and both can move to an adjacent field while spending only P units of energy (where P might be considerably less than B+E, the amount
Bessie and Elsie would have spent individually walking to the adjacent field). If P is very small, the most energy-efficient solution may involve Bessie and Elsie traveling to a common meeting field, then traveling together piggyback for the rest of the journey
to the barn. Of course, if P is large, it may still make the most sense for Bessie and Elsie to travel separately. On a side note, Bessie and Elsie are both unhappy with the term "piggyback", as they don't see why the pigs on the farm should deserve all the
credit for this remarkable form of transportation. Given B, E, and P, as well as the layout of the farm, please compute the minimum amount of energy required for Bessie and Elsie to reach the barn.
给定一个N个点M条边的无向图,其中Bessie在1号点,Elsie在2号点,它们的目的地为N号点。Bessie每经过一条边需要消耗B点能量,Elsie每经过一条边需要消耗E点能量。当它们相遇时,它们可以一起行走,此时它们每经过一条边需要消耗P点能量。求它们两个到达N号点时最少消耗多少能量?
Input
The first line of input contains the positive integers B, E, P, N, and M. All of these are at most 40,000. B, E, and P aredescribed above. N is the number of fields in the farm (numbered 1..N, where N >= 3), and M is the number of connections between fields. Bessie and Elsie start in fields 1 and 2, respectively. The barn resides in field N. The next M lines in the input each
describe a connection between a pair of different fields, specified by the integer indices of the two fields. Connections are bi-directional. It is always possible to travel from field 1 to field N, and field 2 to field N, along a series of such connections.
Output
A single integer specifying the minimum amount of energy Bessie andElsie collectively need to spend to reach the barn. In the example
shown here, Bessie travels from 1 to 4 and Elsie travels from 2 to 3
to 4. Then, they travel together from 4 to 7 to 8.
Sample Input
4 4 5 8 81 4
2 3
3 4
4 7
2 5
5 6
6 8
7 8
Sample Output
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