P1144 最短路计数 (bfs)

Posted 2021-03-13 feiief

题目链接

 

 方法1：

　　bfs

Code:

#include <bits/stdc++.h>
# define LL long long
using namespace std;

const int maxn=1000000+10;
int N,M;
vector<int> adj[maxn];
int res[maxn];
int dis[maxn];
const int mod=100003;

void bfs(){
    memset(dis,-1,sizeof(dis));
    queue<int> q;
    q.push(1);
    res[1]=1;
    dis[1]=0;
    while(!q.empty()){
        int cur=q.front();
        q.pop();
        for(int i=0;i<adj[cur].size();++i){
            int v=adj[cur][i];
            if(dis[v]==-1){
                dis[v]=1+dis[cur];
                q.push(v);
            }
            if(dis[v]==1+dis[cur]){
                res[v]=(res[v]+res[cur])%mod;
            }
        }
    }
}

int main(){
    scanf("%d %d", &N, &M);
    for(int i=1;i<=M;++i){
        int a,b;
        scanf("%d %d", &a, &b);
        adj[a].push_back(b);
        adj[b].push_back(a);
    }

    bfs();
    for(int i=1;i<=N;++i){
        printf("%d
", res[i]);
    }
    return 0;
}

方法2:

　　SPFA

Code:

#include <bits/stdc++.h>
# define LL long long
using namespace std;

const int maxn=1000000+10;
int N,M;
vector<int> adj[maxn];
int res[maxn];
int dis[maxn];
const int mod=100003;

void spfa(){
    memset(dis,127,sizeof(dis));
    dis[1]=0;
    res[1]=1;
    queue<int> q;
    int inqueue[maxn];
    q.push(1);
    inqueue[1]=1;
    while(!q.empty()){
        int cur=q.front();
        q.pop();
        inqueue[cur]=0;
        for(int i=0;i<adj[cur].size();++i){
            int v=adj[cur][i];
            if(dis[cur]+1<dis[v]){
                res[v]=res[cur];
                dis[v]=1+dis[cur];
                if(inqueue[v]==0){
                    inqueue[v]=1;
                    q.push(v);
                }
            }else if(dis[cur]+1==dis[v]){
                res[v]=(res[v]+res[cur])%mod;
            }
        }
    }
}

int main(){
    scanf("%d %d", &N, &M);
    for(int i=1;i<=M;++i){
        int a,b;
        scanf("%d %d", &a, &b);
        adj[a].push_back(b);
        adj[b].push_back(a);
    }

    spfa();
    for(int i=1;i<=N;++i){
        printf("%d
", res[i]);
    }
    return 0;
}

 

以上方法只能针对无权图（权值为1）的情况，如若权值不为1，则先用spfa找最短路径，再Memo求路径条数。

Code:

#include <bits/stdc++.h>
# define LL long long
using namespace std;

const int maxn=1000000+10;
int N,M;
vector<int> adj[maxn];
int res[maxn];
int dis[maxn];
const int mod=100003;

void spfa(){
    memset(dis,127,sizeof(dis));
    dis[1]=0;
    queue<int> q;
    int inqueue[maxn];
    q.push(1);
    inqueue[1]=1;
    while(!q.empty()){
        int cur=q.front();
        q.pop();
        inqueue[cur]=0;
        for(int i=0;i<adj[cur].size();++i){
            int v=adj[cur][i];
            if(dis[cur]+1<dis[v]){
                dis[v]=1+dis[cur];
                if(inqueue[v]==0){
                    inqueue[v]=1;
                    q.push(v);
                }
            }
        }
    }
}

int dfs(int u){
    if(u==1){
        return res[1]=1;
    }
    if(res[u]!=-1) return res[u];
    res[u]=0;
    for(int i=0;i<adj[u].size();++i){
        int v=adj[u][i];
        if(dis[v]+1==dis[u]){
            res[u]=(res[u]+dfs(v))%mod;
        }
    }
    return res[u];
}

int main(){
    scanf("%d %d", &N, &M);
    for(int i=1;i<=M;++i){
        int a,b;
        scanf("%d %d", &a, &b);
        adj[a].push_back(b);
        adj[b].push_back(a);
    }

    spfa();
    memset(res,-1,sizeof(res));
    for(int i=1;i<=N;++i){
        dfs(i);
    }

    for(int i=1;i<=N;++i){
        printf("%d
", res[i]);
    }
    return 0;
}