Cow Relays
| Time Limit: 1000MS | Memory Limit: 65536K | |
| Total Submissions: 7825 | Accepted: 3068 |
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
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
题意
给出一张图，求k边最短路，即经过k条边的最短路。
分析
思考一下：如果一个矩阵，表示走k条边后，一张图的点与点的最短路径，(a,b)表示从a到b的最短路径，然后我们把它与自己，按照矩阵乘法的格式“相乘”，把其中的乘改为取min，c.a[i][j] = min(c.a[i][j],x.a[i][k]+y.a[k][j]);看不懂先看下面。
这样得到的是走k+k条边的矩阵。有点抽象，下面详细解释下：
c中的一个点(a,b)，当我们用x矩阵和y矩阵求它时，我们枚举了x矩阵的a行所有数，与y矩阵的b列所有数，并且他们的坐标只能是相对应的，比如x矩阵的(a,2)这个点，相应的y矩阵点就是(2,b)，那么放到图上去理解，即从a点经过2点到b点的距离，类似的点不只有2，把所有点枚举完后，c.a[a][b]就是从a到b的最短距离。（意会一下）
这样下来，会得到走k+k条边的最短路径，对于其他的矩阵这样操作，得到的是他们两个，经过的边数相加的结果。（一个经过a条边后的矩阵 与 一个经过b条边后的矩阵这样操作后，是经过a+b条边后的矩阵，矩阵中存的是最短路径）。解释一下：向上面的例子一样，(a,2)(2,b)，是即从a点经过2点到b点的距离,因为x矩阵和y矩阵都是走k条边后的最短路径，那么x矩阵中的(a,2)是走k步后的最短路径，(2,b)也是，那么他们相加不就是走k+k条边后的最短路径吗？其他的矩阵一样。
然后，就可以套用快速幂的模板了，只不过将以前的乘改成加了，也就是倍增的思想的，比如对于走10条边，它的二进制是1010，那么我们就让在走2（10）边时的矩阵 乘以 8（1000）边的矩阵，得到走10条边的矩阵即开始时由1->2->4->8->16……即倍增中的2次幂。
code
1 #include<cstdio> 2 #include<algorithm> 3 #include<map> 4 #include<cstring> 5 6 using namespace std; 7 const int MAXN = 210; 8 9 int Hash[1010]; 10 int n,k,q,z,tn; 11 12 int read() 13 { 14 int x = 0,f = 1;char ch = getchar(); 15 while (ch<'0'||ch>'9') {if (ch='-') f=-1;ch = getchar(); } 16 while (ch>='0'&&ch<='9'){x = x*10+ch-'0';ch = getchar(); } 17 return x*f; 18 } 19 20 struct Matrix{ 21 int a[MAXN][MAXN]; 22 Matrix operator * (const Matrix &x) const 23 { 24 Matrix c; 25 memset(c.a,0x3f,sizeof(c.a)); 26 for (int k=1; k<=tn; k++) 27 for (int i=1; i<=tn; ++i) 28 for (int j=1; j<=tn; ++j) 29 c.a[i][j] = min(c.a[i][j],a[i][k]+x.a[k][j]); 30 return c; 31 } 32 }s,ans; 33 void ksm() 34 { 35 ans = s; 36 k--; 37 for (; k; k>>=1) 38 { 39 if (k&1) ans = ans*s; 40 s = s*s; 41 } 42 } 43 int main() 44 { 45 k = read();n = read();q = read();z = read(); 46 memset(s.a,0x3f,sizeof(s.a)); 47 for (int x,y,w,i=1; i<=n; ++i) 48 { 49 w = read();x = read();y = read(); 50 if (!Hash[x]) Hash[x] = ++tn; 51 if (!Hash[y]) Hash[y] = ++tn; 52 s.a[Hash[x]][Hash[y]] = s.a[Hash[y]][Hash[x]] = w; 53 } 54 ksm(); 55 printf("%d",ans.a[Hash[q]][Hash[z]]); 56 return 0; 57 }