| Time Limit: 2000MS | Memory Limit: 65536K | |
| Total Submissions: 8416 | Accepted: 3027 |
Description
Today, ACM is rich enough to pay historians to study its history. One thing historians tried to find out is so called derivation plan -- i.e. how the truck types were derived. They defined the distance of truck types as the number of positions with different letters in truck type codes. They also assumed that each truck type was derived from exactly one other truck type (except for the first truck type which was not derived from any other type). The quality of a derivation plan was then defined as
1/Σ(to,td)d(to,td)
where the sum goes over all pairs of types in the derivation plan such that to is the original type and td the type derived from it and d(to,td) is the distance of the types.
Since historians failed, you are to write a program to help them. Given the codes of truck types, your program should find the highest possible quality of a derivation plan.
Input
Output
Sample Input
4 aaaaaaa baaaaaa abaaaaa aabaaaa 0
Sample Output
The highest possible quality is 1/3.
Source
题目要求距离最小，所以生成的图中有环的存在，则一定可以去掉一条边把距离变得更小，所以最后连接完了的图形，一定是一棵树。。(1)
所以题目转化为求一棵树，且让这棵树的距离之和最小，很显然的最小生成树题目。。(2)
中途居然把prim的实现搞忘了。。很囧。。
my ugly code:
Source Code
| Problem: 1789 | User: bingshen | |
| Memory: 15648K | Time: 360MS | |
| Language: C++ | Result: Accepted |
- Source Code
#include<stdio.h> #include<algorithm> #include<string.h> #define inf 99999999 using namespace std; struct node { int v1; int v2; int dis; }; node e[2005]; int dis[2005][2005]; char truck[2005][10]; int dist(int a,int b) { int i,sum=0; for(i=0;i<7;i++) { if(truck[a][i]!=truck[b][i]) sum++; } return sum; } int prim(int n) { int i,j,min,minl,vx,vy,leng; int res=0; for(i=1;i<=n-1;i++) { e[i].v1=1; e[i].v2=i+1; e[i].dis=dis[1][i+1]; } for(i=1;i<=n-1;i++) { minl=inf; min=-1; for(j=i;j<=n-1;j++) { if(minl>e[j].dis) { minl=e[j].dis; min=j; } } if(min==-1) break; res=res+minl; swap(e[i],e[min]); vx=e[i].v2; for(j=i+1;j<=n-1;j++) { vy=e[j].v2; leng=dis[vx][vy]; if(leng<e[j].dis) { e[j].dis=leng; e[j].v1=vx; } } } return res; } int main() { int i,j,n,ans; scanf("%d",&n); while(scanf("%d",&n)!=EOF) { if(n==0) break; for(i=1;i<=n;i++) scanf("%s",truck[i]); for(i=1;i<=n;i++) for(j=1;j<=n;j++) dis[i][j]=dist(i,j); ans=prim(n); printf("The highest possible quality is 1/%d./n",ans); } return 0; }