Acdream 1415 Shortest circuit + bridge

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Test instructions: For a picture, ask you how many bridges are on the shortest path from 1 to N.

Idea: first to meet the conditions of the shortest circuit, then judge each edge is not the shortest side of the edge, how to judge is also very simple, starting from 1 to find the shortest and starting from N to find the fastest way, for the edge U to V, if the most short-circuit of 1 to u plus the shortest short of N to V in addition to the weight of this edge if equal to 1 to N, So this side is what we want, that's the condition if (dis1[u[i]]+cost[i]+dis2[v[i]]==maxdis| | Dis1[v[i]]+cost[i]+dis2[u[i]]==maxdis) Maxdis is 1 to n the shortest short length, and then the topic said there may be heavy side, good processing, before the teammate told me test instructions said no heavy side, WA for a long time, Then the ID and the number of edges to find the output, just the application of the template, it is not difficult ~ ~ ~

#include <queue> #include <vector> #include <stdio.h> #include <stdlib.h> #include <string.h > #include <iostream> #include <algorithm>using namespace Std;typedef long Long ll;const ll inf=    0x3f3f3f3f3f3f3f3fll;const int maxn=20010;struct edge{int to,cost; Edge () {} edge (int a,int b) {to=a;cost=b;}}; typedef pair<int,long LONG&GT;P;VECTOR&LT;EDGE&GT;G[MAXN];LL dis1[maxn],dis2[maxn];void Dijkstra (int st,ll dis[])    {priority_queue<p,vector<p>,greater<p> >que;    Fill (Dis,dis+maxn,inf);    dis[st]=0;    Que.push (P (0,st));        while (!que.empty ()) {P p=que.top (); Que.pop ();        int V=p.second;        if (Dis[v]<p.first) continue;            for (unsigned int i=0;i<g[v].size (); i++) {Edge e=g[v][i];                if (dis[e.to]>dis[v]+ (LL) e.cost) {dis[e.to]=dis[v]+ (ll) e.cost;            Que.push (P (dis[e.to],e.to));   }}}}int u[100010],v[100010],cost[100010];struct edge1{ int to,id,num; Edge1 (int a,int b,int c) {to=a;id=b;num=c;}}; Vector<edge1>gg[maxn];int l[maxn],e[maxn],ans[maxn],vis[maxn],stack1[maxn];int k,kk,cnt,n,m;void dfs (int x, int fa) {vis[x]=1; L[x]=k;    E[x]=k++;stack1[kk++]=x;    int flag=0;        for (unsigned int i=0;i<gg[x].size (); i++) {edge1 e=gg[x][i];                if (E.TO!=FA) {if (!vis[e.to]) {DFS (e.to,x);                L[x]=min (l[x],l[e.to]);            if (l[e.to]>e[x]) gg[x][i].num=1;        }else l[x]=min (l[x],e[e.to]);            }else{if (flag) l[x]=min (l[x],e[e.to]);        flag++;            }} if (L[x]==e[x]) {while (stack1[kk]!=x&&kk>0) {l[stack1[kk-1]]=l[x];            kk--;        vis[stack1[kk]]=0;    }}}void Tarjan () {Kk=0;k=1;dfs ();    cnt=0;            for (int i=1;i<=n;i++) {for (unsigned int j=0;j<gg[i].size (); j + +) {edge1 e=gg[i][j]; if (l[i]!=l[e.to]&&e.num==1) {ANS[CNT++]=e.id;        }}}}int Main () {while (scanf ("%d%d", &n,&m)!=-1) {for (int i=0;i<maxn;i++) g[i].clear ();        for (int i=0;i<maxn;i++) gg[i].clear ();            for (int i=0;i<m;i++) {scanf ("%d%d%d", &u[i],&v[i],&cost[i]);            G[u[i]].push_back (Edge (v[i],cost[i));        G[v[i]].push_back (Edge (u[i],cost[i));        } Dijkstra (1,DIS1);        Dijkstra (N,DIS2);        ll Maxdis=dis1[n]; for (int i=0;i<m;i++) {if (dis1[u[i]]+cost[i]+dis2[v[i]]==maxdis| |                Dis1[v[i]]+cost[i]+dis2[u[i]]==maxdis) {gg[u[i]].push_back (Edge1 (v[i],i+1,0));            Gg[v[i]].push_back (Edge1 (u[i],i+1,0));        }} Tarjan ();        printf ("%d\n", CNT);        Sort (ans,ans+cnt);        for (int i=0;i<cnt-1;i++) printf ("%d", ans[i]);    printf ("%d\n", ans[cnt-1]); } return 0;}

Acdream 1415 Shortest circuit + bridge