[原]武大预选赛F题-（裸并查集+下标离散化+floyd最短路）

Problem 1542 - F - Countries

Time Limit: 1000MS Memory Limit: 65536KB

Total Submit: 266 Accepted: 36 Special Judge: No

Description

There are n countries at planet X on which Xiao Ming was born.

Some countries, which have been sharing fine bilateral relations, form a coalition and thus all of their citizens will benefit from a policy for which all the travels between these countries will become totally free.

But it is not easy to travel between countries distributed in different coalitions. If possible, it must cost some money called YZB(yu zhou bi) which is always positive.

Help Xiao Ming determine the minimum cost between countries.

Input

The input consists of one or more test cases.

First line of each test case consists two integers n and m. (1<=n<=10^5, 1<=m<=10^5)

Each of the following m lines contains: x y c, and c indicating the YZB traveling from x to y or from y to x. If it equals to zero, that means x and y are in the same coalition. (1<=x, y<=n, 0<=c<=10^9)

You can assume that there are no more than one road between two countries.

Then the next line contains an integer q, the number of queries.(1<=q<=200)

Each of the following q lines contains: x y. (1<=x, y<=n)

It is guaranteed that there are no more 200 coalitions.

Input is terminated by a value of zero (0) for n.

Output

For each test case, output each query in a line. If it is impossible, output “-1”.

Sample Input

6 5

1 2 0

2 4 1

3 5 0

1 4 1

1 6 2

3

4 2

1 3

4 6

0

Sample Output

1

-1

AC代码如下：

#include<iostream>

#include<cstring>

#include<cstdio>

#include<algorithm>

using namespace std;

#define INF1 1000000000000000001

#define INF2 1000000009

const int maxn = 100010;

struct node{

   int s,e,w;

}map[maxn];

long long d[201][201];

long long fa[maxn];

int n, m;

int h[maxn];

void init(int a)

{

  for(int i = 0; i <= a; i++)

  {

    fa[i] = i;

    h[i] = 0;

  }

  for(int i = 0; i <= a; i++){

     map[i].w = INF2;

  }

}

int Find(int x)

{

  if(fa[x] == x)

   return x;

  else

   return fa[x] = Find(fa[x]);

}

void unite(int x,int y)

{

  x = Find(fa[x]);

  y = Find(fa[y]);

  if(x == y) return;

  else

  {

    if(h[x] > h[y]) fa[y] = fa[x];

    else

    {

      fa[x] = fa[y];

      if(h[x] == h[y]) h[y]++;

    }

  }

}

bool M[maxn];

int s[maxn];

int r;

void discretize(){

    memset(M, false, sizeof(M));

    //memset(s, 0, sizeof(s));

    int k;

    for(int i = 1; i <= n; i++){

      k = Find(i);

      if(!M[k]){

       s[k] = r++;

       M[k] = true;

      }

    }

    for(int i = 1; i < r; i++){

     for(int j = 1; j < r; j++){

       if(i == j) d[i][j] = 0;

       else d[i][j] = INF1;

     }

    }

    for(int j = 1; j <= m; j++){

      if(map[j].w != 0){

       int dx = Find(map[j].s);

       int dy = Find(map[j].e);

       dx = s[dx];

       dy = s[dy];

       d[dx][dy] = d[dy][dx] = min(d[dx][dy],(long long)map[j].w);

       //cout<<"*"<<d[dx][dy]<<"*"<<endl;

      }

    }

}

void floyd(int r){

    for(int k = 1; k < r; k++)

      for(int i = 1; i < r; i++)

        for(int j = 1; j < r; j++){

          if(d[i][k] < INF1 && d[k][j] < INF1) d[i][j] = min(d[i][k] + d[k][j], d[i][j]);

        }

}

void input(int m){

 for(int i = 1; i <= m; i++){

     int a , b;

     scanf("%d%d",&map[i].s, &map[i].e);

     scanf("%d",&map[i].w);

     if(map[i].w == 0) unite(map[i].s, map[i].e);

    }

}

int main(){

   while(cin>>n>>m&&n){

    init(n);

    input(m);

    r = 1;

    discretize();

    floyd(r);

    int q;

    cin>>q;

    while(q--){

     int a, b;

     scanf("%d%d",&a, &b);

     a = Find(a);

     b = Find(b);

     a = s[a];

     b = s[b];

     if(d[a][b] == INF1) printf("-1\n");

     else

       printf("%lld\n",d[a][b]);

    }

   }

}

作者：u011652573 发表于2014-4-1 15:55:37 原文链接

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[原]武大预选赛F题-（裸并查集+下标离散化+floyd最短路）

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