最短路变形题目 HDU多校7

Mr.Quin love fishes so much and Mr.Quin’s city has a nautical system,consisiting of N ports and M shipping lines. The ports are numbered 1 to N. Each line is occupied by a Weitian. Each Weitian has an identification number.

The i-th (1≤i≤M) line connects port Ai and Bi (Ai≠Bi) bidirectionally, and occupied by Ci Weitian (At most one line between two ports).

When Mr.Quin only uses lines that are occupied by the same Weitian, the cost is 1 XiangXiangJi. Whenever Mr.Quin changes to a line that is occupied by a different Weitian from the current line, Mr.Quin is charged an additional cost of 1 XiangXiangJi. In a case where Mr.Quin changed from some Weitian A's line to another Weitian's line changes to Weitian A's line again, the additional cost is incurred again.

Mr.Quin is now at port 1 and wants to travel to port N where live many fishes. Find the minimum required XiangXiangJi (If Mr.Quin can’t travel to port N, print −1

instead)

InputThere might be multiple test cases, no more than 20. You need to read till the end of input.

For each test case,In the first line, two integers N (2≤N≤100000) and M (0≤M≤200000), representing the number of ports and shipping lines in the city.

In the following m lines, each contain three integers, the first and second representing two ends Ai and Bi of a shipping line (1≤Ai,Bi≤N) and the third representing the identification number Ci (1≤Ci≤1000000) of Weitian who occupies this shipping line.OutputFor each test case output the minimum required cost. If Mr.Quin can’t travel to port N, output −1 instead.Sample Input

3 3 
1 2 1
1 3 2
2 3 1
2 0
3 2
1 2 1
2 3 2

Sample Output

1
-1
2

题意 ： 给你 n 个点，m 条边，在给你一些两点间的路径值，让你求1 - n的最小花费，当你改变航线以后所消耗的权值就会 + 1；

思路分析 ： 就是一个正常的最短路变形题目，但是好卡时限啊， 2449ms ，

　　　　基本就是正常的最短路更新，从一个点更新到另一个点时，同时用set记录一下有哪些状态到达了这个点，一旦可以更新，就清空此集合中的全部元素

代码示例 ：

　　dij

using namespace std;
#define ll long long
const int maxn = 1e5+5;
const int inf = 0x3f3f3f3f;

int n, m;
struct node
{
    int to, pt, cost, fa;
    bool operator< (const node &v)const{
        return cost > v.cost;
    }
};
priority_queue<node>que;
vector<node>ve[maxn];
bool vis[maxn];
int dis[maxn];
set<int>s[maxn];

void solve(){
    while(!que.empty()) que.pop();
    for(int i = 1; i <= 100000; i++) s[i].clear();
    memset(vis, false, sizeof(vis));
    memset(dis, inf, sizeof(dis));
    que.push({1, 0, 0, 0});
    dis[1] = 0;
    
    while(!que.empty()){
        node v = que.top(); que.pop();
        
        int x = v.to;
        if (v.cost > dis[x]) continue;
        for(int i = 0; i < ve[x].size(); i++){
            int to = ve[x][i].to;
            int pt = ve[x][i].pt;
            if (to == v.fa) continue;
            int cost = 0;
            if (s[x].count(pt) == 0) cost++;
            
            if (dis[x]+cost < dis[to]) {
                dis[to] = dis[x]+cost;
                s[to].clear();
                s[to].insert(pt);
                que.push({to, pt, dis[to], x});
            }
            else if (dis[x]+cost == dis[to] && s[to].count(pt) == 0){
                s[to].insert(pt); 
                que.push({to, pt, dis[to], x});
            }
        }
    }
    if (dis[n] == inf) puts("-1");
    else 
        printf("%d\n", dis[n]);
}

int main() {
    int u, v, w;
    
    while(~scanf("%d%d", &n, &m)){
        for(int i = 1; i <= 100000; i++) ve[i].clear();
        for(int i = 1; i <= m; i++){
            scanf("%d%d%d", &u, &v, &w);    
            ve[u].push_back({v, w, 0, 0});            
            ve[v].push_back({u, w, 0, 0});
        }
        solve();
    }
    return 0;
}

spfa

using namespace std;
#define ll long long
const int maxn = 1e5+5;
const int mod = 1e9+7;
const double eps = 1e-9;
const double pi = acos(-1.0);
const int inf = 0x3f3f3f3f;

int n, m;
struct node
{
    int to, pt, fa;
    //bool operator< (const node& v)const{
        //return cost > v.cost;
    //}
};
queue<node>que;
vector<node>ve[maxn];
bool vis[maxn];
int dis[maxn];
set<int>s[maxn];

void solve(){
    while(!que.empty()) que.pop();
    for(int i = 1; i <= 100000; i++) s[i].clear();
    memset(vis, false, sizeof(vis));
    memset(dis, inf, sizeof(dis));
    que.push({1, 0, 0});
    dis[1] = 0; vis[1] = 1;
    
    while(!que.empty()){
        node v = que.front(); que.pop();
        
        int x = v.to;
        vis[x] = 0;
        for(int i = 0; i < ve[x].size(); i++){
            int to = ve[x][i].to;
            int pt = ve[x][i].pt;
            if (to == v.fa) continue;
            int cost = 0;
            if (s[x].count(pt) == 0) cost++;
            
            if (dis[x]+cost < dis[to]) {
                dis[to] = dis[x]+cost;
                s[to].clear();
                s[to].insert(pt);
                if (!vis[to]) {
                    que.push({to, pt, x});
                    vis[to] = 1;
                } 
            }
            else if (dis[x]+cost == dis[to] && s[to].count(pt) == 0){
                s[to].insert(pt);
                //que.push({to, pt, dis[to], x});
                if (!vis[to]) {
                    que.push({to, pt, x});
                    vis[to] = 1;
                }
            }
        }
    }
    if (dis[n] == inf) puts("-1");
    else 
        printf("%d\n", dis[n]);
}

int main() {
    //freopen("in.txt", "r", stdin);
    //freopen("out.txt", "w", stdout);
    int u, v, w;
    
    while(~scanf("%d%d", &n, &m)){
        for(int i = 1; i <= 100000; i++) ve[i].clear();
        for(int i = 1; i <= m; i++){
            scanf("%d%d%d", &u, &v, &w);    
            ve[u].push_back({v, w, 0});            
            ve[v].push_back({u, w, 0});
        }
        solve();
    }
    return 0;
}