poj 1789 Truck History

 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 d(t_o,t_d) is the distance of the types

1/Σ_(to,td)d(t_o,t_d)

派生关系有向边，边权为两个串的不同字符的个数，要求的是找出一种用到所有串的派生关系，使得Q最小，（因为是1/Q），所以这题就是求最小生成树

最朴素的最小生成树AC

#include<stdio.h>
const int INF = 0x3f3f3f3f;
const int MAXN = 2000;
char word[MAXN][8];
int map[MAXN][MAXN];
int n, lowcost[MAXN], cloest[MAXN];
bool used[MAXN];
inline int countDis(int i, int j){
    int cnt = 0;
    for(int k = 0; k < 7; ++k){
        if(word[i][k] != word[j][k])
            cnt++;
    }
    return cnt;
}
int prim(){
    int ans = 0, min, minidx;
    for(int i = 0; i < n; ++i){
        lowcost[i] = map[0][i];
        cloest[i] = 0;
        used[i] = false;
    }
    used[0] = true;
    for(int i = 1; i < n; ++i){
        min = INF;
        for(int j = 0; j < n; ++j){
            if(!used[j] && lowcost[j] < min)
                min = lowcost[j], minidx = j;
        }
        used[minidx] = true;
        ans += min;
        for(int j = 0; j < n; ++j){
            if(!used[j] && lowcost[j] > map[minidx][j])
                lowcost[j] = map[minidx][j], cloest[j] = minidx;
        }
    }
    return ans;
}
int main(){
    while(scanf("%d",&n), n){
        for(int i = 0; i < n; ++i){
            scanf("%s",word[i]);
        }
        for(int i = 0; i < n; ++i){
            for(int j = 0; j < n; ++j){
                map[i][j] = countDis(i, j);
            }
        }
        printf("The highest possible quality is 1/%d.\n",prim());
    }
    return 0;
}