bzoj3891: [Usaco2014 Dec]Piggy Back
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3891: [Usaco2014 Dec]Piggy Back
Time Limit: 10 Sec Memory Limit: 128 MBSubmit: 151 Solved: 120
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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 barn 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.
Input
Output
Sample Input
1 4
2 3
3 4
4 7
2 5
5 6
6 8
7 8
Sample Output
HINT
Source
Silver
枚举相遇的点。
<span style="font-size:18px;">#include<iostream>#include<cstdio>#include<cstdlib>#include<cstring>#include<algorithm>#include<cmath>#include<queue>#define ll long long#define maxn 40400using namespace std;int read(){int x=0,f=1;char ch=getchar();while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}return x*f;}struct Node{int to,next;}e[maxn<<1];queue<int> q;int head[maxn],f[maxn][3],A,B,C,tot=0,n,m;bool v[maxn];void add(int x,int y){e[++tot]=(Node){y,head[x]};head[x]=tot;}void bfs(int s,int k){q.push(s);while(!q.empty()){int x=q.front();q.pop();for(int i=head[x];i;i=e[i].next)if(!v[e[i].to]){v[e[i].to]=1;f[e[i].to][k]=f[x][k]+1;q.push(e[i].to);}}}int main(){ A=read();B=read();C=read();n=read();m=read(); int x,y; for(int i=1;i<=m;i++) { x=read();y=read();add(x,y);add(y,x); } memset(v,0,sizeof(v));v[1]=1;f[1][0]=0;bfs(1,0); memset(v,0,sizeof(v));v[2]=1;f[2][1]=0;bfs(2,1); memset(v,0,sizeof(v));v[n]=1;f[n][2]=0;bfs(n,2); ll ans=0x7fffffff; for(int i=1;i<=n;i++) { ans=min(ans,(ll)A*f[i][0]+B*f[i][1]+C*f[i][2]); } cout<<ans<<endl; return 0;}</span>
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