| Time Limit: 1000MS | Memory Limit: 65536K | |
| Total Submissions: 5611 | Accepted: 2209 |
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
#include <cstdio>
#include <cstring>
int h[205], cnt;
struct matrix
{
int m[205][205];
matrix()
{
memset(m, 0x3f, sizeof(m));
}
};
matrix Floyd(matrix a, matrix b)
{
matrix ans;
for(int k = 1; k <= cnt; k++)
for(int i = 1; i <= cnt; i++)
for(int j = 1; j <= cnt; j++)
if(ans.m[i][j] > a.m[i][k] + b.m[k][j])
ans.m[i][j] = a.m[i][k] + b.m[k][j];
return ans;
}
matrix quickmod(matrix a, int k)
{
matrix ans = a;
while(k)
{
if(k & 1)
ans = Floyd(ans, a);
k >>= 1;
a = Floyd(a, a);
}
return ans;
}
int main()
{
int n, t, s, e;
cnt = 1;
matrix ans;
scanf("%d %d %d %d", &n, &t, &s, &e);
memset(h, 0, sizeof(h));
for(int i = 0; i < t; i++)
{
int u, v, w;
scanf("%d %d %d", &w, &u, &v);
if(!h[u])
h[u] = cnt++;
if(!h[v])
h[v] = cnt++;
if(ans.m[h[u]][h[v]] > w)
ans.m[h[u]][h[v]] = ans.m[h[v]][h[u]] = w;
}
ans = quickmod(ans, n - 1);
printf("%d\n", ans.m[h[s]][h[e]]);
}POJ 3613 Cow Relays (Floyd + 矩阵快速幂 + 离散化 神题！)
原文地址：http://blog.csdn.net/tc_to_top/article/details/43925123
