AtCoder Beginner Contest 106 2018/08/18

Problem Statement

In Takahashi Kingdom, there is a east-west railroad and N cities along it, numbered 1, 2, 3, ..., N from west to east. A company called AtCoder Expresspossesses M trains, and the train i runs from City Li to City Ri (it is possible that Li=Ri). Takahashi the king is interested in the following Q matters:

• The number of the trains that runs strictly within the section from City pi to City qi, that is, the number of trains j such that pi≤Lj and Rj≤qi.

Although he is genius, this is too much data to process by himself. Find the answer for each of these Q queries to help him.

Constraints

• N is an integer between 1 and 500 (inclusive).
• M is an integer between 1 and 200 000 (inclusive).
• Q is an integer between 1 and 100 000 (inclusive).
• 1≤Li≤Ri≤N (1≤i≤M)
• 1≤pi≤qi≤N (1≤i≤Q)

Sample Input 1

2 3 1
1 1
1 2
2 2
1 2

Sample Output 1

3

As all the trains runs within the section from City 1 to City 2, the answer to the only query is 3.

Sample Input 2

10 3 2
1 5
2 8
7 10
1 7
3 10

Sample Output 2

1
1

The first query is on the section from City 1 to 7. There is only one train that runs strictly within that section: Train 1. The second query is on the section from City 3 to 10. There is only one train that runs strictly within that section: Train 3.

Sample Input 3

10 10 10
1 6
2 9
4 5
4 7
4 7
5 8
6 6
6 7
7 9
10 10
1 8
1 9
1 10
2 8
2 9
2 10
3 8
3 9
3 10
1 10

Sample Output 3

7
9
10
6
8
9
6
7
8
10
简单的区间dp。
代码：

#include<iostream>
#include<cstdio>
#include<algorithm>
using namespace std;
int n,m,q;
int dp[502][502];
int main() {
    int a,b;
    scanf("%d%d%d",&n,&m,&q);
    for(int i = 0;i < m;i ++) {
        scanf("%d%d",&a,&b);
        dp[a][b] ++;
    }
    for(int j = 1;j < n;j ++) {
        for(int i = 1;i + j <= n;i ++) {
            int k = i + j;
            dp[i][k] += dp[i + 1][k] + dp[i][k - 1];
            if(j > 1)dp[i][k] -= dp[i + 1][k - 1];
        }
    }
    for(int i = 0;i < q;i ++) {
        scanf("%d%d",&a,&b);
        printf("%d\n",dp[a][b]);
    }
    return 0;
}

 树状数组，区间向两侧更新，最后查询两侧向内查询。

代码：

#include <iostream>
#include <cstdio>
using namespace std;
int n;
int sum[501][501];
int lowbit(int x) {
    return x&-x;
}
void update(int x,int y) {
    for(int i = x;i > 0;i -= lowbit(i)) {
        for(int j = y;j <= n;j += lowbit(j)) {
            sum[i][j] ++;
        }
    }
}
int get(int x,int y) {
    int ans = 0;
    for(int i = x;i <= y;i += lowbit(i)) {
        for(int j = y;j >= x;j -= lowbit(j)) {
            ans += sum[i][j];
        }
    }
    return ans;
}
int main() {
    int m,q;
    scanf("%d%d%d",&n,&m,&q);
    int a,b;
    for(int i = 0;i < m;i ++) {
        scanf("%d%d",&a,&b);
        update(a,b);
    }
    for(int i = 0;i < q;i ++) {
        scanf("%d%d",&a,&b);
        printf("%d\n",get(a,b));
    }
}

 也可以用邻接表加二分遍历，不过就比较麻烦了。