PATA1030题解

最短路径模版题

//
//  main.cpp
//  PATA1030
//
//  Created by Phoenix on 2018/2/8.
//  Copyright © 2018年 Phoenix. All rights reserved.
//

#include <iostream>
#include <cstdio>
#include <vector>
#include <queue>
using namespace std;
const int maxn = 510;
const int inf = 1000000000;
int G[maxn][maxn], C[maxn][maxn], n, m;
int d[maxn], st, ed;
bool vis[maxn] = {false};
vector<int> pre[maxn];

void dijkstra(int s) {
    fill(d, d + maxn, inf);
    d[s] = 0;
    for(int i = 0; i < n; i++) {
        int u = -1, MIN = inf;
        for(int j = 0; j < n; j++) {
            if(vis[j] == false && d[j] < MIN) {
                u = j;
                MIN = d[j];
            }
        }
        if(u == -1) return;
        vis[u] = true;
        for(int v = 0; v < n; v++) {
            if(vis[v] == false && G[u][v] != inf) {
                if(G[u][v] + d[u] < d[v]){
                    pre[v].clear();
                    d[v] = G[u][v] + d[u];
                    pre[v].push_back(u);
                } else if(G[u][v] + d[u] == d[v]) {
                    pre[v].push_back(u);
                }
            }
        }
    }
}

vector<int> tempPath, path;
int optvalue = inf;
void DFS(int v) {
    if(v == st) {
        tempPath.push_back(v);
        int value = 0;
        for(int i = 0; i < tempPath.size() - 1; i++) {
            value += C[tempPath[i]][tempPath[i+1]];
        }
        if(value < optvalue) {
            path = tempPath;
            optvalue = value;
        }
        tempPath.pop_back();
        return;
    }
    tempPath.push_back(v);
    for(int i = 0; i < pre[v].size(); i++) {
        DFS(pre[v][i]);
    }
    tempPath.pop_back();
}

int main(int argc, const char * argv[]) {
    scanf("%d %d %d %d", &n, &m, &st, &ed);
    fill(G[0], G[0] + maxn * maxn, inf);
    for(int i = 0; i < m; i++) {
        int a, b, len, cost;
        scanf("%d %d %d %d", &a, &b, &len, &cost);
        G[a][b] = G[b][a] = len;
        C[a][b] = C[b][a] = cost;
    }
    dijkstra(st);
    DFS(ed);
    for(int i = path.size() - 1; i >= 0; i--) {
        printf("%d ", path[i]);
    }
    printf("%d %d", d[ed], optvalue);
    return 0;
}