ACDream 1734 Can you make a water problem?(贪心)

Can you make a water problem?

Time Limit: 2000/1000MS (Java/Others) Memory Limit: 128000/64000KB (Java/Others)

Submit Statistic Next Problem

Problem Description

Losanto want to make a water problem. But he have no idea….
Then he thought a problem: A bus arrived at a station, and there were x persons got off the bus, then y persons got on the bus. And there were r persons(include the driver) on the bus before the bus arrived at next station. How many person on the bus initially? The driver can’t get on or off the bus. There is only 1 driver in the bus.
But if the x=8 , y=5 r=4, there are something wrong in the problem. After got on 5 persons and there are 4 persons rest…What a pity! Losanto give you x, y, r want know whether the problem is OK.
Of course few buses has only one station, so there are n stations.
Now give you x and y for every station, and there is only one person (the driver) rest after the last station. Can you find a suitable order for the station to make the problem OK?

Input

There are several cases.

For each case, First line there is a n, indicate n stations. Then n lines followed, each line has two numbers xi and yi. To make the problem easier, we guarantee that xi>=yi.(0<n<100000  0<=xi,yi<1000000)

Output

For each cases, output one line with “Yes” (if you can make an OK order) or “No” (if you can’t).

Sample Input

2
5 0
7 4
2
5 2
3 3

Sample Output

Yes
No

Manager

rihkddd

 

题意：有n个站，给出每个站的下车人数xi和上车人数yi，问怎么把站排序使得符合客观条件。

分析：把站按照上车人数从多到少排序，然后判断下是否符合客观条件，即，车上的人数应该不小于下车的人数。

#include<cstdio>
#include<iostream>
#include<cstring>
#include<cmath>
#include<stdlib.h>
#include<vector>
#include<queue>
#include<stack>
#include<algorithm>
#define LL long long
using namespace std;
const int MAXN=100000+5;
LL tot;
int n;
struct node
{
    int x,y;
    bool operator<(const node A)const
    {
        return x-y>A.x-A.y;
    }
}a[MAXN];
bool cheak()
{
    for(int i=0;i<n;i++)
    {
        if(tot<a[i].x) return false;
        tot-=a[i].y;
    }
    return true;
}
int main()
{
    while(scanf("%d",&n)!=EOF)
    {
        tot=0;
        for(int i=0;i<n;i++)
        {
            scanf("%d %d",&a[i].x,&a[i].y);
            tot+=(a[i].x-a[i].y);
            a[i].y=(a[i].x-a[i].y);
        }
        sort(a,a+n);
        puts(cheak()?"Yes":"No");
    }
    return 0;
}