uva 12717 2013Dhaka I

题意：有一张无向图，要求你只能改变变得权值，且边的权值只为1-m每个数只能出现一次。要求最短路树即为最小生成树。

思路：bfs顺序递增的给买一条边赋值，最后再把不在树上的边赋值即为答案。

代码如下:

/**************************************************
 * Author     : xiaohao Z
 * Blog     : http://www.cnblogs.com/shu-xiaohao/
 * Last modified : 2014-07-05 17:22
 * Filename     : uva_12717.cpp
 * Description     : 
 * ************************************************/

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <cmath>
#include <algorithm>
#include <queue>
#include <stack>
#include <vector>
#include <set>
#include <map>
#define MP(a, b) make_pair(a, b)
#define PB(a) push_back(a)

using namespace std;
typedef long long ll;
typedef pair<int, int> pii;
typedef pair<unsigned int,unsigned int> puu;
typedef pair<int, double> pid;
typedef pair<ll, int> pli;
typedef pair<int, ll> pil;

const int INF = 0x3f3f3f3f;
const double eps = 1E-6;
const int LEN = 3010;
struct E{
    int u, v, val;
}edge[LEN*10];
int n, m, s, top;
vector<int> Map[LEN];

void bfs(){
    queue<int> q;
    int vis[LEN] = {0};
    vis[s] = 1;
    q.push(s);
    while(!q.empty()){
        int nv = q.front(); q.pop();
        for(int i=0; i<Map[nv].size(); i++){
            int pos = Map[nv][i];
            E &e = edge[pos]; int y = e.u+e.v-nv;
            if(!vis[y]){
                e.val = top++;
                vis[y] = 1;
                q.push(y);
            }
        }
    }
}

int main()
{
//    freopen("in.txt", "r", stdin);

    ios::sync_with_stdio(false);
    int T, a, b, c, kase = 1;
    cin >> T;
    while(T--){
        top = 1;
        cin >> n >> m >> s;
        for(int i=0; i<LEN; i++) Map[i].clear();
        for(int i=0; i<m; i++){
            cin >> a >> b >> c;
            edge[i].u = a;edge[i].v = b;edge[i].val = -1;
            Map[a].PB(i);
            Map[b].PB(i);
        }
        bfs();
        cout << "Case " << kase ++ << ":" << endl;
        for(int i=0; i<m; i++){
            if(edge[i].val == -1)edge[i].val = top ++;
            cout << edge[i].u << ' ' << edge[i].v << ' ' << edge[i].val << endl;
        }
    }
    return 0;
}