LightOJ - 1041 Road Construction

LightOJ - 1041 Road Construction Time Limit: 2000MS Memory Limit: 32768KB 64bit IO Format: %lld & %llu Submit Status Description There are several cities in the country, and some of them are connected by bidirectional roads. Unfortunately, some of the roads are damaged and cannot be used right now. Your goal is to rebuild enough of the damaged roads that there is a functional path between every pair of cities. You are given the description of roads. Damaged roads are formatted as "city1 city2 cost" and non-damaged roads are formatted as "city1 city2 0". In this notation city1 and city2 are the case-sensitive names of the two cities directly connected by that road. If the road is damaged, cost represents the price of rebuilding that road. Every city in the country will appear at least once in roads. And there can be multiple roads between same pair of cities. Your task is to find the minimum cost of the roads that must be rebuilt to achieve your goal. If it is impossible to do so, print "Impossible". Input Input starts with an integer T (≤ 50), denoting the number of test cases. Each case begins with a blank line and an integer m (1 ≤ m ≤ 50) denoting the number of roads. Then there will be m lines, each containing the description of a road. No names will contain more than 50 characters. The road costs will lie in the range [0, 1000]. Output For each case of input you have to print the case number and the desired result. Sample Input 2   12 Dhaka Sylhet 0 Ctg Dhaka 0 Sylhet Chandpur 9 Ctg Barisal 9 Ctg Rajshahi 9 Dhaka Sylhet 9 Ctg Rajshahi 3 Sylhet Chandpur 5 Khulna Rangpur 7 Chandpur Rangpur 7 Dhaka Rajshahi 6 Dhaka Rajshahi 7   2 Rajshahi Khulna 4 Kushtia Bhola 1 Sample Output Case 1: 31 Case 2: Impossible

这题是最小生成树的变形没有给城市数量，需要自己根据城市出现的次数自己判断，把城市名字转化成数字非常巧妙

#include <iostream>

#include <cstdio>

#include <cstring>

#include <algorithm>

using namespace std;

const int N = 110;

int a
;

char str

, s1
, s2
;

struct node

{

    int x, y, w;

}p
;

int merg(int x,int y);

int Find(int x);

bool cmp(node a,node b)

{

    return a.w<b.w;

}

int main()

{

    int t, ncase=1;

    scanf("%d", &t);

    while(t--)

    {

        memset(str,0,sizeof(str));

        int m, w,  j, l=0;

        scanf("%d", &m);

        for(int i=0;i<m;i++)

        {

            scanf(" %s %s %d", s1, s2, &w);

            for( j=0;j<l;j++)

            {

                if(strcmp(str[j],s1)==0)

                {

                    p[i].x=j;

                    break;

                }

            }

            if(j==l)

            {

                strcpy(str[l],s1);

                p[i].x=l;

                l++;

            }

            for( j=0;j<l;j++)

            {

                if(strcmp(str[j],s2)==0)

                {

                    p[i].y=j;

                    break;

                }

            }

            if(j==l)

            {

                strcpy(str[l],s2);

                p[i].y=l;

                l++;

            }

            p[i].w=w;

        }

        sort(p,p+m,cmp);

        for(int i=0;i<l;i++)

        {

            a[i]=i;

        }

        int sum=0;

        for(int i=0;i<m;i++)

        {

            int x=p[i].x, y=p[i].y;

            if(merg(x,y)==0)

            {

                sum+=p[i].w;

            }

        }

        int flag=0;

        for(int i=0;i<l;i++)

        {

            if(a[i]==i)

            {

                flag++;

            }

        }

        if(flag!=1)

        {

            printf("Case %d: Impossible\n",ncase++);

        }

        else

        {

            printf("Case %d: %d\n",ncase++,sum);

        }

    }

    return 0;

}

int merg(int x,int y)

{

    int t1=Find(x);

    int t2=Find(y);

    if(t1==t2)

    {

        return 1;

    }

    else

    {

        a[t1]=t2;

        return 0;

    }

}

int Find(int x)

{

    if(x==a[x])

    {

        return x;

    }

    else

    {

        a[x]=Find(a[x]);

        return a[x];

    }

}