POJ3613 Cow Relays
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
USACO 2007 November Gold
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)
#include<iostream> #include<cstdio> #include<algorithm> #include<cmath> #include<cstring> using namespace std; const int mxn=200; int mp[2000],cnt=0; int m; struct edge{ int x,y; int len; }e[mxn]; struct M{ int v[200][200]; M(){memset(v,50,sizeof v);} friend M operator *(M a,M b){ M c; int i,j,k; for(i=0;i<=m;i++) for(j=0;j<=m;j++) for(k=0;k<=m;k++){ c.v[i][j]=min(c.v[i][j],a.v[i][k]+b.v[k][j]); } return c; } }a[120]; int n,t,S,E; int main(){ scanf("%d%d%d%d",&n,&t,&S,&E); int i,j; memset(mp,-1,sizeof mp); int u,v; for(i=1;i<=t;i++){ scanf("%d%d%d",&e[i].len,&u,&v); if(mp[u]==-1)mp[u]=++cnt; if(mp[v]==-1)mp[v]=++cnt; e[i].x=mp[u];e[i].y=mp[v]; } for(i=1;i<=t;i++){ int x=e[i].x;int y=e[i].y; a[0].v[x][y]=a[0].v[y][x]=e[i].len; } if(mp[S]==-1)mp[S]=++cnt; if(mp[E]==-1)mp[E]=++cnt; m=cnt; S=mp[S];E=mp[E]; for(i=1;i<=20;i++){//倍增 a[i]=a[i-1]*a[i-1]; } M ans=a[0]; n--; for(i=20;i>=0;i--){ if((1<<i)&n)ans=ans*a[i]; } printf("%d\n",ans.v[S][E]); return 0; }
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