poj 3255 次短路+dijkstra+挑战模板

题目：

Roadblocks

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 14688		Accepted: 5163

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

Hint

Two routes: 1 -> 2 -> 4 (length 100+200=300) and 1 -> 2 -> 3 -> 4 (length 100+250+100=450)

Source

USACO 2006 November Gold

代码：

#include<iostream>
#include<stdio.h>
#include<algorithm>
#include<queue>
using namespace std;
const int maxn=5e3+30;
const int inf=0x7f7f7f7f;
//const __int64 inf=(__int64)1<<62;//0x3f3f3f3f有时太小了.... 

typedef pair<__int64,int> p;//<a,b> 起点到b的最短距离为a 

struct edge {
	int to;
	int cost;
	edge(int tto=0,int ccost=0):to(tto),cost(ccost){}
};

int n,m;
int d[maxn],d2[maxn];
vector<edge> g[maxn];

void dijkstra(int s) {
	priority_queue<p,vector<p>,greater<p> >qu;//堆按照p的first（最短距离）排序，小顶堆 
	fill(d,d+n+1,inf);
	fill(d2,d2+n+1,inf);
//	memset(prev,sizeof(prev),-1);//混用会报错 
	d[s]=0;
	qu.push(p(0,s));//从起点出发到顶点s的最短距离为0
	while(!qu.empty()) {
		p P=qu.top();
		qu.pop();
		int tmpv=P.second,tmpd=P.first;//该顶点编号以及S到该顶点的一种距离 
		if(tmpd>d2[tmpv]) continue;//若该距离大于到该顶点的次短距离，舍弃 
		for(int i=0;i<g[tmpv].size();++i) {//遍历该顶点连出的每条边 
			edge e=g[tmpv][i];
			int tmpd2=tmpd+e.cost;//走这条路到达e.to的距离 
			if(d[e.to]>tmpd2) {//优先更新最短距离 
				swap(d[e.to],tmpd2); //原最短距离可能会变成新的次短距离 
				qu.push(p(d[e.to],e.to));
			}
			if(d2[e.to]>tmpd2&&d[e.to]<tmpd2){//此时tmpd2应该用于更新次短距离 
				d2[e.to]=tmpd2;
				qu.push(p(d2[e.to],e.to)); 
			}
		}
	}
}

int main() {//3792K	204MS
	int a,b,c;
	scanf("%d%d",&n,&m);
	for(int i=0;i<=n;++i) {
		g[i].clear();
	}
	for(int i=1;i<=m;++i) {
		scanf("%d%d%d",&a,&b,&c);
		g[a].push_back(edge(b,c));
		g[b].push_back(edge(a,c));
	}
	dijkstra(1);
	printf("%d\n",d2[n]);
	return 0;
}