《挑战程序设计竞赛》2.5 最短路 AOJ0189 2249 2200 POJ3255 2139 3259 3268（5）
AOJ0189
http://judge.u-aizu.ac.jp/onlinejudge/description.jsp?id=0189
题意
求某一办公室到其他办公室的最短距离。
多组输入，n表示n条关系，下面n次每次输入 x y d表示x到y的距离是d。需要注意的是n没有给定，需要根据输入来求。
输出办公室的编号和距离。
思路
任意两点之间的最短距离用floyd算法比较合适。
代码
#include <iostream>
#include <cstdio>
#include <algorithm>
using namespace std;
const int N = 10;
const int M = 10;
const int INF = 0x3f3f3f3f;
int d[N][N];
int n;
void input()
{
for (int i = 0; i < N; i ++)
fill(d[i], d[i]+N, INF);
int m;
cin >> m;
n = 0;
while ( m-- ) {
int a, b, c;
scanf("%d%d%d", &a, &b, &c);
d[a][b] = d[b][a] = c;
n = max(max(n, a+1), b+1);
}
}
void solve()
{
for (int k = 0; k < n; k ++) {
for (int i = 0; i < n; i ++) {
for (int j = 0; j < n; j ++) {
d[i][j] = min(d[i][j], d[i][k]+d[k][j]);
}
}
}
int sum, msum = INF;
int id = 0;
for (int i = 0; i < n; i ++) {
sum = 0;
for (int j = 0; j < n; j ++) {
if (i == j) continue;
sum += d[i][j];
}
if (sum < msum) {
msum = sum;
id = i;
}
}
printf("%d %d\n", id, msum);
}
int main(void)
{
int m;
while ( cin >> m && m ) {
for (int i = 0; i < N; i ++)
fill(d[i], d[i]+N, INF);
n = 0;
while ( m-- ) {
int a, b, c;
scanf("%d%d%d", &a, &b, &c);
d[a][b] = d[b][a] = c;
n = max(max(n, a+1), b+1);
}
solve();
}
return 0;
}POJ2139
http://poj.org/problem?id=2139
题意
奶牛们最近要拍电影了……
1、若两个的奶牛一起工作则，他们相互的度（degrees）为；
2、若两只奶牛a、b不一起工作，但与另有一只奶牛都和他们工作，则a、b的相互的度为2。
求奶牛的与其他奶牛的度的平均值的100的整数。
思路
本题题意可以变换的理解为如果N个点在一个集合中，则这些点之间的距离为1。然后由此建立一个无向图。在这N个点中，每一个点与其他的所有点都有一个连接的路径长度，将这些长度都加起来，然后除以N-1，就求出了平均长度。题目所求为这些平均长度中的最小值，然后将最小值乘以100输出。
由于所有点到其他点的距离都要求，这个题用floyd算法最为合适，但也可用dijkstra算法计算N次求解，其中dij算法的实现又分使用邻接矩阵和邻接表两种实现方式。
因此解法有三种，我都实现了。
代码1（floyd）
Source Code
Problem: 2139 User: liangrx06
Memory: 368K Time: 32MS
Language: C++ Result: Accepted
Source Code
#include <iostream>
#include <cstdio>
#include <algorithm>
using namespace std;
const int N = 300;
const int INF = 0x3f3f3f3f;
int d[N+1][N+1];
int n;
void input()
{
int m;
cin >> n >> m;
for (int i = 1; i <= n; i ++)
fill(d[i]+1, d[i]+1+n, INF);
while ( m-- ) {
int k, movie[N];
cin >> k;
for (int i = 0; i < k; i++) {
scanf("%d", &movie[i]);
}
for (int i = 0; i < k; i++) {
for (int j = i+1; j < k; j++) {
d[movie[i]][movie[j]] = 1;
d[movie[j]][movie[i]] = 1;
}
}
}
}
void solve()
{
for (int k = 1; k <= n; k ++) {
for (int i = 1; i <= n; i ++) {
for (int j = 1; j <= n; j ++) {
d[i][j] = min(d[i][j], d[i][k]+d[k][j]);
}
}
}
int sum;
int msum = INF;
for (int i = 1; i <= n; i ++) {
sum = 0;
for (int j = 1; j <= n; j ++) {
if (i == j) continue;
sum += d[i][j];
}
msum = (sum < msum) ? sum : msum;
}
double res = (double)msum / (n-1) * 100;
printf("%d\n", (int)res);
}
int main(void)
{
input();
solve();
return 0;
}代码2（dijkstra-邻接矩阵）
Source Code
Problem: 2139 User: liangrx06
Memory: 336K Time: 16MS
Language: C++ Result: Accepted
Source Code
#include <iostream>
#include <cstdio>
using namespace std;
const int N = 300;
const int INF = 0x3f3f3f3f;
int cost[N+1][N+1];
int d[N+1];
bool v[N+1];
int n;
void input()
{
int m;
cin >> n >> m;
for (int i = 1; i <= n; i ++)
fill(cost[i]+1, cost[i]+1+n, INF);
while ( m-- ) {
int k, movie[N];
cin >> k;
for (int i = 0; i < k; i++) {
scanf("%d", &movie[i]);
}
for (int i = 0; i < k; i++) {
for (int j = i+1; j < k; j++) {
cost[movie[i]][movie[j]] = 1;
cost[movie[j]][movie[i]] = 1;
}
}
}
}
int dijkstra(int s)
{
fill(d+1, d+1+n, INF);
fill(v+1, v+1+n, false);
d[s] = 0;
while ( true ) {
int u = -1;
for (int i = 1; i <= n; i ++) {
if ( !v[i] && ( u == -1 || d[i] < d[u] ) )
u = i;
}
if ( u == -1 )
break;
v[u] = true;
for (int i = 1; i <= n; i ++) {
if ( !v[i] && d[u] + cost[u][i] < d[i] )
d[i] = d[u] + cost[u][i];
}
}
int sum = 0;
for (int i = 1; i <= n; i ++)
sum += d[i];
return sum;
}
void solve()
{
int sum;
int msum = INF;
for (int i = 1; i <= n; i ++) {
sum = dijkstra(i);
msum = (sum < msum) ? sum : msum;
}
double res = (double)msum / (n-1) * 100;
printf("%d\n", (int)res);
}
int main(void)
{
input();
solve();
return 0;
}代码3（dijkstra-邻接表）
Source Code
Problem: 2139 User: liangrx06
Memory: 236K Time: 32MS
Language: C++ Result: Accepted
Source Code
#include <iostream>
#include <cstdio>
#include <queue>
#include <vector>
using namespace std;
const int N = 300;
const int INF = 0x3f3f3f3f;
struct Edge {
int to, cost;
Edge(int t, int c) {
to = t;
cost = c;
}
};
typedef pair<int, int> P;
int n;
int d[N+1];
vector<Edge> G[N+1];
void input()
{
int m;
cin >> n >> m;
while ( m-- ) {
int k, movie[N];
cin >> k;
for (int i = 0; i < k; i++) {
scanf("%d", &movie[i]);
}
for (int i = 0; i < k; i++) {
for (int j = i+1; j < k; j++) {
G[movie[i]].push_back(Edge(movie[j], 1));
G[movie[j]].push_back(Edge(movie[i], 1));
}
}
}
}
int dijkstra(int s) {
priority_queue<P, vector<P>, greater<P> > que;
fill(d+1, d+1+n, INF);
d[s] = 0;
que.push(P(0, s));
while ( !que.empty() ) {
P p = que.top(); que.pop();
int u = p.second;
if (d[u] < p.first) continue;
for (int i = 0; i < G[u].size(); i++) {
Edge e = G[u][i];
if ( d[e.to] > d[u] + e.cost) {
d[e.to] = d[u] + e.cost;
que.push(P(d[e.to], e.to));
}
}
}
int sum = 0;
for (int i = 1; i <= n; i ++)
sum += d[i];
return sum;
}
void solve()
{
int sum;
int msum = INF;
for (int i = 1; i <= n; i ++) {
sum = dijkstra(i);
msum = (sum < msum) ? sum : msum;
}
double res = (double)msum / (n-1) * 100;
printf("%d\n", (int)res);
}
int main(void)
{
input();
solve();
return 0;
}
编程算法爱好者。
