ZOJ 3378 求割边+最短路

来源：互联网发布：mac charles 编辑：程序博客网时间：2024/06/04 19:38

题目大意：给你一个地图，A君要从0走到n-1，B君想要埋伏A君，问A的必经之路有哪些

题目思路：必经之路首先想到割边，但是割边不一定都是解，因为割边是去掉这条边后形成两个字图（应该是这样的吧。。），但是有可能不需要经过那部分，所以在求下最短路，这样就把不需要经过的去掉了。

注意：即使没有满足条件的边也要输出一个回车，不要问我怎么知道的

代码：

#include<cstdio>#include<iostream>#include<algorithm>#include<cstring>#include<map>#include<queue>#include<set>using namespace std;const int maxn = 10010;const int maxm = 200200;const int inf = 0x3f3f3f3f;struct Side{    int v,next,id;}side[maxm];int top;int node[maxn];map<int,int>num[maxn];void add_side(int u,int v,int id){    side[top] = (Side){v,node[u],id};    node[u] = top ++;}int vis[maxn],low[maxn],dfn[maxn],cnt;set<int>ss;set<int>ans;void dfs(int u,int fa){    vis[u] = 1;    dfn[u] = low[u] = cnt++;    for(int i = node[u];i != -1;i = side[i].next){        int v = side[i].v;        if(v != fa&&vis[v] == 1)            low[u] = min(low[u],dfn[v]);        if(vis[v] == 0){            dfs(v,u);            low[u] = min(low[u],low[v]);            if(low[v] > dfn[u]&&num[u][v] == 1){                ss.insert(side[i].id);            }        }    }}int get_id(int u,int v){    for(int i = node[u];i != -1;i = side[i].next){        if(v == side[i].v){return side[i].id;}    }}int fa[maxn];void spfa(int s,int t){    memset(vis,0,sizeof(vis));    memset(dfn,inf,sizeof(dfn));    queue<int>q;    q.push(s);    dfn[s] = 0;    while(!q.empty()){        int now = q.front();q.pop();        vis[now] = 0;        for(int i = node[now];i != -1;i = side[i].next){            int v = side[i].v;            if(dfn[v] > dfn[now] + 1){                dfn[v] = dfn[now] + 1;                fa[v] = now;                if(vis[v] == 0){                    q.push(v);                    vis[v] = 1;                }            }        }    }    return ;}int main(){    int n,m;    //freopen("in.txt","r",stdin);    while(~scanf("%d%d",&n,&m)){        memset(node,-1,sizeof(node));        top = 0;        for(int i = 0;i < maxn;i ++)num[i].clear();        ss.clear();        ans.clear();        for(int i = 0;i < m;i ++){            int a,b;            scanf("%d%d",&a,&b);            add_side(a,b,i);            add_side(b,a,i);            num[a][b] ++;            num[b][a] ++;        }        memset(vis,0,sizeof(vis));        memset(low,0,sizeof(low));        memset(dfn,-1,sizeof(dfn));        memset(fa,-1,sizeof(fa));        cnt = 0;        for(int i = 0;i < n;i ++){            if(dfn[i] == -1)dfs(i,-1);        }        spfa(0,n-1);        int now = n-1;        int tot = 0;        while(1){            int f = fa[now];            if(f == -1)break;            int id = get_id(f,now);            if(ss.count(id)){ans.insert(id);tot++;}            now = f;        }        printf("%d\n",tot);        set<int>::iterator it;        vector<int>pri;        for(it = ans.begin();it != ans.end();it ++){            pri.push_back((*it));        }        int siz = pri.size();        for(int i = 0;i < siz;i ++){            if(i == siz-1)printf("%d",pri[i]);            else printf("%d ",pri[i]);        }        printf("\n");    }    return 0;}

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