1672: [Usaco2005 Dec]Cleaning Shifts 清理牛棚
Time Limit: 5 Sec Memory Limit: 64 MBSubmit: 975 Solved: 419
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Description
Farmer John's cows, pampered since birth, have reached new heights of fastidiousness. They now require their barn to be immaculate. Farmer John, the most obliging of farmers, has no choice but hire some of the cows to clean the barn. Farmer John has N (1 <= N <= 10,000) cows who are willing to do some cleaning. Because dust falls continuously, the cows require that the farm be continuously cleaned during the workday, which runs from second number M to second number E during the day (0 <= M <= E <= 86,399). Note that the total number of seconds during which cleaning is to take place is E-M+1. During any given second M..E, at least one cow must be cleaning. Each cow has submitted a job application indicating her willingness to work during a certain interval T1..T2 (where M <= T1 <= T2 <= E) for a certain salary of S (where 0 <= S <= 500,000). Note that a cow who indicated the interval 10..20 would work for 11 seconds, not 10. Farmer John must either accept or reject each individual application; he may NOT ask a cow to work only a fraction of the time it indicated and receive a corresponding fraction of the salary. Find a schedule in which every second of the workday is covered by at least one cow and which minimizes the total salary that goes to the cows.
Input
* Line 1: Three space-separated integers: N, M, and E. * Lines 2..N+1: Line i+1 describes cow i's schedule with three space-separated integers: T1, T2, and S.
Output
* Line 1: a single integer that is either the minimum total salary to get the barn cleaned or else -1 if it is impossible to clean the barn.
Sample Input
0 2 3 //一号牛,从0号stall打扫到2号,工资为3
3 4 2
0 0 1
INPUT DETAILS:
FJ has three cows, and the barn needs to be cleaned from second 0 to second
4. The first cow is willing to work during seconds 0, 1, and 2 for a total
salary of 3, etc.
Sample Output
HINT
约翰有3头牛，牛棚在第0秒到第4秒之间需要打扫．第1头牛想要在第0，1，2秒内工作，为此她要求的报酬是3美元．其余的依此类推． 约翰雇佣前两头牛清扫牛棚，可以只花5美元就完成一整天的清扫．
Source
将时间视为一个点，相邻两点连长度0的边，将每个牛连u到v+1的边，spfa即可。
#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<queue>
using namespace std;
queue<int> q;
int read()
{
char ch=getchar();int f=0;
while(ch<'0'||ch>'9')ch=getchar();
while(ch>='0'&&ch<='9'){f=(f<<1)+(f<<3)+ch-'0';ch=getchar();}
return f;
}
struct node
{
int from;
int to;
int next;
long long w;
}edge[300005];
int n,m,s,t,head[100005],tot;long long dis[100005];bool vis[100005];
long long inf=9000000000000000000LL;
void add(int u,int v,long long w)
{
edge[tot].from=u;
edge[tot].to=v;
edge[tot].w=w;
edge[tot].next=head[u];
head[u]=tot++;
}
void spfa()
{
q.push(s);
while(!q.empty())
{
int x=q.front();
q.pop();
vis[x]=0;
for(int i=head[x];i!=-1;i=edge[i].next)
{
if(dis[edge[i].to]>dis[x]+edge[i].w)
{
dis[edge[i].to]=dis[x]+edge[i].w;
if(!vis[edge[i].to])
{
vis[edge[i].to]=1;
q.push(edge[i].to);
}
}
}
}
}
int main()
{
memset(head,-1,sizeof(head));
m=read();s=read();t=read()+1;
for(int i=1;i<=m;i++)
{
int u=read(),v=read()+1,w=read();
add(u,v,w);
}
for(int i=1;i<=t;i++)
{
dis[i-1]=inf;
add(i,i-1,0);
}
dis[t]=inf;
dis[s]=0;
spfa();
if(dis[t]==inf)puts("-1");
else printf("%lld",dis[t]);
}#include<iostream>
#include<cstdio>
#include<cstring>
#include<cmath>
#include<algorithm>
#define ll long long
#define inf 10000000000
using namespace std;
inline ll read()
{
int x=0,f=1;char ch=getchar();
while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}
while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}
return x*f;
}
int n,bg,ed;
struct seg{int l,r;ll tag,mn;}t[400005];
struct data{int t1,t2,c;}a[10005];
inline bool operator<(data a,data b)
{
return a.t1<b.t1;
}
void pushdown(int k)
{
if(t[k].l==t[k].r)return;
ll tag=t[k].tag;t[k].tag=inf;
if(tag!=inf)
{
t[k<<1].tag=min(t[k<<1].tag,tag);
t[k<<1|1].tag=min(t[k<<1|1].tag,tag);
t[k<<1].mn=min(t[k<<1].mn,tag);
t[k<<1|1].mn=min(t[k<<1|1].mn,tag);
}
}
void build(int k,int l,int r)
{
t[k].l=l;t[k].r=r;t[k].mn=inf;t[k].tag=inf;
if(l==r)return;
int mid=(l+r)>>1;
build(k<<1,l,mid);build(k<<1|1,mid+1,r);
}
ll query(int k,int x)
{
pushdown(k);
if(x<bg)return 0;
int l=t[k].l,r=t[k].r;
if(l==r)return t[k].mn;
int mid=(l+r)>>1;
if(x<=mid)return query(k<<1,x);
else return query(k<<1|1,x);
}
void update(int k,int x,int y,ll val)
{
pushdown(k);
int l=t[k].l,r=t[k].r;
if(x==l&&y==r)
{
t[k].tag=val;
t[k].mn=min(t[k].mn,val);
return;
}
int mid=(l+r)>>1;
if(y<=mid)update(k<<1,x,y,val);
else if(x>mid)update(k<<1|1,x,y,val);
else
{
update(k<<1,x,mid,val);
update(k<<1|1,mid+1,y,val);
}
}
int main()
{
n=read();bg=read();ed=read();
build(1,bg,ed);
for(int i=1;i<=n;i++)
a[i].t1=read(),a[i].t2=read(),a[i].c=read();
sort(a+1,a+n+1);
for(int i=1;i<=n;i++)
{
int t1=a[i].t1,t2=a[i].t2,c=a[i].c;
ll tmp=query(1,t1-1);
if(tmp==inf){puts("-1");return 0;}
update(1,t1,t2,tmp+c);
}
ll ans=query(1,ed);
if(ans<inf)printf("%lld",ans);
else puts("-1");
return 0;
}