poj-2449-Remmarguts' Date(A*算法+Dijkstra)

题目就是求两点之间第k短路，但如何起点和终点相等的时候，k需要加1

涉及到的算法是Dijkstra和A*寻路算法

#include <iostream>
#include <iomanip>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <string>
#include <cmath>
#include <queue>
#include <vector>
#include <algorithm>
#define ll long long
using namespace std;
const int inf = 0x3fffffff;
const int MAXN = 1003;
const int MAXM = 100005;
int n, m;
struct Edge{
    int u, v, w;
} edge[MAXM];
int head[MAXN], to[MAXM], next[MAXM], length[MAXM], tot;
int dist[MAXN];
bool vis[MAXN];
priority_queue<pair<int, int> > heap;

void init() {
    tot = 0;
    memset(head, -1, sizeof(head));
}

void addEdge(int u, int v, int w) {
    to[tot] = v;
    length[tot] = w;
    next[tot] = head[u];
    head[u] = tot++;
}

bool dijkstra(int s, int t) {
    for (int i = 0; i <= n; i++) {
        dist[i] = inf;
        vis[i] = false;
    }
    dist[s] = 0;
    for (int i = 1; i <= n; i++) {
        int Min = inf;
        int v = -1;
        for (int j = 1; j <= n; j++) {
            if (!vis[j] && Min > dist[j]) {
                Min = dist[j];
                v = j;
            }
        }
        if (v == -1) {
            break;
        }
        vis[v] = true;
        for (int j = head[v]; j != -1; j = next[j]) {
            if (!vis[to[j]] && dist[to[j]] > dist[v] + length[j]) {
                dist[to[j]] = dist[v] + length[j];
            }
        }
    }
    return dist[t] != inf;
}

int solve(int a, int b, int k) {
    init();
    for (int i = 0; i < m; i++) {
        //建图的时候把边都变成反向边 
        addEdge(edge[i].v, edge[i].u, edge[i].w);
    }
    if (!dijkstra(b, a)) {
        //dijkstra算法求的是从b到a的最短路 
        return -1;
    }
    init();
    for (int i = 0; i < m; i++) {
        // 这时候再把边变成正向边 
        addEdge(edge[i].u, edge[i].v, edge[i].w);
    }
    while(!heap.empty()) {
        heap.pop();
    }
    heap.push(pair<int, int>(-dist[a], a));
    // heap里 第一个数是从第二个数到b的距离的负值
    // heap会先按照第一个int降序排列，第一个相等的就按照第二个int降序排列
    // 为了是最近的放在第一个位置，就把距离变成负值 
    while(!heap.empty()) {
        pair<int, int> ret = heap.top();
        heap.pop();
        int real = -ret.first - dist[ret.second];
        // real是从a到ret.second 的真正距离 
        if (ret.second == b) {
            if (!--k) {
                return real;
            }
        }
        for (int i = head[ret.second]; i != -1; i = next[i]) {
            heap.push(pair<int, int>(-(real + length[i] + dist[to[i]]), to[i]));
            // 放到heap的是从a到ret.second再到to[i]再到b的最短距离的负值 
        }
    }
    return -1;
}

int main()
{
#ifndef ONLINE_JUDGE
    freopen("1.txt", "r", stdin);
#endif
    int a, b, k, i, j;
    while(cin >> n >> m) {
        for (i = 0; i < m; i++) {
            scanf("%d%d%d", &edge[i].u, &edge[i].v, &edge[i].w);
        }
        scanf("%d%d%d", &a, &b, &k);
        if (a == b) {
            // a和b相等的时候k一定要加1 
            k++;
        }
        cout << solve(a, b, k) << endl;
    }
    return 0;
}