poj3255 次短路裸题

Description

Bessie has moved to a small farm and sometimes enjoys returning to visit one of her best friends. She does not want to get to her old home too quickly, because she likes the scenery along the way. She has decided to take the second-shortest rather than the shortest path. She knows there must be some second-shortest path.

The countryside consists of R (1 ≤ R ≤ 100,000) bidirectional roads, each linking two of the N (1 ≤ N ≤ 5000) intersections, conveniently numbered 1..N. Bessie starts at intersection 1, and her friend (the destination) is at intersection N.

The second-shortest path may share roads with any of the shortest paths, and it may backtrack i.e., use the same road or intersection more than once. The second-shortest path is the shortest path whose length is longer than the shortest path(s) (i.e., if two or more shortest paths exist, the second-shortest path is the one whose length is longer than those but no longer than any other path).

Input

Line 1: Two space-separated integers: N and R

Lines 2.. R+1: Each line contains three space-separated integers: A, B, and D that describe a road that connects intersections A and B and has length D (1 ≤ D ≤ 5000)

Output

Line 1: The length of the second shortest path between node 1 and node N

Sample Input

4 4

1 2 100

2 4 200

2 3 250

3 4 100

Sample Output

450

次短路其实就是最短路的板子改改…

当新的路比最短路还短那么更新最短路，次短路更新成原来的最短路

当新的路比最短路要长比次短路要短那么更新次短路

所有的边全部扔进优先队列…

新的路不管是次短路还是最短路全部都扔进去更新

#include<iostream>
#include<vector>
#include<algorithm>
#include<cstdio>
#include<cstring>
#include<queue>
using namespace std;
vector<int>tu[5001],juli[5001];
int yijing[5001][2];
long long daon[5001][2];
int inf=0x3f3f3f3f;
struct p
{
long long dian,chang,biaoji;
p (long long diann = 0, long long changg = 0,long long biaojii=0) : dian(diann), chang(changg),biaoji(biaojii) {}
bool operator< (const p&a)const{
return chang>a.chang;
}
};
int main()
{
#define int long long
int n,m;
cin>>n>>m;
int q,w,e;
memset(daon,0x3f,sizeof(daon));
memset(yijing,0,sizeof(yijing));
for(int a=1;a<=m;a++)
{
scanf("%d%d%d",&q,&w,&e);
tu[q].push_back(w);
tu[w].push_back(q);
juli[q].push_back(e);
juli[w].push_back(e);
}
priority_queue<p>qq;
qq.push(p(1,0,0));
daon[1][0]=0;
while(!qq.empty())
{
p zas=qq.top();
qq.pop();
if(yijing[zas.dian][zas.biaoji])continue;
yijing[zas.dian][zas.biaoji]=1;
for(int a=0;a<tu[zas.dian].size();a++)
{
if(!yijing[tu[zas.dian][a]][0]&&daon[tu[zas.dian][a]][0]>daon[zas.dian][zas.biaoji]+juli[zas.dian][a])
{
daon[tu[zas.dian][a]][1]=daon[tu[zas.dian][a]][0];
daon[tu[zas.dian][a]][0]=daon[zas.dian][0]+juli[zas.dian][a];
qq.push(p(tu[zas.dian][a],daon[tu[zas.dian][a]][0],0));
qq.push(p(tu[zas.dian][a],daon[tu[zas.dian][a]][1],1));
}
else if(!yijing[tu[zas.dian][a]][1]&&daon[tu[zas.dian][a]][1]>daon[zas.dian][zas.biaoji]+juli[zas.dian][a])
{
daon[tu[zas.dian][a]][1]=daon[zas.dian][zas.biaoji]+juli[zas.dian][a];
qq.push(p(tu[zas.dian][a],daon[tu[zas.dian][a]][1],1));
}
}
}
cout<<daon
[1]<<endl;
return 0;
}