堵塞

堵塞

随着卡司阵容就必须求去影院撸一发,笔者还是能有这种执念,也是不轻松。真的比小编预想的还要风趣,脑子里翻出了让子弹飞,太极,钢的琴,里面包车型的士湖北明星们正是让自家不止欣喜,第一是因为看黑龙江影视和偶像剧还是广大,所以熟识,第二是他们的演技真的放进大荧光屏也还是很给力。舒淇女士,王千源先生,张孝全先生,真的都太棒了,而舒淇(Shu Qi)的微表情,真的太迷人了!满分!太爱!配乐也是十一分有趣,阿卡Bella用到表演者情景音乐融入,逗乐的同临时间,扎了你一针,是当真不痛不痒不首要?谈及洗脑,大胆啊,明目张胆啊,容小编思念,还应该有何样想说的。

Two Paths

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 153428/153428
K (Java/Others)
Total Submission(s): 613    Accepted Submission(s): 312

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

2 3 3 1 2 1 2 3 4 1 3 3 2 1 1 2 1

 

 

Sample Output

5 3

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

 

才讲的K短路,第二天多校就考到这些。=-=

 

比较裸,直接用A*算法+Dij堆优化,不加堆优化的Dij会超时的,并且以此题数据边的w值会十分大,所以具有的边的变量,中间累加的变量,数组存下的长短,最终的A*函数重返结果(一上马WA了众数次,就在此地,再次来到值也要long
long,只顾改了变量了,白白WA了几发)都要设置成long
long,第贰个正是无向边了,建图和反向建图的八个数组都要push一回。那也是一开始反向建图少一个就WA。

 

#include <iostream>
#include <algorithm>
#include <vector>
#include <cstring>
#include <queue>
#include <cstdio>
#define MAXN 100010
#define INF (1LL<<62)
using namespace std;
typedef long long ll;
typedef pair<int, ll> P;
int N, M, S, T, K;
ll dist[MAXN];
ll tdist[MAXN];
int cnt[MAXN];
bool f[MAXN];
vector<P> Adj[MAXN];
vector<P> Rev[MAXN];
struct Edge {
    int to;
    ll len;
    Edge(){}
    Edge(int t, ll l):to(t), len(l){}
};
priority_queue<Edge> q;

bool operator<(const Edge &a, const Edge &b) {
    return (a.len + dist[a.to]) > (b.len + dist[b.to]);
}

void dijkstra() {
    memset(dist, 0, sizeof(dist));
    fill(tdist, tdist+MAXN, INF);
    tdist[T] = 0;
    while(!q.empty()) q.pop();
    q.push(Edge(T, 0));
    while (!q.empty()) {
        int x = q.top().to;
        ll d = q.top().len;
        q.pop();
        if (tdist[x] < d) continue;
        for (int i = 0; i < Rev[x].size(); i++) {
            int y = Rev[x][i].first;
            ll len = Rev[x][i].second;
            if (d+ len < tdist[y]) {
                tdist[y] = d + len;
                q.push(Edge(y, tdist[y]));
            }
        }
    }
    for (int i = 1; i <= N; i++){
        dist[i] = tdist[i];
    }
}

//注意这里是long long的返回结果啊!
ll aStar() {
    if (dist[S] == INF) return -1;
    while (!q.empty()) q.pop();
    q.push(Edge(S, 0));
    memset(cnt, 0, sizeof(cnt));
    while (!q.empty()) {
        int x = q.top().to;
        ll d = q.top().len;
        q.pop();
        cnt[x]++;
        if (cnt[T] == K) return d;
        if (cnt[x] > K) continue;
        for (int i = 0; i < Adj[x].size(); i++) {
            int y = Adj[x][i].first;
            ll len = Adj[x][i].second;
            q.push(Edge(y, d+len));
        }
    }

    return -1;
}

int main() {
//    freopen("in.txt","r",stdin);
    std::ios::sync_with_stdio(false),cin.tie(0),cout.tie(0);
    int tt;
    cin>>tt;
    while(tt--) {
    //初始化,忘了就WA
    for(int i = 0; i <= N; i++) {
        Adj[i].clear();
        Rev[i].clear();
    }
    int a, b;
    ll t;
    cin >> N >> M;
    for (int i = 0; i < M; i++) {
        cin >> a >> b >> t;
        Adj[a].push_back(make_pair(b, t));
        Adj[b].push_back(make_pair(a, t));
        Rev[b].push_back(make_pair(a, t));
        Rev[a].push_back(make_pair(b, t));
    }
    S = 1;
    T = N;
    K = 2;
//    cin >> S >> T >> K;

    if (S == T) K++;

    dijkstra();
    cout << aStar() << endl;
    }
    return 0;
}

 

次短路做法,用的旁人模版:

 

规律便是在存dis最短路的时候还要用另外多个数组存次短路,在最短路调换边和dis[]值得时候,次短路存下新交流的没最短边长的值就可以。

#include<cstdio>
#include<cstring>
#include<queue>
#include<functional>
#include<vector>
#include<algorithm>
using namespace std;
typedef long long int ll;
#define MAXM  350010
#define MAXN 250010
#define INF 1223372036854775807
typedef pair<ll,int> pp;

typedef struct node
{
    int v;
    ll w;
    node(int tv=0,ll tw=0):
        v(tv),w(tw){};
}node;

int n,m;
vector<node> edge[MAXM];
ll dis[MAXN],dis2[MAXN];   //dis[i]表示最短路,dis2[i]表示次短路


void solve()
{
    fill(dis+1,dis+n+1,INF);
    fill(dis2+1,dis2+n+1,INF);


    priority_queue<pp, vector<pp>, greater<pp> > q;  //用优先队列加速搜索
    dis[1]=0;
    q.push(pp(dis[1],1));   //second是该边指向(虽然是无向的)的顶点,first是这条边的权值
    while(q.size())
    {
        pp p=q.top(); q.pop();
        int v=p.second;
        ll d=p.first;
        if(dis2[v]<d) continue;  //如果当前取出的值不是到v的最短路或次短路就contniue,因为v->e的最短边和次短边一定是由->v的最短边和次短边+edge(v,e)得到
        for(int i=0;i<edge[v].size();i++)
        {
            int e=edge[v][i].v;
            ll d2=d+edge[v][i].w;
            if(dis[e]>d2)
            {
                swap(dis[e],d2);
                q.push(pp(dis[e],e));
            }
            if(dis2[e]>d2&&d2>dis[v])  //d2>dis[v]防止d2小于dis[v],这样v->e就变成负权边了,但可有可无
            {
                dis2[e]=d2;
                q.push(pp(dis2[e],e));
            }
        }
    }
    printf("%lld\n",dis2[n]);


}

int main()
{
    int t;
    scanf("%d",&t);
    while(t--)
    {
        scanf("%d%d",&n,&m);
        for(int i=1;i<=n;i++) edge[i].clear();
        for(int i=0;i<m;i++)
        {
            int a,b;
            ll c;
            scanf("%d%d%lld",&a,&b,&c);
            edge[a].push_back(node(b,c));
            edge[b].push_back(node(a,c));
        }
        solve();
    }
}

 

 另个本子,那些版本INF必需求long long 工夫过,为啥啊??

#include <iostream>
#include <cstdio>
#include <cstring>
#include <vector>
#include <queue>
#define INF (1LL<<62)
using namespace std;
typedef long long ll;
typedef pair<ll,int> P;
struct edge{
    int to;
    ll v;
    edge(int to,ll v):to(to),v(v){}
    edge(){}
};
const int maxn = 100010;
const int maxe = 100010;
int V,E;
vector<edge> g[maxn];
ll d[maxn],d2[maxn];//最短距离和次短距离
void dijkstra(int s)
{
    priority_queue<P,vector<P>,greater<P> > pq;
    for(int i=1;i<=V;i++)
    {
        d[i]=INF;
        d2[i]=INF;
    }
    d[s]=0;
    pq.push(P(0,s));
    while(pq.size())
    {
        P nowe=pq.top();pq.pop();
        if(nowe.first>d2[nowe.second]) continue;  //如果这个距离比当前次短路长continue
        for(int v=0;v<(int)g[nowe.second].size();v++)
        {
            edge nexte=g[nowe.second][v];
            ll dis=nowe.first+nexte.v;
            if(d[nexte.to]>dis)
            {
                swap(dis,d[nexte.to]);
                pq.push(P(d[nexte.to],nexte.to));
            }
            if(d2[nexte.to]>dis&&d[nexte.to]<dis)//保证最短路是小于这个次短路的
            {
                d2[nexte.to]=dis;
                pq.push(P(d2[nexte.to],nexte.to));//次短路的点进入pq
            }
        }
    }
}
int main()
{
//    freopen("in.txt","r",stdin);
    int TT;
    scanf("%d",&TT);
    while(TT--) {
    int s;//起点
    scanf("%d%d",&V,&E);
    {
        for(int i=1;i<=V;i++)
            g[i].clear();
        for(int i=1;i<=E;i++)
        {
            int f,t;
            ll v;
            scanf("%d%d%I64d",&f,&t,&v);
            g[f].push_back(edge(t,v));
            g[t].push_back(edge(f,v));
        }
        s=1;//这题默认起点为1
        dijkstra(s);
        printf("%I64d\n",d2[V]);
    }
    }
    return 0;
}

 

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