POJ 1258 prim模版求最小生成树

题意：求连接所有村庄的最短路径。就是求最小生成树。

分析：直接套用prim模版。 POJ 2421在这题上又该进了点。http://www.cnblogs.com/lanjiangzhou/archive/2013/03/25/2981491.html。其实都一样。

View Code

 // 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 int n,m; int edge[maxn][maxn];//邻接矩阵 int lowcost[maxn]; int nearvex[maxn]; int sumweight=0; 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=1;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=1;i<n;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=1;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=1;j<=n;j++){                 if(nearvex[j]!=-1&&edge[v][j]<lowcost[j]){                     lowcost[j]=edge[v][j];                     nearvex[j]=v;                 }             }         }     }     printf("%d\n",sumweight); } int main(){     int w;     while(scanf("%d",&n)!=EOF){         sumweight=0;         //memset(edge,0,sizeof(edge));         for(int i=1;i<=n;i++){             for(int j=1;j<=n;j++){                 edge[i][j]=INF;             }             edge[i][i]=0;         }         for(int i=1;i<=n;i++){             for(int j=1;j<=n;j++){             scanf("%d",&w);             edge[i][j]=w;             }         }//         scanf("%d",&m);//         for(int i=1;i<=m;i++){//             scanf("%d%d",&u,&v);//             edge[u][v]=edge[v][u]=0;//         } //        for(int i=1;i<=n;i++){ //            for(int j=1;j<=n;j++){ //                if(i==j) edge[i][j]=0; //                else if(edge[i][j]==0 ) edge[i][j]=INF; //            } //        }         prim(1);     }     return 0; }