POJ 1789 Truck History Prim

題意：給出一組truck的代號（每個代號是長度為7的字串），規定兩個代號之間的distance是不同的字母個數。將每個代號視作點，代號之間distance視作邊權，題意就是要求該圖的最小產生樹
題解：

#include <iostream>using namespace std;#define INF  0x3fffffffint dis[2001],map[2001][2001];char truck[2001][7];bool check[2001];int n;int prim(){int i, j, k, min, sum = 0;memset(check,false,sizeof(check));for ( i = 0; i < n; i++ )dis[i] = INF;dis[0] = 0;for ( i = 0; i < n; i++ ){min = INF;for ( j = 0; j < n; j++ ){if ( !check[j] && dis[j] < min ){k = j;min = dis[j];}}sum += min;check[k] = true;for ( j = 0; j < n; j++ ){if ( !check[j] && map[k][j] < dis[j] )dis[j] = map[k][j];}}return sum;}int main(){int i, j, k;while ( scanf("%d",&n) && n ){  for ( i = 0; i < n; i++ )scanf("%s", truck[i]);memset(map,0,sizeof(map));for ( i = 0; i < n-1; i++ ){for ( j = i+1; j < n; j++ )for ( k = 0; k < 7; k++ )if ( truck[i][k] != truck[j][k] )map[i][j] = (++map[j][i]);}printf("The highest possible quality is 1/%d.\n", prim());} return 0;}