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 }