HDU 2680 Choose the best route多源点最短路

题目链接

http://acm.hdu.edu.cn/showproblem.php?pid=2680

题意

多源点最短路。。。

思路

源点有多个，将所有源点与超级源点连一条距离为0的边，跑最短路算法就行了

#include<cstdio>#include<queue>#include<iostream>#include<vector>#include<map>#include<cstring>#include<string>#include<set>#include<stack>#include<algorithm>#define cle(a) memset(a,0,sizeof(a))#define inf(a) memset(a,0x3f,sizeof(a))#define ll long long#define Rep(i,a,n) for(int i=a;i<=n;i++)using namespace std;#define INF2 9223372036854775807llconst int INF = ( 2e9 ) + 2;const ll maxn = 1010;const int maxm = 2e4+10;struct edge{    int v,w,next;}e[maxm];int head[maxn],d[maxn],vis[maxn];int tot;void addedge(int u,int v,int w){    e[tot].v=v;    e[tot].w=w;    e[tot].next=head[u];    head[u]=tot++;}int SPFA(int s,int t,int n){    for(int i=0;i<=n;i++)    d[i]=INF;    d[s]=0;    queue<int> q;    q.push(s);    vis[s]=1;    while(!q.empty())    {        int u=q.front();q.pop();        vis[u]=0;        for(int i=head[u];i!=-1;i=e[i].next)        {            int v=e[i].v;            int w=e[i].w;            if(d[v]>d[u]+w)            {                d[v]=d[u]+w;                if(!vis[v])                {                    q.push(v);                    vis[v]=1;                }            }        }    }    if(d[t]==INF)return -1;    else return d[t];}int main(){    int n,m,s;    while(~scanf("%d%d%d",&n,&m,&s))    {        tot=0;        memset(head,-1,sizeof(head));        for(int i=0;i<m;i++)        {            int u,v,w;            scanf("%d%d%d",&u,&v,&w);            addedge(u,v,w);        }        int p,temp;        scanf("%d",&p);        for(int i=0;i<p;i++)        {            scanf("%d",&temp);            addedge(0,temp,0);        }        int ans=SPFA(0,s,n);        printf("%d\n",ans);    }}