POJ 1789 prim求最小生成树

以前写的报告，可以参考下http://blog.sina.com.cn/s/blog_99ca2df501019bkd.html

// I'm the Topcoder
//C
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <ctype.h>
#include <math.h>
#include <time.h>
//C++
#include <iostream>
#include <algorithm>
#include <cstdio>
#include <cstdlib>
#include <cmath>
#include <cstring>
#include <cctype>
#include <stack>
#include <string>
#include <list>
#include <queue>
#include <map>
#include <vector>
#include <deque>
#include <set>
using namespace std;

//*************************OUTPUT*************************
#ifdef WIN32
#define INT64 "%I64d"
#define UINT64 "%I64u"
#else
#define INT64 "%lld"
#define UINT64 "%llu"
#endif

//**************************CONSTANT***********************
#define INF 0x3f3f3f3f
#define eps 1e-8
#define PI acos(-1.)
#define PI2 asin (1.);
typedef long long LL;
//typedef __int64 LL;   //codeforces
typedef unsigned int ui;
typedef unsigned long long ui64;
#define MP make_pair
typedef vector<int> VI;
typedef pair<int, int> PII;
#define pb push_back
#define mp make_pair

//***************************SENTENCE************************
#define CL(a,b) memset (a, b, sizeof (a))
#define sqr(a,b) sqrt ((double)(a)*(a) + (double)(b)*(b))
#define sqr3(a,b,c) sqrt((double)(a)*(a) + (double)(b)*(b) + (double)(c)*(c))

//****************************FUNCTION************************
template <typename T> double DIS(T va, T vb) { return sqr(va.x - vb.x, va.y - vb.y); }
template <class T> inline T INTEGER_LEN(T v) { int len = 1; while (v /= 10) ++len; return len; }
template <typename T> inline T square(T va, T vb) { return va * va + vb * vb; }

// aply for the memory of the stack
//#pragma comment (linker, "/STACK:1024000000,1024000000")
//end

#define maxn 2000+10//卡车类型数目的最大值
#define codelen 10//代码长度
int n;
char codes[maxn][codelen+5];//存储每种卡车类型的编号
int d[maxn][maxn];//没对卡车类型之间的距离(邻接矩阵)
int lowcost[maxn];//
int dis;
int edge[maxn][maxn];
int sumweight=0;
int nearvex[maxn];
void prim(int u0){
    //从顶点u0出发执行普里姆算法
    sumweight=0;//生成树的权值
    for(int i=0;i<n;i++){
        //初始化lowcost[]数组和neartxt数组
        lowcost[i]=edge[u0][i];
        nearvex[i]=u0;
    }
    nearvex[u0]=-1;
    for(int i=0;i<n-1;i++){
        int min=INF;
        int v=-1;
        //在lowcoat数组的nearvex[]值为-1的元素中找最小值
        for(int j=0;j<n;j++){
            if(nearvex[j]!=-1&&lowcost[j]<min){
                v=j;
                min=lowcost[j];
            }
        }
        if(v!=-1){
            //v==-1表示没找到权值最小的边
           // printf("%d %d %d\n",nearvex[v],v,lowcost[v]);
            nearvex[v]=-1;
            sumweight+=lowcost[v];
            for(int j=0;j<n;j++){
                if(nearvex[j]!=-1&&edge[v][j]<lowcost[j]){
                    lowcost[j]=edge[v][j];
                    nearvex[j]=v;
                }
            }
        }
    }
    printf("The highest possible quality is 1/%d.\n",sumweight);
}

int main(){
    while(scanf("%d",&n)!=EOF){
        if(n==0)  break;
        memset(codes,0,sizeof(codes));
        for(int i=0;i<n;i++){
            scanf("%s",codes[i]);
        }
        //初始化
        for(int i=0;i<n;i++){
            for(int j=0;j<n;j++){
                d[i][j]=0;
            }
            d[i][i]=0;
        }
        for(int i=0;i<n;i++){
            for(int j=i+1;j<n;j++){
                dis=0;
                for(int k=0;k<7;k++){
                    dis+=codes[i][k]!=codes[j][k];
                }
                edge[i][j]=edge[j][i]=dis;
            }
        }
        prim(0);
    }
    return 0;
}