Codeforces Beta Round #79 (Div. 1 Only) B. Buses 树状数组
http://codeforces.com/contest/101/problem/B
给定一个数n,起点是0 终点是n,有m两车,每辆车是从s开去t的,我们只能从[s,s+1,s+2....t-1]处上车,从t处下车。,
问能否去到点n,求方案数
设L[x]表示有多少辆车能够到达x处。
只能从t处下车:说明只能单点更新,对于没辆车x,在区间[s,s+1,s+2....t-1]内上车是可以得,那么有多少辆车呢?明显就是∑区间里能到达的点。然后单点更新t即可
数据大,明显不能到达的点是没用的,离散化一下即可。
坑点就是:因为车是无序的,所以应该优先处理最快下车的点。这样就能对后面进行更新
注意树状数组哪里也要取模
#include <cstdio> #include <cstdlib> #include <cstring> #include <cmath> #include <algorithm> using namespace std; #define inf (0x3f3f3f3f) typedef long long int LL; #include <iostream> #include <sstream> #include <vector> #include <set> #include <map> #include <queue> #include <string> const int maxn = 3e5 + 20; const int MOD = 1e9 + 7; int n, to; LL c[maxn]; int lowbit (int x) { return x&(-x); } void add (int pos,LL val) { while (pos<=to + 20) { c[pos] += val; c[pos] %= MOD; pos += lowbit(pos); } return ; } LL get_sum (int pos) { LL ans = 0; while (pos) { ans += c[pos]; ans %= MOD; pos -= lowbit(pos); } return ans; } struct node { int s, t; }a[maxn]; bool cmp (node a, node b) { if (a.t != b.t) { return a.t < b.t; } else { return a.s < b.s; } } set<int>ss; map<int,int>pos; LL ways[maxn]; void work() { cin >> n; int m; cin >> m; int numzero = 0; int numlast = 0; for (int i = 1; i <= m; ++i) { scanf("%d%d", &a[i].s, &a[i].t); if (a[i].s == 0) numzero++; if (a[i].t == n) numlast++; } if (numzero == 0 || numlast == 0) { cout << "0" << endl; return; } sort(a + 1, a + 1 + m, cmp); for (int i = 1; i <= m; ++i) { ss.insert(a[i].s); ss.insert(a[i].t); } for (set<int>::iterator it = ss.begin(); it != ss.end(); ++it) { if (pos[*it] == 0) { pos[*it] = ++to; } } add(1, 1); for (int i = 1; i <= m; ++i) { int toa = pos[a[i].s]; int tob = pos[a[i].t]; LL t = (get_sum(tob - 1) - get_sum(toa - 1) + MOD) % MOD; add(tob, t); } LL ans = get_sum(pos[n]) - get_sum(pos[n] - 1); if (ans < 0) ans += MOD; cout << ans << endl; } int main() { #ifdef local freopen("data.txt","r",stdin); #endif work(); return 0; }
既然选择了远方,就要风雨兼程~
posted on 2016-09-15 15:44 stupid_one 阅读(260) 评论(0) 编辑 收藏 举报