洛谷 P3906 Geodetic集合

题目大意：

$n$个点 $m$条遍的无向图，如果点 $i$在点 $u$到点 $v$的最短路径( $u$到 $v$的边数最少)上，那么记这些点为集合 $I(u,v)$
有 $k$个询问，问集合 $I(u,v)$

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解题思路：

每次询问以u为起点跑一遍 $Dij$，将以 $u$为起点到其他点的最短路记为 $dis1$，再以 $v$为起点跑一遍 $Dij$，记为 $dis2$
$\forall i\epsilon [1,n]$，若 $dis1[i]+dis2[i]=dis1[v](dis2[u])$那么 $i\epsilon I(u,v)$

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$Accepted\ code$

#include<queue>
#include<cstdio>
#include<algorithm>

using namespace std;

const int inf = 1e9;

struct Line {
	int to, w, next;
}e[5005];

priority_queue < pair<int, int> > q;

int n, m, u, v, cnt;
int dis[50][2], vis[50], last[50], a[5005];

inline void addline(int x, int y) {
	e[++cnt] = (Line){y, 1, last[x]}; last[x] = cnt;
}

void dij(int S, int nm) {
	for (int i = 1; i <= n; ++i) dis[i][nm] = inf, vis[i] = 0;
	dis[S][nm] = 0;
	while (q.size()) q.pop();
	q.push(make_pair(0, S));
	while (q.size()) {
		int x = q.top().second; q.pop();
		if (vis[x]) continue;
		vis[x] = 1;
		for (int i = last[x]; i; i = e[i].next) {
			int y = e[i].to;
			if (dis[y][nm] > dis[x][nm] + e[i].w) {
				dis[y][nm] = dis[x][nm] + e[i].w;
				if (!vis[y]) q.push(make_pair(-dis[y][nm], y));
			}
		}
	}
	return;
}

int main() {
	scanf("%d %d", &n, &m);
	for (int i = 1, x = 0, y = 0; i <= m; ++i)
		scanf("%d %d", &x, &y), addline(x, y), addline(y, x);
	int Q = 0; scanf("%d", &Q);
	while (Q--) {
		scanf("%d %d", &u, &v);
		dij(u, 0); dij(v, 1);
		for (int i = 1; i <= n; ++i)
			if (dis[i][0] + dis[i][1] == dis[v][0]) printf("%d ", i);
		printf("\n");
	}
}