UESTC 482 Charitable Exchange (线段树）

Description

Have you ever heard a star charity show called Charitable Exchange? In this show, a famous star starts with a small item which values 11yuan. Then, through the efforts of repeatedly exchanges which continuously increase the value of item in hand, he (she) finally brings back a valuable item and donates it to the needy.

In each exchange, one can exchange for an item of Vi yuan if he (she) has an item values more than or equal to RiRi yuan, with a time cost of TiTi minutes.

Now, you task is help the star to exchange for an item which values more than or equal to MM yuan with the minimum time.

Input

The first line of the input is TT (no more than 2020), which stands for the number of test cases you need to solve.

For each case, two integers NN, MM (1≤N≤1051≤N≤105, 1≤M≤1091≤M≤109) in the first line indicates the number of available exchanges and the expected value of final item. Then NN lines follow, each line describes an exchange with 33 integers ViVi, RiRi, TiTi (1≤Ri≤Vi≤1091≤Ri≤Vi≤109, 1≤Ti≤1091≤Ti≤109).

Output

For every test case, you should output Case #k: first, where kk indicates the case number and counts from 11. Then output the minimum time. Output −1−1 if no solution can be found.

Sample Input

3 
3 10 
5 1 3 
8 2 5 
10 9 2 
4 5 
2 1 1 
3 2 1 
4 3 1 
8 4 1 
5 9 
5 1 1 
10 4 10 
8 1 10 
11 6 1 
7 3 8

Sample Output

Case #1: -1 
Case #2: 4 
Case #3: 10

这题需要对于每个价值离散化，价值只有大小关系有用，具体数值是没用的。离散化后就变成了n条边，最多有200000个点，每个点的最小花费是该点到最大值的点之间的花费得的最小值，从左到右扫一便，更新和询问最小值用线段树维护。

#include<cstdio>#include<cstring>#include<cmath>#include<iostream>#include<map>#include<queue>#include<algorithm>using namespace std;struct node{    int l,r,v,next;}p[200005];long long tree[300005*2];long long inf;void update(int rt,int l,int r,int rr,long long v){    if(l>rr||r<rr) return ;    if(l==rr&&r==rr)    {        tree[rt]=v;        return ;    }    int mid=(l+r)/2;    update(2*rt,l,mid,rr,v);    update(2*rt+1,mid+1,r,rr,v);    tree[rt]=min(tree[2*rt+1],tree[2*rt]);}long long query(int rt,int l,int r,int ll,int rr){    if(ll>r||rr<l) return inf;    if(ll<=l&&rr>=r) return tree[rt];    int mid=(l+r)/2;    long long ret=inf;    ret=min(ret,query(2*rt,l,mid,ll,rr));    ret=min(ret,query(2*rt+1,mid+1,r,ll,rr));    return ret;}int h[300005],vis[300005],a[300005],v[3][100005],cnt,r[300005];int n,m,mx;void add(int l,int r,int t){    p[++cnt].l=l; p[cnt].r=r;    p[cnt].next=h[l];h[l]=cnt;    p[cnt].v=t;}void spfa(int st){    int i,j;    for(i=1;i<=mx;i++)    {        update(1,1,mx,i,inf);        vis[i]=0;        r[i]=i;    }    update(1,1,mx,1,0);    for(i=1;i<=mx;i++)    {        int now=i;        long long mi=query(1,1,mx,i,mx);        for(j=h[now];j>0;j=p[j].next)        {            int r=p[j].r;            long long tp=query(1,1,mx,r,mx);            if(tp>mi+p[j].v)            {                update(1,1,mx,r,mi+p[j].v);            }        }    }}int main(){    int T; scanf("%d",&T);    int i,j,k;    inf=1000000000;    inf*=inf;    for(int Case=1;Case<=T;Case++)    {    scanf("%d%d",&n,&m);    cnt=0;memset(h,0,sizeof(h));    map<int,int> f;    f[m]=1;    a[++cnt]=m;    a[++cnt]=1;    f[1]=1;    for(i=1;i<=n;i++)    {        scanf("%d%d%d",&v[0][i],&v[1][i],&v[2][i]);        for(j=0;j<2;j++)        {            if(f[v[j][i]]==0)            {                f[v[j][i]]=1;                a[++cnt]=v[j][i];            }        }    }    sort(a+1,a+1+cnt);    for(i=1;i<=cnt;i++) f[a[i]]=i;    m=f[m];    mx=cnt;    cnt=0;    for(i=1;i<=n;i++)    {        add(f[v[1][i]],f[v[0][i]],v[2][i]);    }    spfa(f[1]);    long long ans=inf;    ans=query(1,1,mx,m,mx);    printf("Case #%d: ",Case);    if(ans==inf) printf("-1\n");    else printf("%lld\n",ans);    }    return 0;}