POJ 1789 Truck History (Prim算法求最小生成树)
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题目
Description
Advanced Cargo Movement, Ltd. uses trucks of different types. Some trucks are used for vegetable delivery, other for furniture, or for bricks. The company has its own code describing each type of a truck. The code is simply a string of exactly seven lowercase letters (each letter on each position has a very special meaning but that is unimportant for this task). At the beginning of company’s history, just a single truck type was used but later other types were derived from it, then from the new types another types were derived, and so on.
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
where the sum goes over all pairs of types in the derivation plan such that t o is the original type and t d the type derived from it and
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
The input consists of several test cases. Each test case begins with a line containing the number of truck types,
Output
For each test case, your program should output the text “The highest possible quality is
Sample Input
4aaaaaaabaaaaaaabaaaaaaabaaaa0
Sample Output
The highest possible quality is 1/3.
题意
历史上,曾用
分析
输入的字符串需要转换成邻接矩阵,然后直接套
代码
#include<iostream>#include<cstdio>#include<cstring>#include<string>#include<algorithm>#include<queue>#include<stack>#include<vector>#include<cmath>#include<set>#include<cstdlib>#include<functional>#include<climits>#include<cctype>#include<iomanip>using namespace std;typedef long long ll;#define INF 0x3f3f3f3f#define mod 1e9+7#define clr(a,x) memset(a,x,sizeof(a))const double eps = 1e-6;const int MAXN=2010;bool vis[MAXN];int lowc[MAXN];int map[MAXN][MAXN];int Prim(int cost[][MAXN],int n){ int ans=0; clr(vis,0); vis[0]=1; for(int i=1;i<n;i++) { lowc[i]=cost[0][i]; } for(int i=1;i<n;i++) { int minc=INF; int p=-1; for(int j=0;j<n;j++) if(!vis[j]&&minc>lowc[j]) { minc=lowc[j]; p=j; } if(minc==INF) { return -1; } vis[p]=1; ans+=minc; for(int j=0;j<n;j++) if(!vis[j]&&lowc[j]>cost[p][j]) lowc[j]=cost[p][j]; } return ans;}int main(){ int t; char str[MAXN][8]; int edge; while(cin>>t,t) { for(int i=0;i<t;i++) { scanf("%s",&str[i]); } for(int i=0;i<t;i++) //接下来就是建立邻接矩阵 { for(int j=0;j<t;j++) //这里wa了好久 { edge=0; for(int k=0;k<7;k++) { if(str[i][k]!=str[j][k]) edge++; map[i][j]=map[j][i]=edge; } } } int ans; ans=Prim(map,t); //Prim算法 printf("The highest possible quality is 1/%d.\n",ans); } return 0;}
PS:太菜,太菜了哇…
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