记录编号 |
259798 |
评测结果 |
AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAATATTAAAAAAAAA |
题目名称 |
[APIO2016]划艇 |
最终得分 |
96 |
用户昵称 |
APIO棒子出题人 |
是否通过 |
未通过 |
代码语言 |
C++ |
运行时间 |
36.119 s |
提交时间 |
2016-05-11 17:33:24 |
内存使用 |
4.12 MiB |
显示代码纯文本
/*
* Author: Bumsoo Park(zlzmsrhak)
* Time Complexity: O(N^3)
*/
#include<stdio.h>
#include<algorithm>
#include<vector>
using namespace std;
typedef long long ll;
const int MX = 1005;
const long long MM = 1000000007;
// i-th segment : S[i] ~ (E[i]-1)
// D : dp table ; rv[i] * i % MM = 1
// X : coordinate list, L[i] = X[i] - X[i-1]
int S[MX], E[MX], L[MX];
int D[MX][MX], rv[MX];
vector<int> X;
int add(int a, int b){ return a+b >= MM? a+b-MM: a+b; }
int main()
{
freopen("boat.in","r",stdin);
freopen("boat.out","w",stdout);
int N, W;
scanf("%d", &N);
for(int i = 1; i <= N; i++) scanf("%d%d", S+i, E+i), E[i]++;
rv[1] = 1;
for(int i = 2; i <= N; i++) rv[i] = (MM/i) * (MM-rv[MM%i]) % MM;
for(int i = 1; i <= N; i++) X.push_back(S[i]);
for(int i = 1; i <= N; i++) X.push_back(E[i]);
sort(X.begin(), X.end());
X.resize(unique(X.begin(), X.end()) - X.begin());
for(int i = 1; i <= N; i++) S[i] = lower_bound(X.begin(), X.end(), S[i]) - X.begin() + 1;
for(int i = 1; i <= N; i++) E[i] = lower_bound(X.begin(), X.end(), E[i]) - X.begin() + 1;
W = X.size() - 1;
for(int i = 1; i <= W; i++) L[i] = X[i] - X[i-1];
D[0][0] = 1;
for(int i = 1; i <= W; i++){
D[i][0] = 1;
for(int j = 1; j <= N; j++){
ll mul = L[i];
int count = 1;
D[i][j] = D[i-1][j];
if( i < S[j] || i >= E[j] ) continue;
for(int k = j-1; k >= 0; k--){
D[i][j] = add(D[i][j], D[i-1][k] * mul % MM);
if( S[k] <= i && i < E[k] ){
mul = mul * (count + L[i]) % MM * rv[count+1] % MM;
count++;
}
}
}
}
int ans = 0;
for(int i = 1; i <= N; i++) ans = add(ans, D[W][i]);
printf("%d\n", ans);
}