poj2253 frog之最小生成树prim解法

• 分析:对节点间关系进行分析,可以发现,从Freddy到Fiona节点的每次跳跃都是在寻找最短路径,这和prim算法生成最小生成树的方法是一致的.所以可以直接在prim算法上求解答案.

首先看下prim模板

---

int cost[INF][INF];//边
int used[INF];
int mincost[INF];//已参与生成的所有节点到节点v的最小值
int v;

int prim()
{
    for (int i = 0; i < V; i += 1){
        mincost[i] = INF;
        used[i] = false;
    }
    mincost[0] = 0;
    int res = 0;

    while (1){
        int v = -1;
        //从不属于X的顶点中找到从X到其权值最小的顶点
        for (int i = 0; i < V; i += 1){
            if(!used[i] && (v == -1 || mincost[i] < mincost[v])) 
                v = i;
        }

        if(v == -1) break;
        used[v] = true;
        res += mincost[v];
        //在此顶点基础上对X到其他顶点的最小值进行更新
        for (int i = 0; i < V; i += 1){
            mincost[j] = min(mincost[j],cost[v][j]);
        }
    }
    return res;
}

简单变动下就可以求出答案

---

/*
* Filename:    code.cpp
* Created:     2017-10-03
* Author:        wyl6 
*[mail:[email protected]]
* Desciption:  Desciption
*/
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
using namespace std;

#define INF  0x3f3f3f3f
#define MAX 100
typedef long long ll;
int n;
double x[205];
double y[205];
double d[205][205];
double mind[205];
bool used[205];
int times = 0;

void prim()
{
    for (int i = 0; i < n; i += 1){
        used[i] = false;
        mind[i] = INF;
    }
    mind[0] = 0;
    double ans = 0;

    while (1){
        int u = -1;
        for (int i = 0; i < n; i += 1)
            if (!used[i] && (u == -1 || mind[i] < mind[u]))
                u = i;
        //以下是灵活变动部分
        if(u == -1) break;
        ans = max(ans,mind[u]);
        used[u] = true;
        if(u == 1) break;

        for (int i = 0; i < n; i += 1)
            mind[i] = min(mind[i],d[u][i]);
    }
    printf("Scenario #%d\nFrog Distance = %.3lf\n\n",times,ans);
}

int main()
{
    while (cin >> n,n){
        times++;
        memset(d,INF,sizeof(d));
        for (int i = 0; i < n; i += 1)
            cin >> x[i] >> y[i];
        for (int i = 0; i < n; i += 1)
            for (int j = 0; j < n; j += 1)
                d[i][j] = sqrt(pow(x[i]-x[j],2)+pow(y[i]-y[j],2));
        prim();
    }   
    return 0;
}