Time Limit: 1000MS Memory Limit: 65536K Total Submissions: 7665 Accepted: 3008 Description
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.
Input
Line 1: Four space-separated integers: N, T, S, and ELines 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 Input
2 6 6 4 11 4 6 4 4 8 8 4 9 6 6 8 2 6 9 3 8 9 Sample Output
10 Source
USACO 2007 November Gold
/*求从S到T恰好经过K条边的最短路径的数目 01邻接矩阵A的K次方C=A^K,C[i][j]表示i点到j点正好 经过K条边的路径数 对应于这道题,对邻接图进行K次floyd之后, C[i][j]就是点i到j正好经过K条边的最短路 但是K次floyd难免复杂度太高了. 所以可以使用快速幂的方法，二分的往上求解 */ #include<iostream> #include<cstring> #include<cstdio> #include<cmath> #include<cstdlib> #define maxn 1005 #define maxm 500005 #define INF 1000000000 using namespace std; int k,n,m,S,T; int map[maxn][maxn],tmp[maxn][maxn],dis[maxn][maxn]; int num,ans[maxn][maxn],v[maxn],exist[maxn]; void Floyd(int c[][maxn],int a[][maxn],int b[][maxn]){ for(int k=0;k<num;k++) for(int i=0;i<num;i++) for(int j=0;j<num;j++) if(c[v[i]][v[j]]>a[v[i]][v[k]]+b[v[k]][v[j]]) c[v[i]][v[j]]=a[v[i]][v[k]]+b[v[k]][v[j]]; } void copy(int a[][maxn],int b[][maxn]){ for(int i=0;i<num;i++) for(int j=0;j<num;j++){ a[v[i]][v[j]]=b[v[i]][v[j]]; b[v[i]][v[j]]=INF; } } void Solve(int k){ while(k){ if(k&1){ Floyd(dis,ans,map); copy(ans,dis); } Floyd(tmp,map,map); copy(map,tmp); k >>= 1; } } int main(){ scanf("%d%d%d%d",&k,&m,&S,&T); for(int i=0;i<=1000;i++){ for(int j=0;j<=1000;j++){ map[i][j]=INF; dis[i][j]=INF; ans[i][j]=INF; tmp[i][j]=INF; } ans[i][i]=0; } num=0; int y,x,w; for(int i=0;i<m;i++){ scanf("%d%d%d",&w,&x,&y); if(!exist[x]) exist[x]=1,v[num++]=x; if(!exist[y]) exist[y]=1,v[num++]=y; if(map[x][y]>w) map[x][y]=map[y][x]=w; } Solve(k); printf("%d\n",ans[S][T]); return 0; }