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Re:提供一种思路:可以吧1点的值设为无穷,然后其他点赋值为0,应该很多算法都可以跑In Reply To:提供一种思路:可以吧1点的值设为无穷,然后其他点赋值为0,应该很多算法都可以跑 Posted by:swust5120171204 at 2019-06-06 19:59:35 > 我跑的是spfa > #include<stdio.h> > #include<string.h> > #include<iostream> > #include<algorithm> > #include<math.h> > #include<set> > #include<stack> > #include<vector> > #include<map> > #include<queue> > #define myself i,l,r > #define lson i<<1 > #define rson i<<1|1 > #define Lson i<<1,l,mid > #define Rson i<<1|1,mid+1,r > #define half (l+r)/2 > #define inff 0x3f3f3f3f > #define lowbit(x) x&(-x) > #define PI 3.14159265358979323846 > #define min4(a,b,c,d) min(min(a,b),min(c,d)) > #define min3(x,y,z) min(min(x,y),min(y,z)) > #define pii make_pair > #define pr pair<int,int> > const int dir[4][2]={0,1,0,-1,1,0,-1,0}; > typedef long long ll; > const ll inFF=9223372036854775807; > typedef unsigned long long ull; > using namespace std; > const int maxn=1e5+5; > int head[maxn],sign; > int d[maxn],vis[maxn]; > int n,m; > struct node > { > int to,p,val; > }edge[maxn]; > void add(int u,int v,int val) > { > edge[sign]=node{v,head[u],val}; > head[u]=sign++; > } > void init() > { > sign=0; > memset(head,-1,sizeof(head)); > } > int spfa() > { > for(int i=0;i<=n;i++) > { > d[i]=0; > vis[i]=0; > } > d[1]=inff; > queue<int> q; > q.push(1); > while(!q.empty()) > { > int u=q.front(); > q.pop(),vis[u]=0; > for(int i=head[u];~i;i=edge[i].p) > { > int v=edge[i].to; > if(d[v]<min(edge[i].val,d[u])) > { > d[v]=min(edge[i].val,d[u]); > if(!vis[v]) > { > q.push(v); > vis[v]=1; > } > } > } > } > return d[n]; > } > int main() > { > int t,x,y,z,p=0; > cin>>t; > while(t--) > { > scanf("%d %d",&n,&m); > init(); > for(int i=1;i<=m;i++) > { > scanf("%d %d %d",&x,&y,&z); > add(x,y,z),add(y,x,z); > } > printf("Scenario #%d:\n",++p); > cout<<spfa()<<endl; > printf("\n"); > } > return 0; > } Followed by: Post your reply here: |
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