HDU 3790最短路径问题

来源：互联网发布：网络安全大数据分析编辑：程序博客网时间：2024/06/06 09:08

用的spfa的思想写的，我感觉用一个标记数组flag[]来标志是否在队列中，好像不是很有用啊，只要做了松弛操作，我感觉都应该入队才是吧= =。这道题目就是更新路径的时候顺便更新一下花费。

#include <cstdio>#include <iostream>#include <cstring>#include <queue>#include <vector>using namespace std;const int INF = 0x3f3f3f;const int MAX = 1e3 + 5;int d[MAX], p[MAX];//, flag[MAX];struct EDGE{    int to, cost, price;    EDGE(int a = 0, int b = 0, int c = 0)    : to(a), cost(b), price(c)    {}};vector<EDGE> que[MAX];queue<int> coor;int main(){    int n, m;    while (scanf("%d%d", &n, &m))    {        if (n == 0 && m == 0)        {            break;        }        for (int i = 1; i <= n; ++i)            que[i].clear();        int t_from, t_to, t_cost, t_price;        //memset(flag, 0, sizeof(flag));        memset(d, INF, sizeof(d));        memset(p, INF, sizeof(p));        for (int i = 0; i < m; ++i)        {            scanf("%d%d%d%d", &t_from, &t_to, &t_cost, &t_price);            que[t_from].push_back(EDGE(t_to, t_cost, t_price));            que[t_to].push_back(EDGE(t_from, t_cost, t_price));        }        int st, ed;        scanf("%d%d", &st, &ed);       // flag[st] = 1;        d[st] = 0;        p[st] = 0;        coor.push(st);        int res = INF;        while (!coor.empty())        {            int t_coor = coor.front();            coor.pop();            //flag[t_coor] = 0;            int lenth = que[t_coor].size();            for (int i = 0; i < lenth; ++i)            {                EDGE t_edge = que[t_coor][i];                if (d[t_edge.to] > d[t_coor] + t_edge.cost)                {                    d[t_edge.to] = d[t_coor] + t_edge.cost;                   // printf("from(%d) to(%d)\n", t_coor, t_edge.to);                   // printf("shortest path = %d price = %d\n", d[ed], p[ed]);                  //  if (p[t_edge.to] >= p[t_coor] + t_edge.price)                    p[t_edge.to] = p[t_coor] + t_edge.price;                    //printf("now price = %d\n", p[ed]);                   // res = min(res, p[ed]);                    coor.push(t_edge.to);                }                else if (d[t_edge.to] == d[t_coor] + t_edge.cost && p[t_edge.to] > p[t_coor] + t_edge.price)                    p[t_edge.to] = p[t_coor] + t_edge.price;            }        }        printf("%d %d\n", d[ed], p[ed]);    }    return 0;}

0 0