题目链接：http://codeforces.com/problemset/problem/469/B
B. Chat Online
time limit per test
1 second
memory limit per test
256 megabytes
input
standard input
output
standard output
Little X and Little Z are good friends. They always chat online. But both of them have schedules.
Little Z has fixed schedule. He always online at any moment of time between a1 and b1,
between a2 and b2,
..., between ap and bp (all
borders inclusive). But the schedule of Little X is quite strange, it depends on the time when he gets up. If he gets up at time 0, he will be online at any moment
of time between c1 and d1,
between c2 and d2,
..., between cq and dq (all
borders inclusive). But if he gets up at time t, these segments will be shifted by t.
They become [ci?+?t,?di?+?t] (for
all i).
If at a moment of time, both Little X and Little Z are online simultaneosly, they can chat online happily. You know that Little X can get up at an integer moment of time between l and r (both
borders inclusive). Also you know that Little X wants to get up at the moment of time, that is suitable for chatting with Little Z (they must have at least one common moment of time in schedules). How many integer moments of time from the segment [l,?r] suit
for that?
Input
The first line contains four space-separated integers p,?q,?l,?r (1?≤??p,?q?≤?50; 0?≤?l?≤?r?≤?1000).
Each of the next p lines contains two space-separated integers ai,?bi (0?≤?ai?<?bi?≤?1000).
Each of the next q lines contains two space-separated integers cj,?dj (0?≤?cj?<?dj?≤?1000).
It‘s guaranteed that bi?<?ai?+?1 and dj?<?cj?+?1 for
all valid i and j.
Output
Output a single integer — the number of moments of time from the segment [l,?r] which suit for online conversation.
Sample test(s)
input
1 1 0 4 2 3 0 1
output
3
input
2 3 0 20 15 17 23 26 1 4 7 11 15 17
output
20
代码一如下：
#include <cstdio>
#include <cstring>
#define MAXN 2047
int main()
{
int p, q, l, r;
int vis[MAXN];
int a[MAXN], b[MAXN], c[MAXN], d[MAXN];
while(~scanf("%d%d%d%d",&p,&q,&l,&r))
{
memset(vis,0,sizeof(vis));
for(int i = 0; i < p; i++)
{
scanf("%d%d",&a[i],&b[i]);
for(int j = a[i]; j <= b[i]; j++)
{
vis[j] = 1;
}
}
for(int i = 0; i < q; i++)
{
scanf("%d%d",&c[i],&d[i]);
}
int k = 0;
int flag;
for(int i = l; i <= r; i++)
{
flag = 0;
for(int j = 0; j < q; j++)
{
for(int h = c[j]+i; h <= d[j]+i; h++)
{
if(vis[h])
{
k++;
flag = 1;
break;
}
}
if(flag)
break;
}
}
printf("%d\n",k);
}
return 0;
}
代码二如下：
#include <cstdio>
#include <cstring>
#define MAXN 1017
int main()
{
int p, q, l, r;
int a[MAXN], b[MAXN], c[MAXN], d[MAXN];
while(~scanf("%d%d%d%d",&p,&q,&l,&r))
{
for(int i = 0; i < p; i++)
{
scanf("%d%d",&a[i],&b[i]);
}
for(int i = 0; i < q; i++)
{
scanf("%d%d",&c[i],&d[i]);
}
int k = 0;
int flag;
for(int i = l; i <= r; i++)
{
flag = 0;
for(int j = 0; j < q; j++)
{
int t1 = c[j]+i;
int t2 = d[j]+i;
for(int h = 0; h < p; h++)
{
// if((t1>=a[h]&&t1<=b[h]) || (t2>=a[h]&&t2<=b[h]))
for(int f = t1; f <= t2; f++)
{
if((f>=a[h]&&f<=b[h]) || (f>=a[h]&&f<=b[h]))
{
k++;
// printf("ss::%d\n",i);
flag = 1;
break;
}
}
if(flag)
break;
}
if(flag)
break;
}
}
printf("%d\n",k);
}
return 0;
}