CDOJ 482 Charitable Exchange bfs

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Charitable Exchange




　　Time Limit: 20 Sec
　　Memory Limit: 256 MB

题目连接
　　http://acm.uestc.edu.cn/#/problem/show/482

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 $1$ yuan. 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 $R_i$ yuan, with a time cost of $T_i$ minutes.
　　Now, you task is help the star to exchange for an item which values more than or equal to $M$ yuan with the minimum time.
　　Input
　　The first line of the input is $T$ (no more than $20$), which stands for the number of test cases you need to solve.
　　For each case, two integers $N$, $M$ ($1 \leq N \leq 10^5$, $1 \leq M \leq 109$) in the first line indicates the number of available exchanges and the expected value of final item. Then $N$ lines follow, each line describes an exchange with $3$ integers $V_i$, $R_i$, $T_i$ ($1 \leq R_i \leq V_i \leq 10^9$, $1 \leq T_i \leq 109$).
　　Output


　　For every test case, you should output Case #k: first, where $k$ indicates the case number and counts from $1$. Then output the minimum time. Output $-1$ 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

HINT
　　
　　题意

　　每一个物品价值为x，可以由价值为r的物品来换，需要花费t秒
　　然后问你最小花费多少秒能够得到价值至少为m的物品
　　一开始你有价值为1的物品
　　题解：
　　为毛我的dp+线段树优化t了，迷的不行
　　正解用优先队列，然后直接bfs就好了
　　代码



#include <cstdio>
#include <cmath>
#include <cstring>
#include <ctime>
#include <iostream>
#include <algorithm>
#include <set>
#include <vector>
#include <sstream>
#include <queue>
#include <typeinfo>
#include <fstream>
#include <map>
#include <stack>
typedef long long ll;
using namespace std;
//freopen("D.in","r",stdin);
//freopen("D.out","w",stdout);
#define sspeed ios_base::sync_with_stdio(0);cin.tie(0)
#define test freopen("test.txt","r",stdin)
const int maxn=202501;
#define mod 1000000007
#define eps 1e-9
const int inf=0x3f3f3f3f;
const ll infll = 0x3f3f3f3f3f3f3f3fLL;
inline ll read()
{
ll 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;
}
//**************************************************************************************
struct node
{
ll r,v,t;
}a[maxn];
bool cmp(node aa,node  bb)
{
if(aa.r==bb.r&&aa.v==bb.v)
return aa.t<bb.t;
if(aa.r==bb.r)
return aa.v<bb.v;
return aa.r<bb.r;
}
struct node2
{
ll x,t;
bool operator<(const node2 &p1)const{
return t>p1.t||t==p1.t&&x>p1.x;
}
};
ll ans=0;
int n,m;
ll bfs()
{
priority_queue<node2>q;
node2 xx;
xx.x=1;
xx.t=0;
q.push(xx);
int B=0;
int i;
ll anss=-1;
while(!q.empty())
{
xx=q.top();
q.pop();
if(xx.x>=m)
{
anss=xx.t;
break;
}
for(i=B;i<=n;i++)
{
if(a.r>xx.x)
break;
if(a.r<=xx.x&&a.v>xx.x)
{
node2 kk;
kk.x=a.v;
kk.t=xx.t+a.t;
q.push(kk);
}
}
B=i;
}
return anss;
}
int main()
{
int t=read();
for(int cas=1;cas<=t;cas++)
{
n=read(),m=read();
for(int i=0;i<n;i++)
a.v=read(),a.r=read(),a.t=read();
sort(a,a+n,cmp);
ans=0;
ans=bfs();
printf("Case #%d: %lld\n",cas,ans);
}
}