hdu 1875 畅通工程再续(kruskal || prim)

来源：互联网发布：vb入门教程编辑：程序博客网时间：2024/04/28 17:08

这道题目有点变化，条件是每条路的花费不能超过1000也不能小于10，否则不修该条路，所以呢，用kruskal最好，这种方法是检查每一条边，符合情况就加进去，否则就舍去，这样最后判断一下是不是所有的点都有一个共同的祖先就知道是不是连通图。如果用prim算法的话，它每次选的是最小值，得判断一下，实现起来比较麻烦。

代码：（kruskal）

<span style="font-family:Courier New;font-size:18px;">#include<iostream>#include<cstdio>#include<cstdlib>#include<cstring>#include<cmath>#include<vector>#include<set>#include<queue>#include<string>#include<algorithm>#include<utility>#include<functional>#define MAX 0x7fffffff using namespace std;struct node{int i,j;double len;}gra[5500];int p[105];int cmp(const void *a,const void *b){return ((node *)a)->len - ((node *)b)->len >0 ? 1 : -1;}int find(int x){return x == p[x] ? x : p[x] = find(p[x]);}double dist(int a,int b,int c,int d){return sqrt((a-c)*(a-c)+(b-d)*(b-d));}double a[105],b[105];int n; void kruskal(){int i;double sum = 0;for(i=1; i<=n*(n-1)/2; i++){int x = find(gra[i].i);int y = find(gra[i].j);if((x != y) && (gra[i].len >= 10)&&(gra[i].len<=1000)){sum += gra[i].len;p[x] = y;}}int flag = 0;for(i=1; i<=n; i++){if(p[i]!=p[1]){flag = 1;break;}}sum*=100;if(flag)cout << "oh!" << endl;elseprintf("%.1lf\n",sum);}int main(){int T,i,j;cin >> T;while(T--){int k = 1;cin >> n;for(i=1; i<=n; i++)cin >> a[i] >> b[i];for(i=1; i<n; i++)for(j=i+1; j<=n; j++){gra[k].i = i;gra[k].j = j;gra[k].len = dist(a[i],b[i],a[j],b[j]);k++;}qsort(gra+1,n*(n-1)/2,sizeof(gra[0]),cmp);for(i=1; i<=n; i++)p[i] = i;kruskal();}return 0;}</span>

普利姆算法实现了以下，wa了几次，原因竟然是我把MIN定义成了整型。。。

注意判断距离是否不小于10不大于1000

代码：

#include<iostream>#include<cstdio>#include<cstdlib>#include<cstring>#include<cmath>#include<vector>#include<set>#include<queue>#include<string>#include<algorithm>#include<utility>#include<functional>#define MAX 0x7fffffffusing namespace std;double gra[105][105];int a[105],b[105];double dist(int a,int b,int c,int d){return sqrt((a-c)*(a-c)+(b-d)*(b-d));}int n;void prim(){int visit[105];memset(visit,0,sizeof(visit));int now,i,j;double MIN;double d[105];for(i=1; i<=n; i++)d[i] = MAX;d[1] = 0;double sum = 0;now = 1,visit[1] = 1;for(i=1; i<=n; i++){for(j=1; j<=n; j++)if(!visit[j] && d[j] > gra[now][j] &&(gra[now][j]>=10)&& (gra[now][j]<=1000))d[j] = gra[now][j];MIN = MAX;for(j=1; j<=n; j++){if(!visit[j] && MIN > d[j])MIN = d[now = j];}visit[now] = 1;}int flag = 0;for(i=1; i<=n; i++){sum += d[i];if(visit[i] == 0){flag = 1;break;}}sum *= 100;if(flag)cout << "oh!" << endl;elseprintf("%.1lf\n",sum);}int main(){int T,i,j;cin >> T;while(T--){cin >> n;for(i=1; i<=n; i++)for(j=1; j<=n; j++)gra[i][j] = gra[j][i] = MAX;for(i=1; i<=n; i++)cin >> a[i] >> b[i];for(i=1; i<n; i++)for(j=i+1; j<=n; j++)gra[i][j] = gra[j][i] = dist(a[i],b[i],a[j],b[j]);prim();}return 0;}

0 0