题目链接:http://acm./showproblem.php?pid=1535
分析:
题意:求1点到其它点的最短距离之和+其它点到1点的最短距离之和
前面一部分直接用SPFA算法求出,而后一部分可用一数组存放反向边
(所有边的方向都反一下),利用反向边SPFA求出1点到其它点距离即可。
#include <iostream>#include <cstdio>#include <cmath>#include <cstdlib>#include <string>#include <cstring>#include <algorithm>#include <queue>using namespace std;const int inf = 0xfffffff;const int maxn = 1000000+10;bool vis[maxn];int h1[maxn],h2[maxn],dis[maxn];int t1,t2,n,m;struct node{int x,v,next;}f1[maxn],f2[maxn];///f1存放顺向边, f2存放反向边void init(){t1=t2=0;memset(h1,-1,sizeof(h1));memset(h2,-1,sizeof(h2));}void addnode_1(int a,int b,int c){f1[t1].x=b;f1[t1].v=c;f1[t1].next=h1[a];h1[a]=t1++;}void addnode_2(int a,int b,int c){f2[t2].x=b;f2[t2].v=c;f2[t2].next=h2[a];h2[a]=t2++;}int spfa(node F[ ],int H[ ]){memset(vis,false,sizeof(vis));for(int i=1;i<=n;++i)dis[i]=inf;dis[1]=0;vis[1]=true;queue<int>M;M.push(1);while(!M.empty()){int now=M.front(); M.pop();vis[now]=false;for(int i=H[now];i!=-1;i=F[i].next){int next=F[i].x;if(dis[next]>dis[now]+F[i].v){dis[next]=dis[now]+F[i].v;if(!vis[next]){vis[next]=true;M.push(next);}}}}int sum=0;for(int i=2;i<=n;++i)sum+=dis[i];return sum;}int main(){int T; scanf("%d",&T);while(T--){scanf("%d%d",&n,&m);init();while(m--){int a,b,c;scanf("%d%d%d",&a,&b,&c);addnode_1(a,b,c);addnode_2(b,a,c);}int ans=spfa(f1,h1)+spfa(f2,h2);cout<<ans<<endl;}return 0;}