2753: [SCOI2012]滑雪与时间胶囊
Time Limit: 50 Sec Memory Limit: 128 MBSubmit: 1075 Solved: 381
[Submit][Status]
Description
Input
Output
Sample Input
3 3
3 2 1
1 2 1
2 3 1
1 3 10
Sample Output
HINT
【数据范围】
对于30%的数据,保证 1<=N<=2000
对于100%的数据,保证 1<=N<=100000
对于所有的数据,保证 1<=M<=1000000,1<=Hi<=1000000000,1<=Ki<=1000000000。
1 #include<cstdio> 2 3 #include<cstdlib> 4 5 #include<cmath> 6 7 #include<cstring> 8 9 #include<algorithm> 10 11 #include<iostream> 12 13 #include<vector> 14 15 #include<map> 16 17 #include<set> 18 19 #include<queue> 20 21 #include<string> 22 23 #define inf 1000000000 24 25 #define maxn 1000000+1000 26 27 #define maxm 2000000+10000 28 29 #define eps 1e-10 30 31 #define ll long long 32 33 #define pa pair<int,int> 34 35 #define for0(i,n) for(int i=0;i<=(n);i++) 36 37 #define for1(i,n) for(int i=1;i<=(n);i++) 38 39 #define for2(i,x,y) for(int i=(x);i<=(y);i++) 40 41 #define for3(i,x,y) for(int i=(x);i>=(y);i--) 42 43 #define mod 1000000007 44 45 using namespace std; 46 47 inline int read() 48 49 { 50 51 int x=0,f=1;char ch=getchar(); 52 53 while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();} 54 55 while(ch>='0'&&ch<='9'){x=10*x+ch-'0';ch=getchar();} 56 57 return x*f; 58 59 } 60 ll h[maxn]; 61 int n,m,cnt,sum,tot,head[maxn],fa[maxn],a[maxm],b[maxm],w[maxm],q[maxn]; 62 bool can[maxn]; 63 struct edge{int go,next;}e[2*maxm]; 64 struct rec{int x,y;ll w;}c[maxm]; 65 inline bool cmp(rec a,rec b) 66 { 67 return h[a.y]>h[b.y]||(h[a.y]==h[b.y]&&a.w<b.w); 68 } 69 inline void insert(int x,int y) 70 { 71 e[++tot].go=y;e[tot].next=head[x];head[x]=tot; 72 } 73 inline int find(int x){return fa[x]==x?x:fa[x]=find(fa[x]);} 74 int main() 75 76 { 77 freopen("input.txt","r",stdin); 78 freopen("output.txt","w",stdout); 79 n=read();m=read(); 80 for1(i,n)h[i]=read(); 81 for1(i,m) 82 { 83 int x=read(),y=read(); 84 if(h[x]>=h[y])insert(x,y); 85 if(h[y]>=h[x])insert(y,x); 86 a[i]=x;b[i]=y;w[i]=read(); 87 } 88 cnt=0;can[1]=1; 89 int l,r; 90 for(l=0,r=1,q[1]=1;l<=r;l++) 91 for(int i=head[q[l]],y;i;i=e[i].next) 92 if(!can[y=e[i].go])q[++r]=y,can[y]=1; 93 printf("%d ",r); 94 sum=0; 95 for1(i,m)if(can[a[i]]&&can[b[i]]) 96 { 97 if(h[a[i]]>=h[b[i]])c[++sum].x=a[i],c[sum].y=b[i],c[sum].w=w[i]; 98 if(h[b[i]]>=h[a[i]])c[++sum].x=b[i],c[sum].y=a[i],c[sum].w=w[i]; 99 } 100 sort(c+1,c+sum+1,cmp); 101 for1(i,n)fa[i]=i; 102 int j=1; 103 //for1(i,sum)cout<<c[i].x<<' '<<c[i].y<<' '<<c[i].w<<endl; 104 ll ans=0; 105 for1(i,r-1) 106 { 107 while(find(c[j].x)==find(c[j].y))j++; 108 fa[find(c[j].x)]=find(c[j].y); 109 ans+=c[j].w; 110 j++; 111 } 112 printf("%lld\n",ans); 113 114 return 0; 115 116 }
看来有必要学一下朱刘算法,必要的时候骗分。。。