POJ 3613 Cow Relays (Floyd + 矩阵快速幂 + 离散化 神题！)

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 ≤
I_1i ≤ 1,000; 1 ≤ I_2i ≤ 1,000), each of which is the termination for at least two trails. The cows know the
length_i of each trail (1 ≤ length_i  ≤ 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:
length_i , I_1i , and I_2i

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

题目链接：http://poj.org/problem?id=3613

题目大意：求从起点s到终点e经过k条边的最短路径

题目分析：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，做n - 1次Floyd即是在i，j之间经过n - 1 个点时的最短路，i，j之间有n－1个点即有n条边，因为n比较大，考虑采用矩阵快速幂来求，还有就是这题的t比较小，但是l1，l2比较大，所以将其离散化，因为t最大为100，所以离散化后最多有200个点

#include <cstdio>
#include <cstring>
int h[205], cnt;

struct matrix
{
    int m[205][205];
    matrix()
    {
        memset(m, 0x3f, sizeof(m));
    }
};

matrix Floyd(matrix a, matrix b)
{
    matrix ans;
    for(int k = 1; k <= cnt; k++)
        for(int i = 1; i <= cnt; i++)
            for(int j = 1; j <= cnt; j++)
                if(ans.m[i][j] > a.m[i][k] + b.m[k][j])
                    ans.m[i][j] = a.m[i][k] + b.m[k][j];
    return ans;
}

matrix quickmod(matrix a, int k)
{
    matrix ans = a;
    while(k)
    {
        if(k & 1)
            ans = Floyd(ans, a);
        k >>= 1;
        a = Floyd(a, a);
    }
    return ans;
}

int main()
{
    int n, t, s, e;
    cnt = 1;
    matrix ans;
    scanf("%d %d %d %d", &n, &t, &s, &e);
    memset(h, 0, sizeof(h));
    for(int i = 0; i < t; i++)
    {
        int u, v, w;
        scanf("%d %d %d", &w, &u, &v);
        if(!h[u])
            h[u] = cnt++;
        if(!h[v])
            h[v] = cnt++;
        if(ans.m[h[u]][h[v]] > w)
            ans.m[h[u]][h[v]] = ans.m[h[v]][h[u]] = w;
    }
    ans = quickmod(ans, n - 1);
    printf("%d\n", ans.m[h[s]][h[e]]);
}