hdoj 2122 Ice_cream’s world III 【最小生成树】

Ice_cream’s world III

Time Limit: 3000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1074    Accepted Submission(s): 345

Problem Description

ice_cream’s world becomes stronger and stronger; every road is built as undirected. The queen enjoys traveling around her world; the queen’s requirement is like II problem, beautifies the roads, by which there are some ways from every city to the capital. The project’s cost should be as less as better.

Input

Every case have two integers N and M (N<=1000, M<=10000) meaning N cities and M roads, the cities numbered 0…N-1, following N lines, each line contain three integers S, T and C, meaning S connected with T have a road will cost C.

Output

If Wiskey can’t satisfy the queen’s requirement, you must be output “impossible”, otherwise, print the minimum cost in this project. After every case print one blank.

Sample Input

2 10 1 104 0

Sample Output

10impossible

kruskal+prime:

 

kruskal:

 

#include<stdio.h>#include<string.h>#include<algorithm>#define max 10000+10using namespace std;int set[1010];int city,road;struct line{    int start;    int end;    int money;}num[max];bool cmp(line a,line b){    return a.money<b.money;}int find(int p){    int child=p;    int t;    while(p!=set[p])    p=set[p];    while(child!=p)    {        t=set[child];        set[child]=p;        child=t;    }    return p;}void merge(int x,int y){    int fx=find(x);    int fy=find(y);    if(fx!=fy)    set[fx]=fy;}int main(){    int i,j,x,y,c,t,need,exist;    while(scanf("%d%d",&city,&road)!=EOF)    {        for(i=0;i<city;i++)        set[i]=i;        t=0;        while(road--)        {            scanf("%d%d%d",&x,&y,&c);            num[t].start=x;            num[t].end=y;            num[t].money=c;            t++;        }        sort(num,num+t,cmp);        need=0;        for(i=0;i<t;i++)        {            if(find(num[i].start)!=find(num[i].end))            {                merge(num[i].start,num[i].end);                need+=num[i].money;            }        }        exist=0;        for(i=0;i<city;i++)        {            if(set[i]==i)            {                exist++;                if(exist>1)                break;            }        }        if(exist>1)        printf("impossible\n\n");        else        printf("%d\n\n",need);    }    return 0;}

prime:

 

#include<stdio.h>#include<string.h>#define max 1000+10#define INF 0x3f3f3fint visit[max],map[max][max],low[max];int city,road;void prime(){    int i,j,next;    int min,mincost=0;    for(i=0;i<city;i++)    {        visit[i]=0;        low[i]=map[1][i];    }    visit[1]=1;    for(i=1;i<city;i++)    {        min=INF;        next=1;        for(j=0;j<city;j++)        {            if(!visit[j]&&min>low[j])            {                min=low[j];                next=j;            }        }        if(min==INF)//遍历全部无法连通         {            printf("impossible\n\n");            return ;        }        mincost+=min;        visit[next]=1;        for(j=0;j<city;j++)        {            if(!visit[j]&&map[next][j]<low[j])            {                low[j]=map[next][j];            }        }    }    printf("%d\n\n",mincost);}int main(){    int i,j,x,y,c;    while(scanf("%d%d",&city,&road)!=EOF)    {        for(i=0;i<city;i++)        {            for(j=0;j<city;j++)            {                if(i==j)                map[i][j]=0;                else                map[i][j]=INF;            }        }         while(road--)        {            scanf("%d%d%d",&x,&y,&c);            if(map[x][y]>c)            map[x][y]=map[y][x]=c;        }        prime();    }    return 0;}