USACO Apple Delivery
洛谷 P3003 [USACO10DEC]苹果交货Apple Delivery
JDOJ 2717: USACO 2010 Dec Silver 1.Apple Delivery
Description
Bessie has two crisp red apples to deliver to two of her friends
in the herd. Of course, she travels the C (1 <= C <= 200,000)
cowpaths which are arranged as the usual graph which connects P (1
<= P <= 100,000) pastures conveniently numbered from 1..P: no cowpath
leads from a pasture to itself, cowpaths are bidirectional, each
cowpath has an associated distance, and, best of all, it is always
possible to get from any pasture to any other pasture. Each cowpath
connects two differing pastures P1_i (1 <= P1_i <= P) and P2_i (1
<= P2_i <= P) with a distance between them of D_i. The sum of all
the distances D_i does not exceed 2,000,000,000.
What is the minimum total distance Bessie must travel to deliver
both apples by starting at pasture PB (1 <= PB <= P) and visiting
pastures PA1 (1 <= PA1 <= P) and PA2 (1 <= PA2 <= P) in any order.
All three of these pastures are distinct, of course.
Consider this map of bracketed pasture numbers and cowpaths with
distances:
3 2 2
[1]-----[2]------[3]-----[4]
\ / \ /
7\ /4 \3 /2
\ / \ /
[5]-----[6]------[7]
1 2
If Bessie starts at pasture [5] and delivers apples to pastures [1]
and [4], her best path is:
5 -> 6-> 7 -> 4* -> 3 -> 2 -> 1*
with a total distance of 12.
Input
* Line 1: Line 1 contains five space-separated integers: C, P, PB,
PA1, and PA2
* Lines 2..C+1: Line i+1 describes cowpath i by naming two pastures it
connects and the distance between them: P1_i, P2_i, D_i
Output
* Line 1: The shortest distance Bessie must travel to deliver both
apples
Sample Input
9 7 5 1 4 5 1 7 6 7 2 4 7 2 5 6 1 5 2 4 4 3 2 1 2 3 3 2 2 2 6 3
Sample Output
12
题目翻译：
贝西有两个又香又脆的红苹果要送给她的两个朋友。当然她可以走的C（1<=C<=200000）条“牛路”都被包含在一种常用的图中，包含了P（1<=P<=100000）个牧场，分别被标为1..P。没有“牛路”会从一个牧场又走回它自己。“牛路”是双向的，每条牛路都会被标上一个距离。最重要的是，每个牧场都可以通向另一个牧场。每条牛路都连接着两个不同的牧场P1_i和P2_i（1<=P1_i,p2_i<=P)，距离为D_i。所有“牛路”的距离之和不大于2000000000。
现在，贝西要从牧场PB开始给PA_1和PA_2牧场各送一个苹果（PA_1和PA_2顺序可以调换），那么最短的距离是多少呢？当然，PB、PA_1和PA_2各不相同。
题解：
这道题如果用裸的SPFA做会TLE，所以要加优化。
如果有像20分钟之前的我一样对SPFA算法优化一无所知的人，请移步我的上一篇博客：
但是光知道怎么优化是不够的，我们还要就这个题想一想怎么写。
首先，这个题的源点不再是一，所以SPFA的时候要传参数。
其次，最后统计答案的时候要比较一下。
最后，拍一遍SPFA模板（SLF LLL优化版都可以）（本蒟蒻比较喜欢SLF），AC。
注意双向边的问题。
CODE:
#include<cstdio>
#include<cstring>
#include<queue>
#include<algorithm>
using namespace std;
int m,n,s,t1,t2;
int tot,to[400001],val[400001],nxt[400001],head[100001];
int dist[100001],v[100001];
void add(int x,int y,int z)
{
to[++tot]=y;
val[tot]=z;
nxt[tot]=head[x];
head[x]=tot;
}
void spfa(int start)
{
memset(v,0,sizeof(v));
memset(dist,0x3f,sizeof(dist));
deque<int> q;
dist[start]=0;
v[start]=1;
q.push_front(start);
while(!q.empty())
{
int x=q.front();
q.pop_front();
v[x]=0;
for(int i=head[x];i;i=nxt[i])
{
int y=to[i];
if(dist[x]+val[i]<dist[y])
{
dist[y]=dist[x]+val[i];
if(!v[y])
{
if(!q.empty() && dist[y]<dist[q.front()])
q.push_front(y);
else
q.push_back(y);
v[y]=1;
}
}
}
}
}
int main()
{
scanf("%d%d%d%d%d",&m,&n,&s,&t1,&t2);
for(int i=1;i<=m;i++)
{
int x,y,z;
scanf("%d%d%d",&x,&y,&z);
add(x,y,z);
add(y,x,z);
}
int ans1=0;
int ans2=0;
spfa(t1);
ans1+=dist[t2]+dist[s];
spfa(t2);
ans2+=dist[t1]+dist[s];
int ans=min(ans1,ans2);
printf("%d",ans);
return 0;
}
