POJ - 1789 Truck History （最小生成树）

https://vjudge.net/problem/POJ-1789

题意

n个车牌，长度固定为7，车牌与车牌间的距离为不同字母个数。问所有车牌形成一棵树的最小边权和是多少。

分析

最小生成树模板

#include<iostream>
#include<cmath>
#include<cstring>
#include<queue>
#include<vector>
#include<cstdio>
#include<algorithm>
#include<map>
#include<set>
#define rep(i,e) for(int i=0;i<(e);i++)
#define rep1(i,e) for(int i=1;i<=(e);i++)
#define repx(i,x,e) for(int i=(x);i<=(e);i++)
#define X first
#define Y second
#define PB push_back
#define MP make_pair
#define mset(var,val) memset(var,val,sizeof(var))
#define scd(a) scanf("%d",&a)
#define scdd(a,b) scanf("%d%d",&a,&b)
#define scddd(a,b,c) scanf("%d%d%d",&a,&b,&c)
#define pd(a) printf("%d\n",a)
#define scl(a) scanf("%lld",&a)
#define scll(a,b) scanf("%lld%lld",&a,&b)
#define sclll(a,b,c) scanf("%lld%lld%lld",&a,&b,&c)
#define IOS ios::sync_with_stdio(false);cin.tie(0)

using namespace std;
typedef long long ll;
template <class T>
void test(T a){cout<<a<<endl;}
template <class T,class T2>
void test(T a,T2 b){cout<<a<<" "<<b<<endl;}
template <class T,class T2,class T3>
void test(T a,T2 b,T3 c){cout<<a<<" "<<b<<" "<<c<<endl;}
template <class T>
inline bool scan_d(T &ret){
    char c;int sgn;
    if(c=getchar(),c==EOF) return 0;
    while(c!='-'&&(c<'0'||c>'9')) c=getchar();
    sgn=(c=='-')?-1:1;
    ret=(c=='-')?0:(c-'0');
    while(c=getchar(),c>='0'&&c<='9') ret = ret*10+(c-'0');
    ret*=sgn;
    return 1;
}
//const int N = 1e6+10;
const int inf = 0x3f3f3f3f;
const ll INF = 0x3f3f3f3f3f3f3f3fll;
const ll mod = 1000000000;
int T;

void testcase(){
    printf("Case %d:",++T);
}

const int MAXN = 2500 ;
const int MAXM = 250;
const double eps = 1e-8;
const double PI = acos(-1.0);

bool vis[MAXN];
int lowc[MAXN];
int cost[MAXN][MAXN];

int Prim(int n){
    int ans=0;
    mset(vis,false);
    vis[0]=true;
    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;
        }
        ans+=minc;
        vis[p]=true;
        for(int j=0;j<n;j++){
            if(!vis[j]&&lowc[j]>cost[p][j]){
                lowc[j] = cost[p][j];
            }
        }
    }
    return ans;
}
char s[MAXN][10];
int main() {
#ifdef LOCAL
    freopen("in.txt","r",stdin);
#endif // LOCAL
    int n;
    while(~scanf("%d",&n)&&n){
        for(int i=0;i<n;i++){
            scanf("%s",s[i]);
            for(int j=0;j<i;j++){
                int cnt=0;
                for(int k=0;k<7;k++){
                    cnt += s[i][k]==s[j][k];
                }
                cost[i][j]=cost[j][i]=7-cnt;
            }
        }

        int ans=Prim(n);
        printf("The highest possible quality is 1/%d.\n",ans);
    }
    return 0;
}