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！！！！！！！！！

 

 1 #include <iostream>

 2 #include <cstdio>

 3 #include <cstring>

 4 #include <cmath>

 5 #include <algorithm>

 6 #include <climits>

 7 #include <vector>

 8 #include <queue>

 9 #include <cstdlib>

10 #include <string>

11 #include <set>

12 #include <stack>

13 #define LL long long

14 #define pii pair<int,int>

15 #define INF 0x3f3f3f3f

16 using namespace std;

17 int lisan[2010],n = 0,k,t,s,e;

18 struct Matrix {

19     int m[210][210];

20     Matrix() {

21         for(int i = 0; i < 210; i++)

22             for(int j = 0; j < 210; j++)

23                 m[i][j] = INF;

24     }

25 };

26 Matrix mul(Matrix &x,Matrix &y) {

27     Matrix z;

28     for(int k = 1; k <= n; k++){

29         for(int i = 1; i <= n; i++){

30             for(int j = 1; j <= n; j++)

31                 z.m[i][j] = min(z.m[i][j],x.m[i][k]+y.m[k][j]);

32         }

33     }

34     return z;

35 }

36 Matrix fastPow(Matrix x,int index){

37     Matrix y;

38     for(int i = 0; i <= n; i++) y.m[i][i] = 0;

39     while(index){

40         if(index&1) y = mul(y,x);

41         index >>= 1;

42         x = mul(x,x);

43     }

44     return y;

45 }

46 int main() {

47     Matrix now;

48     int w,u,v;

49     scanf("%d %d %d %d",&k,&t,&s,&e);

50     while(t--){

51         scanf("%d %d %d",&w,&u,&v);

52         if(!lisan[u]) lisan[u] = ++n;

53         if(!lisan[v]) lisan[v] = ++n;

54         now.m[lisan[u]][lisan[v]] = now.m[lisan[v]][lisan[u]] = w;

55     }

56     now = fastPow(now,k);

57     printf("%d\n",now.m[lisan[s]][lisan[e]]);

58     return 0;

59 }

View Code