POJ 3613 Cow Relays

Cow Relays

Time Limit: 1000ms

Memory Limit: 65536KB

This problem will be judged on PKU. Original ID: 3613
64-bit integer IO format: %lld      Java class name: Main

 

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

 

解题：Floyd矩阵快速幂。矩阵大法好，矩阵大法妙，矩阵大法我不知道！！！！！！！！！

 

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <algorithm>
#include <climits>
#include <vector>
#include <queue>
#include <cstdlib>
#include <string>
#include <set>
#include <stack>
#define LL long long
#define pii pair<int,int>
#define INF 0x3f3f3f3f
using namespace std;
int lisan[2010],n = 0,k,t,s,e;
struct Matrix {
    int m[210][210];
    Matrix() {
        for(int i = 0; i < 210; i++)
            for(int j = 0; j < 210; j++)
                m[i][j] = INF;
    }
};
Matrix mul(Matrix &x,Matrix &y) {
    Matrix z;
    for(int k = 1; k <= n; k++){
        for(int i = 1; i <= n; i++){
            for(int j = 1; j <= n; j++)
                z.m[i][j] = min(z.m[i][j],x.m[i][k]+y.m[k][j]);
        }
    }
    return z;
}
Matrix fastPow(Matrix x,int index){
    Matrix y;
    for(int i = 0; i <= n; i++) y.m[i][i] = 0;
    while(index){
        if(index&1) y = mul(y,x);
        index >>= 1;
        x = mul(x,x);
    }
    return y;
}
int main() {
    Matrix now;
    int w,u,v;
    scanf("%d %d %d %d",&k,&t,&s,&e);
    while(t--){
        scanf("%d %d %d",&w,&u,&v);
        if(!lisan[u]) lisan[u] = ++n;
        if(!lisan[v]) lisan[v] = ++n;
        now.m[lisan[u]][lisan[v]] = now.m[lisan[v]][lisan[u]] = w;
    }
    now = fastPow(now,k);
    printf("%d
",now.m[lisan[s]][lisan[e]]);
    return 0;
}