《挑战程序设计竞赛》2.5.1 最短路 AOJ0189 2249 2200 POJ3255 2139 3259 3268（5）

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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 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 CodeProblem: 2139       User: liangrx06Memory: 368K        Time: 32MSLanguage: C++       Result: AcceptedSource 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 CodeProblem: 2139       User: liangrx06Memory: 336K        Time: 16MSLanguage: C++       Result: AcceptedSource 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 CodeProblem: 2139       User: liangrx06Memory: 236K        Time: 32MSLanguage: C++       Result: AcceptedSource 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; }

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