hdu 6181 Two Paths（次短路径长度）POJ 3255 Roadblocks （ 次短路长度）

Two Paths

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 153428/153428 K (Java/Others)
Total Submission(s): 378    Accepted Submission(s): 206

Problem Description

You are given a undirected graph with n nodes (numbered from 1 to n) and m edges. Alice and Bob are now trying to play a game. 
Both of them will take different route from 1 to n (not necessary simple).
Alice always moves first and she is so clever that take one of the shortest path from 1 to n.
Now is the Bob's turn. Help Bob to take possible shortest route from 1 to n.
There's neither multiple edges nor self-loops.
Two paths S and T are considered different if and only if there is an integer i, so that the i-th edge of S is not the same as the i-th edge of T or one of them doesn't exist.

 

Input

The first line of input contains an integer T(1 <= T <= 15), the number of test cases.
The first line of each test case contains 2 integers n, m (2 <= n, m <= 100000), number of nodes and number of edges. Each of the next m lines contains 3 integers a, b, w (1 <= a, b <= n, 1 <= w <= 1000000000), this means that there's an edge between node a and node b and its length is w.
It is guaranteed that there is at least one path from 1 to n.
Sum of n over all test cases is less than 250000 and sum of m over all test cases is less than 350000.

 

Output

For each test case print length of valid shortest path in one line.

 

Sample Input

23 31 2 12 3 41 3 32 11 2 1

 

Sample Output

53
Hint
For testcase 1, Alice take path 1 - 3 and its length is 3, and then Bob will take path 1 - 2 - 3 and its length is 5.For testcase 2, Bob will take route 1 - 2 - 1 - 2  and its length is 3

求与 一条最短路径长度  不同的路径  的最短长度

#include <bits/stdc++.h>using namespace std;typedef long long LL;const int N= 1e5+10;typedef long long LL;struct node{    int to, next;    LL w;} p[N*4];int head[N], cnt;void init(){    memset(head,-1,sizeof(head));    cnt=0;    return ;}void add(int u,int v,LL w){    p[cnt].to=v,p[cnt].next=head[u],p[cnt].w=w;    head[u]=cnt++;    p[cnt].to=u,p[cnt].next=head[v],p[cnt].w=w;    head[v]=cnt++;    return ;}LL dist[N][2], vis[N][2];struct node1{    int v, mark;    LL w;    bool operator <(const node1 &A)const    {        if(w!=A.w) return w>A.w;        return v>v;    }};priority_queue <node1>q;LL dij(int s,int t){    memset(dist,-1,sizeof(dist));    memset(vis,0,sizeof(vis));    while(!q.empty())q.pop();    dist[s][0]=0;    node1 tmp;    tmp.v=s,tmp.w=0,tmp.mark=0;    q.push(tmp);    while(!q.empty())    {        tmp=q.top();q.pop();        int u=tmp.v;        LL w=tmp.w;        if(vis[u][tmp.mark]) continue;        vis[u][tmp.mark]=1;        for(int i=head[u]; i!=-1; i=p[i].next)        {            int v=p[i].to;            if(dist[v][0]==-1||dist[v][0]>w+p[i].w)            {                dist[v][1]=dist[v][0];                dist[v][0]=w+p[i].w;                node1 x;                x.v=v,x.w=dist[v][0],x.mark=0;                q.push(x);            }            else if(dist[v][1]==-1||dist[v][1]>w+p[i].w)            {                dist[v][1]=w+p[i].w;                node1 x;                x.v=v,x.w=dist[v][1],x.mark=1;                q.push(x);            }        }    }    return dist[t][1];}int main(){    int t, ncase=1;    scanf("%d", &t);    while(t--)    {        int n, m;        scanf("%d %d", &n, &m);        init();        for(int i=0; i<m; i++)        {            int x, y;            LL w;            scanf("%d %d %lld", &x, &y, &w);            add(x,y,w);        }        printf("%lld\n", dij(1,n));    }    return 0;}

Roadblocks

Time Limit: 2000MS Memory Limit: 65536KTotal Submissions: 15811 Accepted: 5580

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 intersectionN.

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

Output

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

Sample Input

4 41 2 1002 4 2002 3 2503 4 100

Sample Output

450

求次短路长度（与最短路长度不同）

#include <iostream>#include <cstring>#include <string>#include <queue>#include <vector>#include <map>#include <set>#include <stack>#include <cmath>#include <cstdio>#include <algorithm>using namespace std;typedef long long LL;const int N= 1e5+10;typedef long long LL;struct node{    int to, next;    LL w;} p[N*4];int head[N], cnt;void init(){    memset(head,-1,sizeof(head));    cnt=0;    return ;}void add(int u,int v,LL w){    p[cnt].to=v,p[cnt].next=head[u],p[cnt].w=w;    head[u]=cnt++;    p[cnt].to=u,p[cnt].next=head[v],p[cnt].w=w;    head[v]=cnt++;    return ;}LL dist1[N], dist2[N], vis[N];typedef pair<LL,int>pi;priority_queue <pi,vector<pi>,greater<pi> >q;LL dij(int s,int t){    memset(dist1,-1,sizeof(dist1));    memset(dist2,-1,sizeof(dist2));    memset(vis,0,sizeof(vis));    while(!q.empty())q.pop();    dist1[s]=0;    q.push(make_pair(0,s));    while(!q.empty())    {        pi tmp=q.top();q.pop();        int u=tmp.second;        LL w=tmp.first;        if(dist2[u]!=-1&&dist2[u]<w) continue;        for(int i=head[u]; i!=-1; i=p[i].next)        {            int v=p[i].to;            LL d=w+p[i].w;            if(dist1[v]==-1||dist1[v]>d)            {                swap(dist1[v],d);                q.push(make_pair(dist1[v],v));            }            if((d!=-1&&dist1[v]<d)&&(dist2[v]==-1||dist2[v]>d))            {                dist2[v]=d;                q.push(make_pair(dist2[v],v));            }        }    }    return dist2[t];}int main(){    int t, ncase=1;    int n, m;    while(scanf("%d %d", &n, &m)!=EOF)    {        init();        for(int i=0; i<m; i++)        {            int x, y;            LL w;            scanf("%d %d %lld", &x, &y, &w);            add(x,y,w);        }        printf("%lld\n", dij(1,n));    }    return 0;}