poj 3613 Cow Relays（经过N条边的最短路，floyd + 二分矩阵）

Cow Relays

Time Limit: 1000MS Memory Limit: 65536KTotal Submissions: 5028 Accepted: 2001

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 6 411 4 64 4 88 4 96 6 82 6 93 8 9

Sample Output

10

题意：给出一个无向图，T条边，给出N，S，E，求S到E经过N条边的最短路径长度。

思路：floyd + 倍增思想。参考http://hi.baidu.com/lxxstar1226/item/9119a40de25c55faa1103462

AC代码：

#include <iostream>#include <cmath>#include <cstdlib>#include <cstring>#include <cstdio>#include <queue>#include <stack>#include <ctime>#include <vector>#include <algorithm>#define ll long long#define L(rt) (rt<<1)#define R(rt)  (rt<<1|1)#define eps 1e-6;using namespace std;const int INF = 1e9 + 7;const int maxn = 205;struct Mat{    int mat[maxn][maxn];}A, ans;int hash[1005];int n, t, s, e, cnt;void init(){    for(int i = 0; i < maxn; i++)    for(int j = 0; j < maxn; j++)    A.mat[i][j] = ans.mat[i][j] = INF;    cnt = 0;}Mat floyd(Mat a, Mat b){    Mat tmp;    for(int i = 1; i <= cnt; i++)    for(int j = 1; j <= cnt; j++)    tmp.mat[i][j] = INF;    for(int k = 1; k <= cnt; k++)    for(int i = 1; i <= cnt; i++)    for(int j = 1; j <= cnt; j++)    if(tmp.mat[i][j] > a.mat[i][k] + b.mat[k][j])    tmp.mat[i][j] = a.mat[i][k] + b.mat[k][j];    return tmp;}void pow(){    while(n)    {        if(n & 1) ans = floyd(ans, A);        A = floyd(A, A);        n >>= 1;    }}int main(){    int a, b, c;    while(~scanf("%d%d%d%d", &n, &t, &s, &e))    {        init();        memset(hash, 0, sizeof(hash));        while(t--)        {            scanf("%d%d%d", &c, &a, &b);            if(!hash[a]) hash[a] = ++cnt;            a = hash[a];            if(!hash[b]) hash[b] = ++cnt;            b = hash[b];            A.mat[a][b] = A.mat[b][a] = min(A.mat[a][b], c);        }        for(int i = 1; i <= cnt; i++) ans.mat[i][i] = 0;        pow();        printf("%d\n", ans.mat[hash[s]][hash[e]]);    }    return 0;}