UVALive 6885 Flowery Trails 最短路
Flowery Trails
题目连接：
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;
}
}
