poj 3613Cow Relays

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DescriptionFor 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.Input* 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 I2iOutput* Line 1: A single integer that is the shortest distance from intersection S to intersection E that traverses exactly N cow trails.Sample Input2 6 6 411 4 64 4 88 4 96 6 82 6 93 8 9Sample Output10SourceUSACO 2007 November Gold

題面

用一個矩陣a(i, j)來表示i到j經過若干條邊的最短路，
初始化a為i到j邊的長度，沒有則是正無窮。
比如a矩陣表示經過n條邊，b矩陣表示經過m條邊，
那麼a * b得到的矩陣表示經過m + n條邊，
採用Floyd的思想進行更新。

 

 1 #include<iostream>  2 #include<cstring> 3 #include<cstdio> 4 #include<algorithm> 5 #include<cmath> 6 #include<queue> 7 #include<string>  8 #include<map> 9 #define ll long long10 using namespace std;11 ll n,m,S,T,l;12 map<ll,ll>id; 13 struct node{14     ll a[202][202];15     friend node operator *(node x,node y)16     {17          node z;18          memset(z.a,0x3f,sizeof(z.a));19          for(ll k=1;k<=l;k++)20           for(ll i=1;i<=l;++i)21            for(ll j=1;j<=l;++j)22             z.a[i][j]=min(z.a[i][j],x.a[i][k]+y.a[k][j]);23          return z;24     } 25 }s,ans; 26 void ksm()27 {28     ans=s;29     n--;30     while(n)31     {32         if(n&1) ans=ans*s;33         s=s*s;34         n>>=1;35     }36 }37 int main()38 {39     freopen("run.in","r",stdin);40     freopen("run.out","w",stdout);41     memset(s.a,0x3f,sizeof(s.a));42     scanf("%lld%lld%lld%lld",&n,&m,&S,&T);43     for(ll i=1,x,y,z;i<=m;++i)44     {45        scanf("%lld%lld%lld",&z,&x,&y);46        if(id[x]) x=id[x];47        else l++,id[x]=l,x=l;48        if(id[y]) y=id[y];49        else l++,id[y]=l,y=l;50        s.a[x][y]=s.a[y][x]=z;51     }52     S=id[S];T=id[T];53     ksm();54     printf("%lld",ans.a[S][T]);55     return 0;56 } 57 /*58 2 6 6 459 11 4 660 4 4 861 8 4 962 6 6 863 2 6 964 3 8 965 1066 */

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poj 3613Cow Relays