UVa1112 - Mice and Maze(Dijkstra和Floyd

题意：给出n个点和m个边及起始点（相当于此问题中的出口点），求从起始点可到达的点的个数

思路：因为出口点为从其它点通过一些边后到达的终点，因此在添加边关系时，起点转为终点，终点转为起点。

代码如下：

#include<iostream>
#include <vector>
#include <queue>
#include <set>
#include <fstream>
#include <cstring>using namespace std;const int N = 110;
const int INF = 0x3fffffff;class Solution
{
public:void init(int n, int e, int t){this->n = n;this->e = e;this->t = t;for (int i = 0; i < n; i++) adjList[i].clear();edges.clear();}void addEdge(int u, int v, int t){Edge edge = {v, u, t};edges.push_back(edge);int size = edges.size();adjList[v - 1].push_back(size - 1);}int dijkstra(int s){for (int i = 0; i < n; i++){d[i] = INF;}d[s - 1] = 0;priority_queue<HeapNode> pq;pq.push({0, s - 1});set<int> done;while (!pq.empty()){HeapNode curNode = pq.top(); pq.pop();if (done.count(curNode.u)) continue;done.insert(curNode.u);int u = curNode.u;for (size_t i = 0; i < adjList[u].size(); i++){Edge& e = edges[adjList[u][i]];if (d[e.from - 1] + e.cost < d[e.to - 1] && d[e.from - 1] + e.cost <= t){d[e.to - 1] = d[e.from - 1] + e.cost;pq.push({d[e.to - 1], e.to - 1});}}}int cnt = 0;for (int i = 0; i < n; i++){if (d[i] != INF) cnt++;}return cnt;}private:class Edge{public:int from, to, cost;};class HeapNode{public:int d, u;bool operator < (const HeapNode& other) const{return d > other.d;}};private:int n, e, t;vector<Edge> edges;vector<int> adjList[N];int d[N];
};int main()
{Solution solver;int cases;#ifndef ONLINE_JUDGEifstream fin("f:\\OJ\\uva_in.txt");streambuf* old = cin.rdbuf(fin.rdbuf());
#endifcin >> cases;while (cases--){int n, e, t, m;cin >> n >> e >> t;solver.init(n, e, t);cin >> m;while (m--){int u, v, t;cin >> u >> v >> t;solver.addEdge(u, v, t);}int cnt = solver.dijkstra(e);cout << cnt << endl;if (cases) cout << endl;}
#ifndef ONLINE_JUDGEcin.rdbuf(old);
#endifreturn 0;
}

floyd_warshall算法实现如下：

#include <cstdio>
#include <fstream>
#include <iostream>
#include <cstring>
#include <algorithm>using namespace std;const int N = 110;
const int INF = 0x3fffffff;class Solution
{
public:void init(int n, int e, int t){this->n = n;this->e = e;this->t = t;for (int i = 0; i < n; i++){for (int j = 0; j < n; j++){if (i != j)f[i][j] = INF;elsef[i][j] = 0;}}}void addEdge(int u, int v, int t){f[u - 1][v - 1] = t;}int floyd_warshall(){memcpy(dp, f, sizeof(f));for (int k = 0; k < n; k++){for (int i = 0; i < n; i++){for (int j = 0; j < n; j++){if (dp[i][k] != INF && dp[k][j] != INF){dp[i][j] = min(dp[i][k] + dp[k][j], dp[i][j]);}}}}int ans = 0;for (int i = 0; i < n; i++){if (dp[i][e - 1] <= t){ans++;}}return ans;}private:int n, e, t;int f[N][N];int dp[N][N];};int main()
{
#ifndef ONLINE_JUDGEifstream fin("D:\\program\\clion\\spoj.txt");streambuf* old = cin.rdbuf(fin.rdbuf());
#endifSolution solver;int cases;cin >> cases;while (cases--){int n, e, t, m;cin >> n  >> e >> t;solver.init(n, e, t);cin >> m;while (m--){int u, v, t;cin >> u >> v >> t;solver.addEdge(u, v, t);}int ans = solver.floyd_warshall();cout << ans << endl;if (cases) cout << endl;}#ifndef ONLINE_JUDGEcin.rdbuf(old);
#endifreturn 0;
}

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