POJ 3613 Cow Relays k步最短路 (floyd + 矩阵快速幂)

Cow Relays
Time Limit: 1000MS Memory Limit: 65536K
Total Submissions: 7738 Accepted: 3033
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
Source

USACO 2007 November Gold

#include <iostream>#include <cstdio>#include <cstring>#include <cmath>#include <algorithm>#define maxn 1010using namespace std;const int inf = 0x3f3f3f3f3f;int n,t,s,e;int num;int v[maxn];int map[maxn][maxn];int tmp[maxn][maxn];int ans[maxn][maxn];int dis[maxn][maxn];int vis[maxn];void floyd(int c[][maxn], int a[][maxn], int b[][maxn]){    for(int k = 0; k < num; k++)        for(int i = 0; i < num; i++)            for(int j = 0; j < num; j++)                if(c[v[i]][v[j]] > a[v[i]][v[k]] + b[v[k]][v[j]])                    c[v[i]][v[j]] = a[v[i]][v[k]] + b[v[k]][v[j]];}void cop(int a[][maxn], int b[][maxn]){    for(int i = 0; i < num; i++)        for(int j = 0; j < num; j++)        {            a[v[i]][v[j]] = b[v[i]][v[j]];            b[v[i]][v[j]] = inf;        }}void quickpower(int k){    while(k)    {        if(k & 1)        {            floyd(dis, ans, map);//相当于dis=ans*map            cop(ans, dis);//ans=dis        }        floyd(tmp, map, map);//map=map*map        cop(map, tmp);        k >>= 1;    }}void init(){ for(int i = 0; i <= 1000; i++)    {        for(int j = 0; j <= 1000; j++)        {            map[i][j] = inf;            tmp[i][j] = inf;            dis[i][j] = inf;            ans[i][j] = inf;        }        ans[i][i] = 0;    }    num = 0;}int main(){    int x, y, w;    scanf("%d%d%d%d", &n, &t, &s, &e);    init();    for(int i = 0; i < t; i++)    {        scanf("%d%d%d", &w, &x, &y);        if(!vis[x])        {            vis[x] = 1;            v[num++] = x;        }        if(!vis[y])        {            vis[y] = 1;            v[num++] = y;        }        if(map[x][y] > w)            map[x][y] = map[y][x] = w;    }    quickpower(n);    printf("%d\n", ans[s][e]);}