poj Cow Relays 矩阵乘法思想与floyd
2012-05-03 19:30
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Cow Relays
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 E
* Lines 2..T+1: Line i+1 describes trail i with three space-separated integers: lengthi , I1i , and I2i
Output
* Line 1: A single integer that is the shortest distance from intersection S to intersection E that traverses exactly N cow trails.
Sample Input
Sample Output
Source
USACO 2007 November Gold
题目求i,j之间边数恰为N的最短路径(边可以重复走),我们知道线性代数中有: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。
而floyd则是每次使用一个中间点k去更新i,j之间的距离,那么更新成功表示i,j之间恰有一个点k时的最短路,如果做N次floyd那么不就是i,j之间借助N个点时的最短路了?
考虑当a[i][k]+a[k][j]<c[i][j]的时候,c[i][j]=a[i][k]+a[k][j],这样c[i][j]保存了i,j之间有一个点的最短路,第二次将c[i][j]拷贝回到a[i][j]当中,并将c[i][j]重新置为INF,再做一次,则是在原来的基础上在i,j之间再用一个点k来松弛,这时候i,j之间实际上已经是两个点了,之后重复这么做就好了,可以利用二进制加速
贴一个网上的code:
Time Limit: 1000MS | Memory Limit: 65536K | |
Total Submissions: 3507 | Accepted: 1388 |
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 E
* Lines 2..T+1: Line i+1 describes trail i with three space-separated integers: lengthi , I1i , and I2i
Output
* 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
题目求i,j之间边数恰为N的最短路径(边可以重复走),我们知道线性代数中有: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。
而floyd则是每次使用一个中间点k去更新i,j之间的距离,那么更新成功表示i,j之间恰有一个点k时的最短路,如果做N次floyd那么不就是i,j之间借助N个点时的最短路了?
考虑当a[i][k]+a[k][j]<c[i][j]的时候,c[i][j]=a[i][k]+a[k][j],这样c[i][j]保存了i,j之间有一个点的最短路,第二次将c[i][j]拷贝回到a[i][j]当中,并将c[i][j]重新置为INF,再做一次,则是在原来的基础上在i,j之间再用一个点k来松弛,这时候i,j之间实际上已经是两个点了,之后重复这么做就好了,可以利用二进制加速
贴一个网上的code:
#include<iostream> #include<cstdio> #include<cstring> using namespace std; #define inf 1000000001 #define maxn 1010 int K,M,S,T; int v[maxn],cnt,map[maxn][maxn],used[maxn]; int ans[maxn][maxn],dis[maxn][maxn],tmp[maxn][maxn]; void floyd(int c[][maxn],int a[][maxn],int b[maxn][maxn]) { int i,j,k; for(k=0;k<cnt;k++){ for(i=0;i<cnt;i++){ for(j=0;j<cnt;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]) { int i,j; for(i=0;i<cnt;i++) { for(j=0;j<cnt;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%2) { floyd(dis,ans,map); //有1的位加到ans copy(ans,dis); } floyd(tmp,map,map); // 1,2,4,8...... copy(map,tmp); k=k/2; } } int main() { int i,j; int x,y,val; while(scanf("%d%d%d%d",&K,&M,&S,&T)==4) { for(i=0;i<=1001;i++) { for(j=0;j<=1001;j++) { map[i][j]=inf; ans[i][j]=inf; tmp[i][j]=inf; dis[i][j]=inf; } ans[i][i]=0; } memset(used,0,sizeof(used)); cnt=0; for(i=0;i<M;i++) { scanf("%d%d%d",&val,&x,&y); if(map[x][y]>val) { map[x][y]=val; map[y][x]=map[x][y]; } if(!used[x]) { used[x]=1; v[cnt++]=x; } if(!used[y]) { used[y]=1; v[cnt++]=y; } } solve(K); printf("%d\n",ans[S][T]); } return 0; }
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