icpc 银川 H. Delivery Route SPFA优化

Problem Description

Pony is the boss of a courier company. The company needs to deliver packages to n offices numbered from 1 to n. Especially, the s-th office is the transfer station of the courier company. There are x ordinary two-way roads and y one-way roads between these offices. The delivery vans will consume ci power if they pass through the i-th road. In general, the power consumption on one road must be non-negative. However, thanks to the experimental charging rail, the consumption may be negative on some one-way roads. Besides, Pony got the following public information. The relevant department promised that if there is a one-way street from ai to bi, it is impossible to return from bi to ai. To avoid the delivery vans anchoring on the road, Xiaodao wants to find these lowest power consumptions from the transfer station to these offices

Input

The first line contains four integers n (1 ≤ n ≤ 25000); x; y (1 ≤ x; y ≤ 50000), and s (1 ≤ s ≤ n). This is followed by x + y lines, each line of which contains three integer ai; bi (1 ≤ ai; bi ≤ n; ai 6= bi) and ci (−10000 ≤ ci ≤ 10000) describing the roads. The first x given roads are ordinary two-way roads, and the last y given roads are one-way roads.

Output

The output should contain n lines, the i-th line represents the minimum energy consumption from s-th to the i-th office if possible, or output “NO PATH” if it is impossible to reach the i-th office.

Sample Input

6 3 3 4

1 2 5

3 4 5

5 6 10

3 5 -100

4 6 -100

1 3 -10

Sample Output

NO PATH

NO PATH

50

-95

-100

Analysis of ideas

n个点,x条双向边,y条单向边(有可能负权),s是起点,求s到各个点的最短距离,每个点最多被访问一次

手写双端队列,并且使用vector一直卡在最后一个数据,然后就用了链式前向星

spfa slf+容错

Accepted code

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <algorithm>
#include <set>
#include <vector>
#include <queue>
using namespace std;
#define mem(a,b) memset(a,b,sizeof(a))
#define cin(a) scanf("%d",&a)
#define pii pair<int,int>
#define ll long long
#define gcd __gcd
const int inf = 0x3f3f3f3f;
const int maxn = 200100;
int n,m,k,t;

inline int read(){
    int date=0,w=1;char c=0;
    while(c<'0'||c>'9'){if(c=='-')w=-1;c=getchar();}
    while(c>='0'&&c<='9'){date=date*10+c-'0';c=getchar();}
    return date*w;
}

ll dis[maxn];
bool vis[maxn];
ll val = 0;
int head[maxn], to[maxn], w[maxn], nex[maxn], e = 1;

void add(int a,int b,int c)
{
    to[e] = b;
    nex[e] = head[a];
    head[a] = e;
    w[e] = c;
    e++;
}

int q[20000000];
void spfa(int s)
{
    for(int i = 1;i <= n; i++) dis[i] = 1e18;
    dis[s] = 0;

    int l = 1e7,r = 1e7;    //l是队首,r是队尾
    q[l] = s;

    while(l <= r)
    {
        s = q[l]; l++;
        vis[s] = 0;         //标记出队

        for(int i = head[s]; i ; i = nex[i])
        {
            int v = to[i];
            if(dis[v] > dis[s]+w[i])
            {
                dis[v] = dis[s]+w[i];
                if(!vis[v])
                {
                    vis[v] = 1;
                    if(l > r) q[++r] = v;
                    else
                    {
                        if(dis[v] < dis[q[l]]+val) q[--l] = v;
                        else q[++r] = v;
                    }

                }
            }
        }
    }
}

int main()
{
#ifdef ONLINE_JUDGE
#else
    freopen("in.txt", "r", stdin);
    freopen("out.txt", "w", stdout);
#endif
    int x,y,s,u,v,w;
    n = read(),x = read(), y = read(),s = read();
    for(int i = 0; i < x; i++)
    {
        u = read(),v = read(), w = read();
        add(u,v,w);
        add(v,u,w);
        val += 2*w;
    }
    for(int i = 0; i < y; i++)
    {
        u = read(),v = read(), w = read();
        add(u,v,w);
        val += w;
    }
    val = sqrt(val)/100;
    spfa(s);
    for(int i = 1; i <= n; i++)
    {
        if(dis[i] == 1e18)
        {
            puts("NO PATH");
        }
        else printf("%lld\n",dis[i]);
    }
    return 0;
}

参考博客:

https://blog.csdn.net/guogai13/article/details/103328697