HDU 3760 Ideal Path (bfs，分层)

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此题的自环不存即可，重边都存起来，并不影响结果

分层的过程也可用bfs，但要注意取每层最小值时，取需取该层全部节点向下边值的最小值

#include<functional>#include<algorithm>#include<iostream>#include<cstring>#include<cstdio>#include<cmath>#include<vector>#include<queue>#include<map>#include<set>#define REP(i, n) for(int i=0; i<n; i++)#define CLR(a, b) memset(a, b, sizeof(a))#define LL long longusing namespace std;const int INF = 0x3f3f3f3f;struct Edge {    int u, v, c, next;    bool vis;    Edge(){}    Edge(int u, int v, int c, bool vis, int next): u(u), v(v), c(c), vis(vis), next(next){}};int tot;Edge adj[100010 * 4];int head[100010];void add_edge(int u, int v, int c){    adj[tot] = Edge(u, v, c, false, head[u]);    head[u] = tot++;}int d[100010];set<int>S[2];set<int>::iterator si;vector<int>ans;int n, m;void bfs(){    queue<int>q;    while(!q.empty()) q.pop();    CLR(d, INF); d[n] = 0;    q.push(n);    while (!q.empty())    {        int u = q.front(); q.pop();        for (int r = head[u]; ~r; r = adj[r].next)        {            int v = adj[r].v;            if (d[v] > d[u] + 1)            {                d[v] = d[u] + 1;                q.push(v);            }        }    }    printf("%d\n", d[1]);}void solve(){    bfs();    int now = 0, next;    REP(i, 2) S[i].clear();    S[0].insert(1);    ans.clear();    int step = d[1];    while (step)    {        next = now ^ 1;        int smin = INF;        for (si = S[now].begin(); si != S[now].end(); si++)        {            int u = (*si);            for (int r = head[u]; ~r; r = adj[r].next)            {                int v = adj[r].v;                if (d[v] == step - 1 && smin > adj[r].c)                    smin = adj[r].c;            }        }        ans.push_back(smin);        for (si = S[now].begin(); si != S[now].end(); si++)        {            int u = (*si);            for (int r = head[u]; ~r; r = adj[r].next)            {                int v = adj[r].v;                if (d[v] == step - 1 && smin == adj[r].c)                    S[next].insert(v);            }        }        S[now].clear();        now ^= 1;        step--;    }    REP(i, ans.size())    {        if (i) printf(" ");        printf("%d", ans[i]);    }    printf("\n");}int main(){    int x, y, z;    int T;    scanf("%d", &T);    while (T--)    {        scanf("%d%d", &n, &m);        CLR(head, -1); tot = 0;        REP(i, m)        {            scanf("%d%d%d", &x, &y, &z);            if (x == y) continue;            add_edge(x, y, z);            add_edge(y, x, z);        }        solve();    }}/*126 61 3 13 5 15 6 21 2 12 4 24 6 1*/

注意下面的Edge中vis的定义和使用，标记双向边

#include<functional>#include<algorithm>#include<iostream>#include<cstring>#include<cstdio>#include<cmath>#include<vector>#include<queue>#include<map>#include<set>#define REP(i, n) for(int i=0; i<n; i++)#define CLR(a, b) memset(a, b, sizeof(a))#define LL long longusing namespace std;const int INF = 0x3f3f3f3f;struct Edge {    int u, v, c, next;    bool vis;    Edge(){}    Edge(int u, int v, int c, bool vis, int next): u(u), v(v), c(c), vis(vis), next(next){}};int tot;Edge adj[100010 * 4];int head[100010];void add_edge(int u, int v, int c){    adj[tot] = Edge(u, v, c, false, head[u]);///    head[u] = tot++;}int d[100010];set<int>S[2];set<int>::iterator si;vector<int>ans;int n, m;void bfs(){    queue<int>q;    while(!q.empty()) q.pop();    CLR(d, INF); d[n] = 0;    q.push(n);    while (!q.empty())    {        int u = q.front(); q.pop();        for (int r = head[u]; ~r; r = adj[r].next)        {            int v = adj[r].v;            if (d[v] >= d[u] + 1)            {                if (d[v] > d[u] + 1) q.push(v);                d[v] = d[u] + 1;                adj[r ^ 1].vis = true; ///            }        }    }    printf("%d\n", d[1]);}void solve(){    bfs();    int now = 0, next;    REP(i, 2) S[i].clear();    S[0].insert(1);    ans.clear();    int step = d[1];    while (step)    {        next = now ^ 1;        int smin = INF;        for (si = S[now].begin(); si != S[now].end(); si++)        {            int u = (*si);            for (int r = head[u]; ~r; r = adj[r].next)            {                int v = adj[r].v;                if (adj[r].vis && smin > adj[r].c)///                    smin = adj[r].c;            }        }        ans.push_back(smin);        for (si = S[now].begin(); si != S[now].end(); si++)        {            int u = (*si);            for (int r = head[u]; ~r; r = adj[r].next)            {                int v = adj[r].v;                if (adj[r].vis && smin == adj[r].c)///                    S[next].insert(v);            }        }        S[now].clear();        now ^= 1;        step--;    }    REP(i, ans.size())    {        if (i) printf(" ");        printf("%d", ans[i]);    }    printf("\n");}int main(){    int x, y, z;    int T;    scanf("%d", &T);    while (T--)    {        scanf("%d%d", &n, &m);        CLR(head, -1); tot = 0;        REP(i, m)        {            scanf("%d%d%d", &x, &y, &z);            if (x == y) continue;            add_edge(x, y, z);            add_edge(y, x, z);        }        solve();    }}/*126 61 3 13 5 15 6 21 2 12 4 24 6 1*/