606第九周周六赛 B - War

B - War

Time Limit:1000MS     Memory Limit:32768KB     64bit IO Format:%I64d & %I64u

Submit  Status  Practice  HDU 2600

Description

      War is the reciprical and violent application of force between hostile political entities aimed at bringing about a desired political end-state via armed conflict. War is an interaction in which two or more militaries have a “struggle of wills”. When qualified as a civil war, it is a dispute inherent to a given society, and its nature is in the conflict over modes of governance rather than sovereignty. War is not considered to be the same as mere occupation,murder or genocide because of the reciprical nature of the violent struggle, and the organized nature of the units involved. 
      War is also a cultural entity, and its practice is not linked to any single type of political organisation or society. Rather, as discussed by John Keegan in his “History Of Warfare”, war is a universal phenomenon whose form and scope is defined by the society that wages it. 

      Today we don't want to talk about war, rather than , I want you to tell me which year there was no war.

 

Input

For each test case, first input one integer N(1<= N <= 100), then two integers p and q (-6000000 <= p <= q <= 6000000) represent the starting year and the ending year in this case. Followed by N wars. 
Each war described as follows: 
A  _i B  _i N  _i (A,B is integer, p <= A, B <= q, N  _i is a String which is the ith war's name ) 
Represent that the ith War took place between year A and year B. 

 

Output

Output one number represent which year there was no war in the range p and q, if there are not only one year, output the maximum year which there was no war. If all the year had war, just output "Badly!". 

 

Sample Input

      
      
      
      

       
       
       
       3
100 200
100 120 RtWar
110 140 WeWar
141 199 QqWar 
1
-600 600
-600 600 Cool War 
      
      
      
      

 

Sample Output

      
      
      
      

       
       
       
       200
Badly!
      
      
      
      

典型的贪心思想，选择一个不在这些区间里且最大的点！

首先把这些点按照区间右端点由大到小排列，相等的情况下按照左端点由大到小排列

然后拿需要判断的那个区间的右端点和各个区间的右端点比较，如果在这个区间就把左端点的值减一赋给这个值，如果大于右端点就输出这个值

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#include<stdio.h>
#include<algorithm>
using namespace std;
struct war
{
    int a,b;
} w[110];
int cmp(war x,war y)
{
    if(x.b==y.b)
        return x.a>y.a;
    else
        return x.b>y.b;
}
int main()
{
    int n,p,q,i,j,f;
    char k[110];
    while(scanf("%d",&n)!=EOF)
    {
        f=0;
        scanf("%d%d",&p,&q);
        for(i=0; i<n; i++)
        {
            scanf("%d%d",&w[i].a,&w[i].b);
            gets(k);
        }
        sort (w,w+n,cmp);
        if(q>w[0].b)
        {
            printf("%d\n",q);
            continue;
        }
        q=w[0].a-1;
        for(i=1; i<n; i++)
        {
            if(q>w[i].a&&q<w[i].b)
                q=w[i].a-1;
            else
            {
                f=1;
                printf("%d\n",q);
                break;
            }
        }
        if(f==0)
            printf("Badly!\n");
    }

    return 0;
}