求[S,T]间长度为k的最短路长度。
考虑Floyd算法。发现Floyd每枚举一个中间点,最短路长度就可能翻倍。
利用这个性质,使用类似矩阵乘法的方式计算即可。
#include
#include
using namespace std;
#define FOR(i,j,k) for(i=j;i<=k;i++)
int num[1005],size=0;
int get(int i) {
return num[i]?num[i]:num[i]=++size;
}
struct Matrix {
int a[105][105];
Matrix(){int i,j;FOR(i,0,104)FOR(j,0,104)a[i][j]=0x7ffffff;}
} root;
Matrix floyd(Matrix a, Matrix b) {
Matrix c; int i, j, k;
FOR(i,1,size) FOR(j,1,size) FOR(k,1,size)
c.a[i][j]=min(c.a[i][j],a.a[i][k]+b.a[k][j]);
return c;
}
Matrix quick(Matrix a, int b) {
Matrix ans; int c=0;
for(;b;b/=2,a=floyd(a,a)) if(b&1)
if(!c) { ans=a;c=1; }
else ans=floyd(ans,a);
return ans;
}
int main() {
int n, t, s, e, i, z, x, y;
scanf("%d%d%d%d", &n, &t, &s, &e);
memset(num,0,sizeof num);
s=get(s),e=get(e);
FOR(i,1,t) {
scanf("%d%d%d", &z, &x, &y);
x=get(x),y=get(y);
root.a[x][y]=root.a[y][x]=z;
}
Matrix a=quick(root,n);
printf("%d",a.a[s][e]);
return 0;
}
Time Limit: 1000MS Memory Limit: 65536K
Total Submissions: 6499 Accepted: 2536
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.
2 6 6 4
11 4 6
4 4 8
8 4 9
6 6 8
2 6 9
3 8 9
10