社交网络图中结点的“重要性“计算（Dijkstra + SPFA + Floyd + 模板）

题目链接：

无

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题目大意：

求一个点到其他所有点的最短距离和，保证图连通。

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解题过程：

刚开始用 Floyd 水过的，后来用换了几种方法，不错的模板题，Floyd 的时候，要用 vector 存边，否则超内存。

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题目分析

略

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AC代码（Dijkstra + SPFA）

#include<bits/stdc++.h>
using namespace std;const int MAX = 11234, INF = 0x3f3f3f3f;vector<int> edges[MAX];
int dist[MAX], book[MAX];void spfa(int s) {memset(dist, INF, sizeof(dist));memset(book, 0, sizeof(book));queue<int> q;q.push(s);book[s] = 1;dist[s] = 0;while (!q.empty()) {int u = q.front();for (int i = 0; i < edges[u].size(); i++) {int v = edges[u][i];if (dist[v] > dist[u] + 1) {dist[v] = dist[u] + 1;if (!book[v]) {q.push(v);book[v] = 1;}}}q.pop();book[u] = 0;}
}void dijkstra(int s) {memset(dist, INF, sizeof(dist));priority_queue<pair<int, int> > q;dist[s] = 0;q.push(make_pair(-dist[s], s));while (!q.empty()) {int u = q.top().second;q.pop();for (int i = 0; i < edges[u].size(); i++) {int v = edges[u][i];if (dist[v] > dist[u] + 1) {dist[v] = dist[u] + 1;q.push(make_pair(-dist[v], v));}}}
}int main() {int n, m;scanf("%d %d", &n, &m);while (m--) {int u, v;scanf("%d %d", &u, &v);edges[u].push_back(v);edges[v].push_back(u);}int k;scanf("%d", &k);while (k--) {int s;scanf("%d", &s);dijkstra(s);int sum = 0;for (int i = 1; i <= n; i++) {if (i == s)continue;sum += dist[i];}printf("Cc(%d)=%.2f\n", s, (n-1.0)/sum);}
}

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AC代码（Floyd）：

#include<bits/stdc++.h>
using namespace std;
const int INF = 0x3f3f3f3f, MAX = 10001;int main()
{vector<int>edge[MAX];int n, m;scanf("%d %d", &n, &m);for (int i = 0; i <= n; i++) {for (int j = 0; j <= n; j++) {edge[i].push_back(INF);}}for (int i = 1; i <= n; i++) {edge[i][i] = 0;}for (int i = 0; i < m; i++) {int u, v;scanf("%d %d", &u, &v);edge[u][v] = edge[v][u] = 1;}for (int k = 1; k <= n; k++) {for (int i = 1; i <= n; i++) {for (int j = 1; j <= n; j++) {if (edge[i][j] > edge[i][k] + edge[k][j])edge[i][j] = edge[i][k] + edge[k][j];}}}int k;scanf("%d", &k);while (k--) {int c;scanf("%d", &c);double sum = 0;for (int i = 1; i <= n; i++) {if (i == c)continue;sum += edge[c][i];}printf("Cc(%d)=%.2f\n", c, (n-1)/sum);}
}