| Time Limit: 1000MS | Memory Limit: 65536K | |
| Total Submissions: 6726 | Accepted: 2626 |
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
经过n条边的最短路径。
好像可以贪心，先找到一条最短路，然后随便找个最小环绕圈圈，我没有尝试。
使用了标准解法，矩阵乘法+倍增floyd
floyd是枚举k点，更新i到j的最短路径，更新后实际上就是从原本的走一条路到达更新成了走两条路到达。
根据这个思想，我们可以倍增跑floyd，就能得出走2^k条边从i到j的最短路径。
代码如下：（a[i]里存的就是走2^i条边的最短路邻接矩阵)
1 #include<iostream> 2 #include<cstdio> 3 #include<algorithm> 4 #include<cmath> 5 #include<cstring> 6 using namespace std; 7 const int mxn=200; 8 int mp[2000],cnt=0; 9 int m; 10 struct edge{ 11 int x,y; 12 int len; 13 }e[mxn]; 14 struct M{ 15 int v[200][200]; 16 M(){memset(v,50,sizeof v);} 17 friend M operator *(M a,M b){ 18 M c; 19 int i,j,k; 20 for(i=0;i<=m;i++) 21 for(j=0;j<=m;j++) 22 for(k=0;k<=m;k++){ 23 c.v[i][j]=min(c.v[i][j],a.v[i][k]+b.v[k][j]); 24 } 25 return c; 26 } 27 }a[120]; 28 int n,t,S,E; 29 int main(){ 30 scanf("%d%d%d%d",&n,&t,&S,&E); 31 int i,j; 32 memset(mp,-1,sizeof mp); 33 int u,v; 34 for(i=1;i<=t;i++){ 35 scanf("%d%d%d",&e[i].len,&u,&v); 36 if(mp[u]==-1)mp[u]=++cnt; 37 if(mp[v]==-1)mp[v]=++cnt; 38 e[i].x=mp[u];e[i].y=mp[v]; 39 } 40 for(i=1;i<=t;i++){ 41 int x=e[i].x;int y=e[i].y; 42 a[0].v[x][y]=a[0].v[y][x]=e[i].len; 43 } 44 if(mp[S]==-1)mp[S]=++cnt; 45 if(mp[E]==-1)mp[E]=++cnt; 46 m=cnt; 47 S=mp[S];E=mp[E]; 48 for(i=1;i<=20;i++){//倍增 49 a[i]=a[i-1]*a[i-1]; 50 } 51 M ans=a[0]; 52 n--; 53 for(i=20;i>=0;i--){ 54 if((1<<i)&n)ans=ans*a[i]; 55 } 56 printf("%d ",ans.v[S][E]); 57 return 0; 58 }