poj3613Cow Relays

Cow Relays

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 7683		Accepted: 3017

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 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

题意：求从s到e恰好经过n个点的最短路。

分析：给我的第一印象就是暴力，想dp但没啥思路，没想到还用更厉害的做法。

预备知识：《矩阵乘法在信息学中的应用》，主要是看邻接矩阵那一块，可以发现，我们可以把图用邻接矩阵的方式存起来，那么a[i][j]就是i到j的路径长度，这是读入后存起来的矩阵，其中每两个点之间的路径不经过其它的点，如果我们将它和自己相乘，就会得到一个新矩阵，其实a[i][j]就是i到j经过1个点的最短路径，当然，直接乘是不能得到最短路径的，我们需要floyd算法。

     可是这个N这么大，意味着我们要做n-1次floyd，这个复杂度妥妥的TLE啊，但是因为这个矩阵每次都和自己相乘，所以可以想到利用快速幂，这里套用快速幂的算法即可。

     还有一个问题：点数有点大，数组可能会开不下，那么怎么做呢？很简单，离散化即可.

#include <cstring>
#include <cstdio>
#include <iostream>
#include <algorithm>

#define inf 0x7ffffff  

using namespace std;

int n, t, s, e,ans[210][210],a[210][210],d[210][210],cnt,lisan[100000],temp[210][210];

void floyd1()
{
    for (int k = 1; k <= cnt; k++)
        for (int i = 1; i <= cnt; i++)
            for (int j = 1; j <= cnt; j++)
                d[i][j] = min(ans[i][k] + a[k][j], d[i][j]);
    memcpy(ans, d, sizeof(ans));
    memset(d, 0x3f, sizeof(d));
}

void floyd2()
{
    for (int k = 1; k <= cnt; k++)
        for (int i = 1; i <= cnt; i++)
            for (int j = 1; j <= cnt; j++)
                temp[i][j] = min(temp[i][j], a[i][k] + a[k][j]);
    memcpy(a, temp, sizeof(a));
    memset(temp, 0x3f, sizeof(temp));
}

int main()
{
    scanf("%d%d%d%d", &n, &t, &s, &e);
    memset(ans, 0x3f, sizeof(ans));
    memset(a, 0x3f, sizeof(a));
    memset(d, 0x3f, sizeof(d));
    memset(temp, 0x3f, sizeof(temp));
    for (int i = 1; i <= 200; i++)
        ans[i][i] = 0;
    for (int i = 1; i <= t; i++)
    {
        int w, x, y;
        scanf("%d%d%d", &w, &x, &y);
        if (!lisan[x])
            lisan[x] = ++cnt;
        if (!lisan[y])
            lisan[y] = ++cnt;
        a[lisan[x]][lisan[y]] = a[lisan[y]][lisan[x]] = min(a[lisan[x]][lisan[y]], w);
    }
    while (n)
    {
        if (n & 1)
            floyd1();
        floyd2();
        n >>= 1;
    }
    printf("%d\n", ans[lisan[s]][lisan[e]]);

    //while (1);
    return 0;
}

floyd算法初始化弄错了，WA了几次，智障地发现每个点和自己的路径长度竟然初始化成了inf,TAT.