Cow Relays G
題目描述
For their physical fitness program, N (2 ≤ N ≤ 1,000,000) cows have decided to run a relay race using the T (2 ≤ T ≤ 100) cow trails throughout the pasture.
Each trail connects two different intersections (1 ≤ I1i ≤ 1,000; 1 ≤ I2i ≤ 1,000), each of which is the termination for at least two trails. The cows know the lengthi of each trail (1 ≤ lengthi ≤ 1,000), the two intersections the trail connects, and they know that no two intersections are directly connected by two different trails. The trails form a structure known mathematically as a graph.
To run the relay, the N cows position themselves at various intersections (some intersections might have more than one cow). They must position themselves properly so that they can hand off the baton cow-by-cow and end up at the proper finishing place.
Write a program to help position the cows. Find the shortest path that connects the starting intersection (S) and the ending intersection (E) and traverses exactly N cow trails.
輸入格式
* Line 1: Four space-separated integers: N, T, S, and E
* Lines 2..T+1: Line i+1 describes trail i with three space-separated integers: lengthi , I1i , and I2i
輸出格式
* Line 1: A single integer that is the shortest distance from intersection S to intersection E that traverses exactly N cow trails.
樣例 #1
樣例輸入 #1
2 6 6 4
11 4 6
4 4 8
8 4 9
6 6 8
2 6 9
3 8 9
樣例輸出 #1
10
有T條邊連起來的圖最多T+1個點,可以對所有點進行離散化。點的個數是T級別的。
首先有一個暴力的思路就是迴圈n次Floyd,然後輸出s到e的距離。複雜度\(O(nT^3)\).\(T^3\)很難優化,考慮怎麼優化\(n\)
發現Floyd的過程似乎是可以通過倍增來實現的。首先預處理出在\(n=2^i\)時任意兩點之間的距離,然後把n拆成二進位制數一個一個鬆弛上去。複雜度\(O(T^3logn)\),可以通過此題。
#include<bits/stdc++.h>
using namespace std;
int n,t,s,e,u[105],v[105],w[105],m,dp[25][205][205],f[2][205][205],c,lsh[205],fa[1005],x,y;
int find(int x)
{
if(fa[x]==x)
return x;
return fa[x]=find(fa[x]);
}
int main()
{
memset(dp,0x3f,sizeof(dp));
memset(f,0x3f,sizeof(f));
scanf("%d%d%d%d",&t,&m,&s,&e);
for(int i=1;i<=1000;i++)
fa[i]=i;
for(int i=1;i<=m;i++)
{
scanf("%d%d%d",w+i,u+i,v+i);
fa[find(u[i])]=find(v[i]);
}
for(int i=1;i<=1000;i++)
if(find(i)==find(s))
lsh[++n]=i;
for(int i=1;i<=n;i++)
f[1][i][i]=0;
for(int i=1;i<=m;i++)
{
x=lower_bound(lsh+1,lsh+n+1,u[i])-lsh;
y=lower_bound(lsh+1,lsh+n+1,v[i])-lsh;
if(lsh[x]==u[i]&&lsh[y]==v[i])
dp[0][x][y]=dp[0][y][x]=min(dp[0][x][y],w[i]);
}
for(int p=1;(1<<p)<=t;p++)
{
for(int k=1;k<=n;k++)
{
for(int i=1;i<=n;i++)
{
for(int j=1;j<=n;j++)
{
dp[p][i][j]=min(dp[p][i][j],dp[p-1][i][k]+dp[p-1][k][j]);
}
}
}
}
for(int p=0;(1<<p)<=t;p++)
{
if(t&(1<<p))
{
for(int k=1;k<=n;k++)
{
for(int i=1;i<=n;i++)
{
for(int j=1;j<=n;j++)
{
f[c][i][j]=min(f[c][i][j],f[!c][i][k]+dp[p][k][j]);
}
}
}
c^=1;
memset(f[c],0x3f,sizeof(f[c]));
}
}
printf("%d",f[!c][lower_bound(lsh+1,lsh+n+1,s)-lsh][lower_bound(lsh+1,lsh+n+1,e)-lsh]);
return 0;
}