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

Cow Relays

Time Limit: 1000MS Memory Limit: 65536KTotal Submissions: 5611 Accepted: 2209

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 thelength_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 exactlyN 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 intersectionS to intersection E that traverses exactly N cow trails.

Sample Input

2 6 6 411 4 64 4 88 4 96 6 82 6 93 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]]);}