UVALive 6885 Flowery Trails 最短路

Flowery Trails

题目连接：

https://icpcarchive.ecs.baylor.edu/index.php?option=com_onlinejudge&Itemid=8&page=show_problem&problem=4897

Description

In order to attract more visitors, the manager of a national

park had the idea of planting flowers along both

sides of the popular trails, which are the trails used by

common people. Common people only go from the park

entrance to its highest peak, where views are breathtaking,

by a shortest path. So, he wants to know how many

metres of flowers are needed to materialize his idea.

For instance, in the park whose map is depicted in

the figure, common people make only one of the three

following paths (which are the shortest paths from the

entrance to the highest peak).

• At the entrance, some start in the rightmost trail

to reach the point of interest 3 (after 100 metres),

follow directly to point 7 (200 metres) and then pick

the direct trail to the highest peak (620 metres).

• The others go to the left at the entrance and reach

point 1 (after 580 metres). Then, they take one of

the two trails that lead to point 4 (both have 90

metres). At point 4, they follow the direct trail to the highest peak (250 metres).

Notice that popular trails (i.e., the trails followed by common people) are highlighted in yellow. Since

the sum of their lengths is 1930 metres, the extent of flowers needed to cover both sides of the popular

trails is 3860 metres (3860 = 2 × 1930).

Given a description of the park, with its points of interest and (two-way) trails, the goal is to find

out the extent of flowers needed to cover both sides of the popular trails. It is guaranteed that, for the

given inputs, there is some path from the park entrance to the highest peak.

Input

The input file contains several test cases, each of them as described below.

The first line of the input has two integers: P and T. P is the number of points of interest and T

is the number of trails. Points are identified by integers, ranging from 0 to P − 1. The entrance point

is 0 and the highest peak is point P − 1.

Each of the following T lines characterises a different trail. It contains three integers, p1, p2, and

l, which indicate that the (two-way) trail links directly points p1 and p2 (not necessarily distinct) and

has length l (in metres).

Integers in the same line are separated by a single space.

Constraints:

2 ≤ P ≤ 10 000 Number of points.

1 ≤ T ≤ 250 000 Number of trails.

1 ≤ l ≤ 1 000 Length of a trail

Output

For each test case, the output has a single line with the extent of flowers (in metres) needed to cover

both sides of the popular trails.

Sample Input

10 15

0 1 580

1 4 90

1 4 90

4 9 250

4 2 510

2 7 600

7 3 200

3 3 380

3 0 150

0 3 100

7 8 500

7 9 620

9 6 510

6 5 145

5 9 160

4 7

0 1 1

0 2 2

0 3 10

0 3 3

1 3 2

2 3 1

1 1 1

Sample Output

3860

18

Hint

题意

求在最短路上的边的长度和

题解：

枚举边，如果边起点到一端的距离+终点到一端的距离+这条边的长度，那么这条边就在最短路上。

代码

#include<bits/stdc++.h>
using namespace std;
const int maxn = 10005;
const int Maxn = 250005;
int d1[maxn],d2[maxn];
int n,m;
struct node{
    int x,y;
    node(int X,int Y):x(X),y(Y){};
};
vector<node> E[maxn];
int a[Maxn],b[Maxn],c[Maxn];
priority_queue<pair<int,int> >Q;
void init(){
    while(!Q.empty())Q.pop();
    for(int i=0;i<maxn;i++)E[i].clear();
}
int main(){
    while(scanf("%d%d",&n,&m)!=EOF){
        init();
    for(int i=1;i<=m;i++){
        scanf("%d%d%d",&a[i],&b[i],&c[i]);
        E[a[i]].push_back(node{b[i],c[i]});
        E[b[i]].push_back(node(a[i],c[i]));
    }
    for(int i=0;i<maxn;i++)d1[i]=1000000005;
    for(int i=0;i<maxn;i++)d2[i]=1000000005;
    Q.push(make_pair(0,0));
    d1[0]=0;
    while(!Q.empty()){
        int now=Q.top().second;
        Q.pop();
        for(int i=0;i<E[now].size();i++){
            int v=E[now][i].x;
            int sp=E[now][i].y;
            if(d1[v]>d1[now]+sp){
                d1[v]=d1[now]+sp;
                Q.push(make_pair(-d1[v],v));
            }
        }
    }
    Q.push(make_pair(0,n-1));
    d2[n-1]=0;
    while(!Q.empty()){
        int now=Q.top().second;
        Q.pop();
        for(int i=0;i<E[now].size();i++){
            int v=E[now][i].x;
            int sp=E[now][i].y;
            if(d2[v]>d2[now]+sp){
                d2[v]=d2[now]+sp;
                Q.push(make_pair(-d2[v],v));
            }
        }
    }
    long long ans = 0;
    for(int i=1;i<=m;i++){
        if(d1[n-1]==d1[a[i]]+d2[b[i]]+c[i])
            ans+=2ll*c[i];
        else if(d1[n-1]==d2[a[i]]+d1[b[i]]+c[i])
            ans+=2ll*c[i];
    }
    cout<<ans<<endl;
    }
}