POJ 1789 Truck History【最小生成树简单应用】
链接:
Time Limit: 2000MS | Memory Limit: 65536K | |
Total Submissions: 14950 | Accepted: 5714 |
Description
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
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
the codes uniquely describe the trucks, i.e., no two of these N lines are the same. The input is terminated with zero at the place of number of truck types.
Output
Sample Input
4
aaaaaaa
baaaaaa
abaaaaa
aabaaaa
0
Sample Output
The highest possible quality is 1/3.
Source
题意:
算法:最小生成树
思路:
Kruskal:
1789 | Accepted | 22860K | 563MS | C++ | 1378B |
//Accepted 22860 KB 579 ms C++ 1302 B 2013-07-31 09:37:35
#include<stdio.h>
#include<algorithm>
using namespace std; const int maxn = 2000+10;
char map[maxn][10];
int p[maxn];
int n,m; struct Edge{
int u,v;
int w;
}edge[maxn*maxn/2]; int dist(int st, int en)
{
int distance = 0;
for(int i = 0; i < 7; i++)
if(map[st][i] != map[en][i])
distance++;
return distance;
} bool cmp(Edge a, Edge b)
{
return a.w < b.w;
} int find(int x)
{
return x == p[x] ? x : p[x] = find(p[x]);
} int Kruskal()
{
int ans = 0;
for(int i = 1; i <= n; i++) p[i] = i;
sort(edge,edge+m,cmp); for(int i = 0; i < m; i++)
{
int u = find(edge[i].u);
int v = find(edge[i].v); if(u != v)
{
p[v] = u;
ans += edge[i].w;
}
}
return ans;
}
int main()
{
while(scanf("%d", &n) != EOF)
{
if(n == 0) break;
for(int i = 1; i <= n; i++)
scanf("%s", map[i]); m = 0;
for(int i = 1; i <= n; i++)
{
for(int j = i+1; j <= n; j++)
{
edge[m].u = i;
edge[m].v = j;
edge[m++].w = dist(i,j);
}
} int ans = Kruskal();
printf("The highest possible quality is 1/%d.\n", ans);
}
return 0;
}
Prime:
Accepted | 15672K | 454MS | C++ | 1289B | 2013-07-31 09:38:56 |
//Accepted 15672 KB 469 ms C++ 1227 B 2013-07-31 09:37:25
#include<stdio.h>
#include<string.h>
#include<algorithm>
using namespace std; const int maxn = 2000+10;
const int INF = maxn*7; char map[maxn][10];
int w[maxn][maxn];
int d[maxn];
int vis[maxn];
int n; int dist(int st, int en)
{
int distance = 0;
for(int i = 0; i < 7; i++)
if(map[st][i] != map[en][i])
distance++;
return distance;
} int Prime()
{
int ans = 0;
for(int i = 1; i <= n; i++) d[i] = INF;
d[1] = 0;
memset(vis, 0, sizeof(vis)); for(int i = 1; i <= n; i++)
{
int x, m = INF;
for(int y = 1; y <= n; y++) if(!vis[y] && d[y] <= m) m = d[x=y];
vis[x] = 1; ans += d[x];
for(int y = 1; y <= n; y++) if(!vis[y])
d[y] = min(d[y], w[x][y]);
}
return ans;
}
int main()
{
while(scanf("%d", &n) != EOF)
{
if(n == 0) break; for(int i = 1; i <= n; i++) scanf("%s", map[i]); for(int i = 1; i <= n; i++)
{
for(int j = i+1; j <= n; j++)
{
w[i][j] = dist(i,j);
w[j][i] = w[i][j];
}
} int ans = Prime();
printf("The highest possible quality is 1/%d.\n", ans);
}
return 0;
}
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