洛谷 P1342

思路

分別建一張原圖，一張與原圖中所有邊方向相反的反圖。在兩張圖上分別跑一遍SPFA，再計算出兩次dis數組的和即可。

Code

#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<algorithm>
#include<queue>
typedef long long LL;
#define maxn 1000005

using namespace std;

int n,m;
int pre1[maxn],last1[maxn],other1[maxn],len1[maxn],tot1;
 
int pre2[maxn],last2[maxn],other2[maxn],len2[maxn],tot2;
int dis1[maxn],dis2[maxn];
bool vis1[maxn],vis2[maxn];

inline void add1(int x,int y,int z){
    tot1++;
    pre1[tot1]=last1[x];
    last1[x]=tot1;
    other1[tot1]=y;
    len1[tot1]=z;
}

inline void add2(int x,int y,int z){
    tot2
 
++;
    pre2[tot2]=last2[x];
    last2[x]=tot2;
    other2[tot2]=y;
    len2[tot2]=z;
}

void spfa1(int s){
    queue<int>q;
    memset(dis1,0x3f,sizeof(dis1));
    q.push(s);
    dis1[s]=0;
    vis1[s]=1;
    while(!q.empty()){
        int u=q.front();
        q.pop();
        vis1[u]
 
=0;
        for(register int p=last1[u];p;p=pre1[p]){
            int v=other1[p];
            if(dis1[v]>dis1[u]+len1[p]){
                dis1[v]=dis1[u]+len1[p];
                if(!vis1[v]){
                    q.push(v);
                    vis1[v]=1;
                }
            }
        }
    }
}

void spfa2(int s){
    queue<int>q;
    memset(dis2,0x3f,sizeof(dis2));
    q.push(s);
    dis2[s]=0;
    vis2[s]=1;
    while(!q.empty()){
        int u=q.front();
        q.pop();
        vis2[u]=0;
        for(register int p=last2[u];p;p=pre2[p]){
            int v=other2[p];
            if(dis2[v]>dis2[u]+len2[p]){
                dis2[v]=dis2[u]+len2[p];
                if(!vis2[v]){
                    q.push(v);
                    vis2[v]=1;
                }
            }
        }
    }
}

int main(){
    scanf("%d%d",&n,&m);
    for(register int i=1;i<=m;i++){
        int x,y,z;
        scanf("%d%d%d",&x,&y,&z);
        add1(x,y,z);
        add2(y,x,z);
    }
    spfa1(1);
    spfa2(1);
    LL ans=0;
    for(register int i=1;i<=n;i++) ans+=dis1[i]+dis2[i];
    printf("%lld",ans);
}

[題解] P1342 請柬