原題:http://acm.hdu.edu.cn/showproblem.php?pid=1025
題意:有n個窮城市,n個富城市,每個窮城市都要從某個富城市運輸一種物資(窮城市和富城市的物資供需一對一),需要建立道路,但任意兩條路不能交叉;窮城市和富城市分 別位列平行線兩側(均按1 - n 分佈);給出城市個數n,下面n行輸入兩個數字 p 和 r 表示窮城市p要從富城市r運輸物資,即需要在p和r之間建路,問最多可以建幾條路;
思路:按p的順序存入相對應的r,然後對r求最長上升子序列的個數;
#include<cstdio>
#include<algorithm>
using namespace std;
const int maxn = 500000+10;
int n, cas = 0;
int a[maxn], stack[maxn];
int main()
{
while(~scanf("%d", &n))
{
for(int i = 1;i<=n;i++)
{
int u, v;
scanf("%d%d", &u, &v);
a[u] = v;
}
int top = 0;
stack[top] = -1;
for(int i = 1;i<=n;i++)
{
if(a[i] > stack[top])
stack[++top] = a[i];
else
{
int pos = lower_bound(stack+1, stack+top+1, a[i]) - stack;
stack[pos] = a[i];
}
}
printf("Case %d:\n", ++cas);
if(top == 1)
printf("My king, at most 1 road can be built.\n\n");
else
printf("My king, at most %d roads can be built.\n\n", top);
}
return 0;
}