【AtCoder - 4244 】AtCoder Express 2 （区间dp 或暴力枚举，思维）

题干：

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 Express possesses Mtrains, 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)

Input

Input is given from Standard Input in the following format:

N M Q
L1 R1
L2 R2
:
LM RM
p1 q1
p2 q2
:
pQ qQ

Output

Print Q lines. The i-th line should contain the number of the trains that runs strictly within the section from City pi to City qi.

Sample Input 1

Sample Output 1

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

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

Sample Output 3

解题报告：

这题区间dp乱搞一下就可以了。

AC代码1：（区间dp）

#include<cstdio>
#include<iostream>
#include<algorithm>
#include<queue>
#include<map>
#include<vector>
#include<set>
#include<string>
#include<cmath>
#include<cstring>
#define ll long long
#define pb push_back
#define pm make_pair
#define fi first
#define se second
using namespace std;
const int MAX = 500 + 5;
int a[MAX][MAX],n,m,q;
ll dp[MAX][MAX];
int main()
{cin>>n>>m>>q;int l,r;while(m--) {scanf("%d%d",&l,&r);a[l][r]++;}for(int len = 1; len<=n; len++) {for(int l = 1; l+len-1<=n; l++) {//每一个起点 int r = l+len-1;if(l == r) dp[l][r]=a[l][r];else if(r-l==1) dp[l][r] = dp[l][l] + dp[r][r] + a[l][r];else dp[l][r] = dp[l][r-1] + dp[l+1][r] - dp[l+1][r-1]+a[l][r];}}while(q--) {ll ans = 0;scanf("%d%d",&l,&r);printf("%lld\n",dp[l][r]);}return 0 ;}

AC代码2：（区间dp）

#include<cstdio>
#include<iostream>
#include<algorithm>
#include<queue>
#include<map>
#include<vector>
#include<set>
#include<string>
#include<cmath>
#include<cstring>
#define ll long long
#define pb push_back
#define pm make_pair
#define fi first
#define se second
using namespace std;
const int MAX = 500 + 5;
int a[MAX][MAX],n,m,q;
ll dp[MAX][MAX];
int main()
{cin>>n>>m>>q;int l,r;while(m--) {scanf("%d%d",&l,&r);a[l][r]++;}for(int l = n; l>0; l--) {for(int r = 1; r<=n; r++) {dp[l][r] = a[l][r] + dp[l+1][r] + dp[l][r-1] - dp[l+1][r-1];}}while(q--) {ll ans = 0;scanf("%d%d",&l,&r);printf("%lld\n",dp[l][r]);}return 0 ;}

AC代码3：

#include<cstdio>
#include<iostream>
#include<algorithm>
#include<queue>
#include<map>
#include<vector>
#include<set>
#include<string>
#include<cmath>
#include<cstring>
#define ll long long
#define pb push_back
#define pm make_pair
#define fi first
#define se second
using namespace std;
const int MAX = 500 + 5;
int a[MAX][MAX],n,m,q;
int main()
{cin>>n>>m>>q;int maxx = -1,l,r;while(m--) {scanf("%d%d",&l,&r);maxx = max(maxx,r);a[l][r]++;}for(int i = 1; i<=maxx; i++) {for(int j = 1; j<=maxx; j++) {a[i][j] += a[i][j-1];}}while(q--) {ll ans = 0;scanf("%d%d",&l,&r);for(int i = l; i<=r; i++) {ans += a[i][r];}printf("%lld\n",ans);}return 0 ;}

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