POJ 1789-truck History

Title Link: Truck History

Test instructions is the model of n trucks, the development of a generation, two truck models in the number of different letters represents the distance of two trucks, determine a point. Traverse to the full point to make it the smallest distance.

Very obvious minimum spanning tree, dense graph. 1 times AC, water over

PS: Brazen first looked at the discuss to do, otherwise I also miss "." Sins and sins

#include <iostream> #include <cstdio> #include <cstring> #include <cstdlib> #include < algorithm>const int N = 2001;const int INF = 1e8;using namespace Std;int mapp[n][n];int n,ans;int dis[n];bool vis[N];ch    AR a[n][8];void init () {memset (vis,0,sizeof (VIS)); memset (dis,0,sizeof (DIS));}    int sum (int x,int y) {int sum = 0;    for (int i = 0;i<7;i++) {if (a[x][i]!=a[y][i]) sum++; } return sum;}     void Prim () {ans = 0;    int Pos,minn;    for (int i = 0;i<n;i++) dis[i] = Mapp[0][i];    Vis[0] = true;        for (int i = 0;i<n;i++) {minn = INF;                for (int j = 0;j<n;j++) {if (!vis[j] && dis[j]<minn) {pos = j;            Minn = dis[j];        }} if (minn!=inf) ans + = Minn;        Vis[pos] = true; for (int j = 0;j<n;j++) {if (!vis[j] && dis[j] > Mapp[pos][j]) dis[j] = Mapp [POS][J];        }}}int Main () {while (Cin>>n && N) {init ();        for (int i = 0;i<n;i++) cin>>a[i];        int i,j; for (i = 0;i<n;i++) {for (j = 0;j<n;j++) {if (i==j) m                APP[I][J] = 0;            else mapp[i][j] = SUM (I,J);        }} Prim (); cout<< "The highest possible quality is 1/' <<ans<< '. '    <<endl; } return 0;}

POJ 1789-truck History