POJ

先上题目：

Domestic Networks

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 732		Accepted: 204		Special Judge

Description

Alex is a system administrator of Domestic Networks Inc. His network connects apartments and spans over multiple buildings.

The network expands and Alex has to design a new network segment. He has a map that shows apartments to connect and possible links. Each link connects two apartments and for each possible link its length is known. The goal is to make all apartments connected (possibly through other apartments).

Domestic Networks Inc. buys cable in the nearest cable shop. Unfortunately, shop sells only category 5 and 6 cables at price of p₅ and p₆ rubles per meter respectively. Moreover, there are only q₅ meters of category 5 cable and q₆ meters of category 6 cable available in the shop.

Help Alex to solve a hard problem: make a new network construction plan with possible minimal cost. A plan consists of list of links to be made and cable category for each link (each link should be a single piece of cable of either 5 or 6 category). The cost of the plan is the sum of cost of all cables. The total length of cables of each category used in the plan should not exceed the quantity of the cable available in the shop.

Input

The first line of the input file contains two numbers: n — the number of apartments to be connected and m — the number of possible links (1 ≤ n ≤ 1000, 1 ≤ m ≤ 10 000).

Following m lines contain possible link descriptions. Each description consists of three integer numbers: a_i and b_i — apartments that can be connected by the link and l_i — link length in meters (0 ≤ l_i ≤ 100). Apartments are numbered from 1 to n.

The last line of the input file contains four integer numbers: p₅, q₅, p₆ and q₆ — price and quantity of category 5 and 6 cables respectively (1 ≤ p_i, q_i ≤ 10 000).

Output

If all apartments can be connected with the available cable, output n lines — an optimal network construction plan. The first line of the plan must contain plan’s cost. Other lines of the plan must consist of two integer numbers each: a_i — number of the link to make and c_i — the category of the cable to make it of. Links are numbered from 1 to m in the order they are specified in the input file. If there are more than one optimal plans, output any of them.

If there is no plan meeting all requirements, output a single word “Impossible”.

Sample Input

6 7
1 2 7
2 6 5
1 4 8
2 3 5
3 4 5
5 6 6
3 5 3
2 11 3 100

Sample Output

65
1 5
2 6
4 6
5 6
7 5

Source

Northeastern Europe 2007, Northern Subregion

 

　　题意：给你一幅图，图上每一条边有长度，给你一定数量的两种电线，告诉你它们的数量以及单价，问你能不能用这些电线花最小的代价令图上每个点互相连通。如果可以就输出最小代价，并输出选了哪些线路，它们分别用了哪种电线，一条线路上只能用一种电线。

　　做法是先对这个图求一次MST，如果不存在MST，就输出"Impossbile"，否则就有可能存在合法的方案，这里我们可以用背包选电线。我们的目标是让尽量多的电线使用便宜的那一种电线，所以我们对便宜的那种电线进行一次背包，剩下的线路用贵的电线，然后判断一下是否符合输出的电线数量，如果符合就输出所有结果，否者就是非法的。

 

上代码：

 

#include <cstdio>
#include <cstring>
#include <algorithm>
#include <set>
#define MAX 10002
using namespace std;

typedef struct edge{
    int u,v,l,id;

    bool operator < (const edge& o)const{
        return l<o.l;
    }
}edge;

edge e[MAX];
int p[MAX],mst[MAX],tot;
int n,m,p5,q5,p6,q6,i5,i6;
int pag[MAX];
bool f[MAX];
set<int> e5,e6;

int findset(int u){
    return u==p[u] ? p[u] : p[u]=findset(p[u]);
}

void MST(){
    tot=0;
    int u,v;
    for(int i=0;i<m;i++){
        u=findset(e[i].u);   v=findset(e[i].v);
        if(p[u]!=p[v]){
            p[v]=u;
            mst[tot++]=i;
        }
    }
}

int main()
{
    int I5,I6;
    //freopen("data.txt","r",stdin);
    while(scanf("%d %d",&n,&m)!=EOF){
        for(int i=1;i<=n;i++) p[i]=i;
        for(int i=0;i<m;i++){
            scanf("%d %d %d",&e[i].u,&e[i].v,&e[i].l);
            e[i].id=i+1;
        }
        scanf("%d %d %d %d",&p5,&q5,&p6,&q6);
        i5=5;   i6=6;
        if(p5>p6){
            swap(p5,p6);
            swap(q5,q6);
            swap(i5,i6);
        }
        sort(e,e+m);
        MST();
        if(tot!=n-1){
            printf("Impossible
");
            continue;
        }

        /*********pag*********/
        memset(f,0,sizeof(f));
        e5.clear(); e6.clear();
        //memset(pag,0,sizeof(pag));
        f[0]=1;
        for(int i=0;i<tot;i++){
            int l=e[mst[i]].l;
            e6.insert(mst[i]);
            for(int j=q5;j>=l;j--){
                if(!f[j] && f[j-l]){
                    f[j]=1;
                    pag[j]=mst[i];
                }
            }
        }
        int k=q5;
        while(!f[k]) k--;
        while(k>0){
            e5.insert(pag[k]);
            e6.erase(pag[k]);
            k-=e[pag[k]].l;
        }
        I5=I6=0;
        for(set<int>::iterator it = e5.begin();it!=e5.end();it++){
            I5+=e[*it].l;
        }
        for(set<int>::iterator it = e6.begin();it!=e6.end();it++){
            I6+=e[*it].l;
        }
        if(I5<=q5 && I6<=q6){
            printf("%d
",I5*p5+I6*p6);
            for(set<int>::iterator it = e5.begin();it!=e5.end();it++){
                printf("%d %d
",e[*it].id,i5);
            }
            for(set<int>::iterator it = e6.begin();it!=e6.end();it++){
                printf("%d %d
",e[*it].id,i6);
            }
        }else{
            printf("Impossible
");
        }
    }
    return 0;
}