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

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  1 // I'm the Topcoder  2 //C  3 #include <stdio.h>  4 #include <stdlib.h>  5 #include <string.h>  6 #include <ctype.h>  7 #include <math.h>  8 #include <time.h>  9 //C++ 10 #include <iostream> 11 #include <algorithm> 12 #include <cstdio> 13 #include <cstdlib> 14 #include <cmath> 15 #include <cstring> 16 #include <cctype> 17 #include <stack> 18 #include <string> 19 #include <list> 20 #include <queue> 21 #include <map> 22 #include <vector> 23 #include <deque> 24 #include <set> 25 using namespace std; 26  27 //*************************OUTPUT************************* 28 #ifdef WIN32 29 #define INT64 "%I64d" 30 #define UINT64 "%I64u" 31 #else 32 #define INT64 "%lld" 33 #define UINT64 "%llu" 34 #endif 35  36 //**************************CONSTANT*********************** 37 #define INF 0x3f3f3f3f 38 #define eps 1e-8 39 #define PI acos(-1.) 40 #define PI2 asin (1.); 41 typedef long long LL; 42 //typedef __int64 LL;   //codeforces 43 typedef unsigned int ui; 44 typedef unsigned long long ui64; 45 #define MP make_pair 46 typedef vector<int> VI; 47 typedef pair<int, int> PII; 48 #define pb push_back 49 #define mp make_pair 50  51 //***************************SENTENCE************************ 52 #define CL(a,b) memset (a, b, sizeof (a)) 53 #define sqr(a,b) sqrt ((double)(a)*(a) + (double)(b)*(b)) 54 #define sqr3(a,b,c) sqrt((double)(a)*(a) + (double)(b)*(b) + (double)(c)*(c)) 55  56 //****************************FUNCTION************************ 57 template <typename T> double DIS(T va, T vb) { return sqr(va.x - vb.x, va.y - vb.y); } 58 template <class T> inline T INTEGER_LEN(T v) { int len = 1; while (v /= 10) ++len; return len; } 59 template <typename T> inline T square(T va, T vb) { return va * va + vb * vb; } 60  61 // aply for the memory of the stack 62 //#pragma comment (linker, "/STACK:1024000000,1024000000") 63 //end 64  65 #define maxn 2000+10//卡车类型数目的最大值 66 #define codelen 10//代码长度 67 int n; 68 char codes[maxn][codelen+5];//存储每种卡车类型的编号 69 int d[maxn][maxn];//没对卡车类型之间的距离(邻接矩阵) 70 int lowcost[maxn];// 71 int dis; 72 int edge[maxn][maxn]; 73 int sumweight=0; 74 int nearvex[maxn]; 75 void prim(int u0){ 76     //从顶点u0出发执行普里姆算法 77     sumweight=0;//生成树的权值 78     for(int i=0;i<n;i++){ 79         //初始化lowcost[]数组和neartxt数组 80         lowcost[i]=edge[u0][i]; 81         nearvex[i]=u0; 82     } 83     nearvex[u0]=-1; 84     for(int i=0;i<n-1;i++){ 85         int min=INF; 86         int v=-1; 87         //在lowcoat数组的nearvex[]值为-1的元素中找最小值 88         for(int j=0;j<n;j++){ 89             if(nearvex[j]!=-1&&lowcost[j]<min){ 90                 v=j; 91                 min=lowcost[j]; 92             } 93         } 94         if(v!=-1){ 95             //v==-1表示没找到权值最小的边 96            // printf("%d %d %d\n",nearvex[v],v,lowcost[v]); 97             nearvex[v]=-1; 98             sumweight+=lowcost[v]; 99             for(int j=0;j<n;j++){100                 if(nearvex[j]!=-1&&edge[v][j]<lowcost[j]){101                     lowcost[j]=edge[v][j];102                     nearvex[j]=v;103                 }104             }105         }106     }107     printf("The highest possible quality is 1/%d.\n",sumweight);108 }109 110 int main(){111     while(scanf("%d",&n)!=EOF){112         if(n==0)  break;113         memset(codes,0,sizeof(codes));114         for(int i=0;i<n;i++){115             scanf("%s",codes[i]);116         }117         //初始化118         for(int i=0;i<n;i++){119             for(int j=0;j<n;j++){120                 d[i][j]=0;121             }122             d[i][i]=0;123         }124         for(int i=0;i<n;i++){125             for(int j=i+1;j<n;j++){126                 dis=0;127                 for(int k=0;k<7;k++){128                     dis+=codes[i][k]!=codes[j][k];129                 }130                 edge[i][j]=edge[j][i]=dis;131             }132         }133         prim(0);134     }135     return 0;136 }