【UVA1057】Routing

【UVA1057】Routing

题面

洛谷

题解

有一个比较好想的dp就是\(f_{i,j}\)表示第一个点在\(i\)，第二个点在\(j\)的最小点数，但是直接搞不好转移。
考虑建出反图，那么\(j\)表示在反图上的点\(j\)其实是和正图上的是一样的。

这样子的话我们枚举出边转移：

\[f[v][u2]=f[u1][u2]+[u2!=v],((u1,v)\in G)\\ f[u1][v]=f[u1][u2]+[u1!=v],((u2,v)\in G') \]

然而我们交换两条路径时发现点数会算多，这种情况我们用另一种方式转移：

\[f[u2][u1]=\min(f[u2][u1],f[u1][u2]+dis[u1][u2]-1) \]

其中\(dis[u1][u2]\)表示\(u1,u2\)间的最短路，可以用\(floyd\)求出。

代码

#include <iostream> 
#include <cstdio> 
#include <cstdlib> 
#include <cstring> 
#include <cmath> 
#include <algorithm>
#include <queue>
#include <vector> 
using namespace std; 
const int INF = 1e9; 
const int MAX_N = 105; 
int N, M; 
int f[MAX_N][MAX_N], dis[MAX_N][MAX_N]; 
vector<int> G[MAX_N], E[MAX_N]; 
bool inq[MAX_N][MAX_N]; 
void spfa() { 
	queue<pair<int, int> > que; 
	for (int i = 1; i <= N; i++) 
		for (int j = 1; j <= N; j++) f[i][j] = INF; 
	f[1][1] = 1, inq[1][1] = 1, que.push(make_pair(1, 1)); 
	while (!que.empty()) { 
		pair<int, int> p = que.front(); que.pop(); 
		int x1 = p.first, x2 = p.second; 
		for (auto v : G[x1]) 
			if (f[v][x2] > f[x1][x2] + (v != x2)) { 
				f[v][x2] = f[x1][x2] + (v != x2); 
				if (!inq[v][x2]) inq[v][x2] = 1, que.push(make_pair(v, x2)); 
			}
		for (auto v : E[x2]) 
			if (f[x1][v] > f[x1][x2] + (v != x1)) { 
				f[x1][v] = f[x1][x2] + (v != x1); 
				if (!inq[x1][v]) inq[x1][v] = 1, que.push(make_pair(x1, v)); 
			} 			
		if (x1 != x2 && f[x2][x1] > f[x1][x2] + dis[x1][x2] - 1) { 
			f[x2][x1] = f[x1][x2] + dis[x1][x2] - 1; 
			if (!inq[x2][x1]) inq[x2][x1] = 1, que.push(make_pair(x2, x1)); 
		} 
		inq[x1][x2] = 0; 
	} 
} 
int main () { 
	int Case = 0; 
	while (scanf("%d %d", &N, &M) != EOF) { 
		if (!N && !M) break;
		printf("Network %d\n", ++Case); 
		for (int i = 1; i <= N; i++) G[i].clear(), E[i].clear(); 
		for (int i = 1; i <= N; i++) 
			for (int j = 1; j <= N; j++) dis[i][j] = INF; 
		for (int i = 1; i <= N; i++) dis[i][i] = 0; 
		for (int i = 1; i <= M; i++) { 
			int u, v; scanf("%d %d", &u, &v); 
			dis[u][v] = 1, G[u].push_back(v), E[v].push_back(u); 
		} 
		for (int k = 1; k <= N; k++) 
			for (int i = 1; i <= N; i++) 
				for (int j = 1; j <= N; j++) 
					dis[i][j] = min(dis[i][j], dis[i][k] + dis[k][j]); 
		if (dis[1][2] == INF || dis[2][1] == INF) puts("Impossible"); 
		else spfa(), printf("Minimum number of nodes = %d\n", f[2][2]); 
		putchar('\n'); 
	} 
    return 0; 
} 

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