Poj 1751 Highways(prim)

来源：互联网发布：nginx lua set cookie 编辑：程序博客网时间：2024/04/24 09:58

题意：求最小生成树，给出了每个城市的坐标，和已经修建好的公路，求出除已经修建好的路的最小生成树的路径。 special judge~~~~~~

虽然是个模版题，但改了很久才A的，主要是一开始不予考虑给出的公路。酿成大错......关键还是对prim不理解.........慢慢来吧。

#include <iostream>#include <algorithm>#include <cmath>#include <cstdio>#include <cstdlib>#include <cstring>#include <string>#include <vector>#include <set>#include <queue>#include <stack>#include <climits>//形如INT_MAX一类的#define MAX 1000#define INF 0x7FFFFFFF# define eps 1e-5using namespace std;struct Point{    int x,y;} point[MAX];int dist[MAX],edge[MAX][MAX];int index[MAX],n,m;//index记录路径int distance(int x1,int y1,int x2,int y2)//给出的坐标算距离{    return ((x1-x2)*(x1-x2)+(y1-y2)*(y1-y2));}void prim(int u)//prim模版{    int i,j,v;    for(i=1; i<=n; i++)    {        dist[i] = edge[u][i];        index[i] = u;    }    index[u] = -1;    for(j=1; j<n; j++)    {        int min = INF;        v = -1;        for(i=1; i<=n; i++)        {            if(index[i] != -1 && min > dist[i])            {                min = dist[i];                v = i;            }        }        if(v != -1)        {            if(min != 0)//考虑已经给出的公路，所以给出的公路不输出            printf("%d %d\n",index[v],v);            index[v] = -1;            for(i=1; i<=n; i++)//更新值            {                if(index[i] != -1 && edge[v][i] < dist[i])                {                    dist[i] = edge[v][i];                    index[i] = v;                }            }        }    }}int main(){    int t,i,j,a,b;    scanf("%d",&n);    for(i=1; i<=n; i++)    {        scanf("%d%d",&point[i].x,&point[i].y);    }    for(i=1; i<=n; i++)    {        edge[i][i] = 0;    }    for(i=1; i<=n; i++)//构图    {        for(j=1; j<=n; j++)        {            if(i != j)            edge[i][j] = distance(point[i].x,point[i].y,point[j].x,point[j].y);        }    }    scanf("%d",&m);    for(i=1; i<=m; i++)    {        scanf("%d%d",&a,&b);        edge[a][b] = edge[b][a] = 0;//给出的公路归零    }    prim(1);    return 0;}