ZOJ 3888 Twelves Monkeys

Twelves Monkeys

Time Limit: 5000ms

Memory Limit: 32768KB

This problem will be judged on ZJU. Original ID: 3888
64-bit integer IO format: %lld      Java class name: Main

 

James Cole is a convicted criminal living beneath a post-apocalyptic Philadelphia. Many years ago, the Earth's surface had been contaminated by a virus so deadly that it forced the survivors to move underground. In the years that followed, scientists had engineered an imprecise form of time travel. To earn a pardon, Cole allows scientists to send him on dangerous missions to the past to collect information on the virus, thought to have been released by a terrorist organization known as the Army of the Twelve Monkeys.

The time travel is powerful so that sicentists can send Cole from year x[i] back to year y[i]. Eventually, Cole finds that Goines is the founder of the Army of the Twelve Monkeys, and set out in search of him. When they find and confront him, however, Goines denies any involvement with the viruscan. After that, Cole goes back and tells scientists what he knew. He wants to quit the mission to enjoy life. He wants to go back to the any year before current year, but scientists only allow him to use time travel once. In case of failure, Cole will find at least one route for backup. Please help him to calculate how many years he can go with at least two routes.

Input

The input file contains multiple test cases.

The first line contains three integers n,m,q(1≤ n ≤ 50000, 1≤ m ≤ 50000, 1≤ q ≤ 50000), indicating the maximum year, the number of time travel path and the number of queries.

The following m lines contains two integers x,y(1≤ y ≤ x ≤ 50000) indicating Cole can travel from year x to year y.

The following q lines contains one integers p(1≤ p ≤ n) indicating the year Cole is at now

Output

For each test case, you should output one line, contain a number which is the total number of the year Cole can go.

Sample Input

9 3 3
9 1
6 1
4 1
6
7
2

Sample Output

5
0
1

Hint

6 can go back to 1 for two route.

One is 6-1, the other is 6-7-8-9-1. 6 can go back to 2 for two route.

One is 6-1-2, the other is 6-7-8-9-1-2.

 

Source

ZOJ Monthly, July 2020

Author

GAN, Tiansheng

 

解题：离线二分BIT

 

#include <bits/stdc++.h>
using namespace std;
const int maxn = 100001;
struct QU {
    int id,year;
    bool operator<(const QU &t)const {
        if(year == t.year) return id < t.id;
        return year > t.year;
    }
} Q[maxn];
int c[maxn];
vector<int>g[maxn];
void add(int i,int val) {
    while(i < maxn) {
        c[i] += val;
        i += i&-i;
    }
}
int sum(int i,int ret = 0) {
    while(i > 0) {
        ret += c[i];
        i -= i&-i;
    }
    return ret;
}
int query(int low = 1,int high = maxn-1,int ret = -1) {
    while(low <= high) {
        int mid = (low + high)>>1;
        if(sum(mid) >= 2) {
            ret = mid;
            high = mid-1;
        } else low = mid + 1;
    }
    return ret;
}
int ans[maxn];
int main() {
    int n,m,q,x,y;
    while(~scanf("%d%d%d",&n,&m,&q)) {
        for(int i = 0; i < maxn; ++i) g[i].clear();
        memset(c,0,sizeof c);
        memset(ans,0,sizeof ans);
        for(int i = 0; i < m; ++i) {
            scanf("%d%d",&x,&y);
            g[x].push_back(y);
        }
        for(int i = 0; i < q; ++i) {
            scanf("%d",&Q[i].year);
            Q[i].id = i;
        }
        sort(Q,Q+q);
        int now = 0;
        for(int i = n; i >= 1; --i) {
            for(int j = g[i].size()-1; j >= 0; --j)
                add(g[i][j],1);
            if(Q[now].year == i) {
                int idx = query(1,i-1);
                if(idx == -1) ans[Q[now].id] = 0;
                else ans[Q[now].id] = i - idx;
                if(++now == q) break;
            }
        }
        for(int i = 0; i < q; ++i)
            printf("%d\n",ans[i]);
    }
    return 0;
}