poj 1789（最小生成树）

题意：。。。

 

思路：朴素的最短路算法，没什么可说的。

 

参考代码：

#include <cstdlib>#include <iostream>#include <queue>using namespace std;#define MAXN 2002int n,visited[MAXN],closedge[MAXN],edge[MAXN][MAXN];char str[MAXN][9];int res;#define MAX 1220000struct Node{int u,d;bool operator< (const struct Node a)const{return d>a.d;}}t1,t2;int dist(char* s1,char* s2){ int i,j=0; for(i=0;i<7;i++) if(s1[i]!=s2[i]) j++; return j;}void prim(){int i,j,k;memset(visited,0,sizeof(visited));for(i=1;i<=n;i++){closedge[i]=MAX;}closedge[1]=0;t1.u=1;t1.d=0;priority_queue<struct Node> que;que.push(t1);while(!que.empty()){t1=que.top();que.pop();int u=t1.u;if(visited[u]){continue;}res+=closedge[u];visited[u]=true;for(i=1;i<=n;i++){if(!visited[i]&&edge[u][i]<closedge[i]){closedge[i]=edge[u][i];t2.u=i;t2.d=closedge[i];que.push(t2);}}}}int main(int argc, char *argv[]){ // freopen("in.txt","r",stdin); int i,j,k; while(scanf("%d",&n)!=EOF&&n) { for(i=1;i<=n;i++) { scanf("%s",str[i]); } memset(edge,0,sizeof(edge)); for(i=1;i<=n;i++) { for(j=i+1;j<=n;j++) { edge[i][j]=dist(str[i],str[j]); edge[j][i]=edge[i][j]; } } res=0; prim(); printf("The highest possible quality is 1/%d./n",res); } //while(1); //system("PAUSE"); return EXIT_SUCCESS;}