LUOGU P1342 请柬(最短路)

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解题思路

又是一道语文题，弄清楚题意之后其实就能想出来了，从1跑一遍最短路，把$dis[n]$加入答案。在建个反图跑一遍最短路，把$dis[n]_$加入最短路就行了。第一遍是去的时候，第二遍是回的时候。

#include<iostream>
#include<cstdio>
#include<cstring>
#include<cmath>
#include<queue>

using namespace std;
const int MAXN = 1000005;
typedef long long LL;

inline int rd(){
    int x=0,f=1;char ch=getchar();
    while(!isdigit(ch)) {f=ch=='-'?0:1;ch=getchar();}
    while(isdigit(ch))  {x=(x<<1)+(x<<3)+ch-'0';ch=getchar();}
    return f?x:-x;
}

int n,m,head[MAXN],cnt;
int to[MAXN],nxt[MAXN],val[MAXN];
int head_[MAXN],cnt_,val_[MAXN],nxt_[MAXN],to_[MAXN];
LL dis[MAXN],ans;
bool vis[MAXN];

struct Node{
    int w,id;
    friend bool operator<(const Node A,const Node B){
        return A.w>B.w;
    }
};

priority_queue<Node> Q;

inline void add(int bg,int ed,int w){
    to[++cnt]=ed,val[cnt]=w,nxt[cnt]=head[bg],head[bg]=cnt;
}

inline void add_(int bg,int ed,int w){
    to_[++cnt_]=ed,val_[cnt_]=w,nxt_[cnt_]=head_[bg],head_[bg]=cnt_;
}

void dijkstra(){
    memset(dis,0x3f,sizeof(dis));
    Node now;now.id=1;now.w=0;dis[1]=0;Q.push(now);
    while(!Q.empty()){
        Node zz=Q.top();Q.pop();int x=zz.id;
        if(dis[x]!=zz.w || vis[x]) continue;vis[x]=1;
        for(register int i=head[x];i;i=nxt[i]){
            int u=to[i];
            if(dis[u]<=dis[x]+val[i]) continue;
            dis[u]=dis[x]+val[i];
            now.id=u;now.w=dis[u];Q.push(now);
        }
    }
    for(register int i=2;i<=n;i++) ans+=dis[i];
}

void dijkstra_(){
    memset(vis,false,sizeof(vis));
    memset(dis,0x3f,sizeof(dis));
    Node now;now.id=1;now.w=0;dis[1]=0;Q.push(now);
    while(!Q.empty()){
        Node zz=Q.top();Q.pop();int x=zz.id;
        if(dis[x]!=zz.w || vis[x]) continue;vis[x]=1;
        for(register int i=head_[x];i;i=nxt_[i]){
            int u=to_[i];
            if(dis[u]<=dis[x]+val_[i]) continue;
            dis[u]=dis[x]+val_[i];
            now.id=u;now.w=dis[u];Q.push(now);
        }
    }
    for(register int i=2;i<=n;i++) ans+=dis[i];
}


int main(){
    n=rd(),m=rd();int x,y,z;
    for(register int i=1;i<=m;i++){
        x=rd(),y=rd(),z=rd();
        add(x,y,z),add_(y,x,z);
    }
    dijkstra(),dijkstra_();cout<<ans;
    return 0;
}