POJ 3613 Cow Relays k步最短路 二分优化

Cow Relays

Time Limit: 1000MS Memory Limit: 65536K
Total Submissions: 7003 Accepted: 2757

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：
在m条边的无向图中求经过k条边的从s到t的最短路。
注意，输入是k,m,s,t
接下来m行是
边权，起点，终点。

不过，值得注意的是点并不是从1~n编号的。

所以要稍微处理一下（离散化）；
算法很简单，就是k次Floyd，每一次把答案加起来。
但是会超时，于是二分优化，大概方法和快速幂相似。

#include <iostream>#include <cstdio>#include <cstdlib>#include <cstring>using namespace std ;const int maxn = 210, zhf = 0x3fffffff;int n = maxn , m, numth, st, end ;int p[1000010] ;struct Matrix {    int s[maxn][maxn] ;    Matrix () {        for ( int i = 0 ; i < maxn ; i ++ )             for ( int j = 0 ; j < maxn ; j ++ )                 s[i][j] = zhf ;    }    friend Matrix operator * ( Matrix x, Matrix y ) {        Matrix a ;        for ( int k = 1 ; k <= n ; k ++ )             for ( int i = 1 ; i <= n ; i ++ )                 for ( int j = 1 ; j <= n ; j ++ )                     if ( a.s[i][j] > x.s[i][k] + y.s[k][j] )                         a.s[i][j] = x.s[i][k] + y.s[k][j] ;        return a ;    }    friend Matrix operator ^ ( Matrix a, int b ) {        if ( b == 1 ) return a ;        Matrix c ;        for ( -- b , c = a ; b ; b >>= 1 , a = a * a )             if ( b & 1 ) c = a * c ;         return c ;    }} yyc, ans;int hash ( int x ) {    if ( p[x] ) return p[x] ;    return p[x] = ++n ;}int main () {    n = 0 ;    int i, j, k, u, v, w ;    scanf ( "%d%d%d%d", &numth, &m, &st, &end ) ;    for ( i = 1 ; i <= m ; i ++ ) {        scanf ( "%d%d%d", &w, &u, &v ) ;        u = hash(u) ; v = hash(v) ;         if ( yyc.s[u][v] > w )            yyc.s[u][v] = yyc.s[v][u] = w ;    }    ans=yyc^numth;    printf ( "%d\n", ans.s[hash(st)][hash(end)] ) ;    return 0 ;}