CodeForces-1257D Monster Killing Problem

传送门：CodeForces-1257D

思路分析

肯定优先选择耐力值大且能力值大的
设mx[i]表示耐力值大于等于i的最大能力值

样例输入

2
6
2 3 11 14 1 8
2
3 2
100 1
5
3 5 100 2 3
2
30 5
90 1

样例输出

5
-1

AC代码

#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
#include <bits/stdc++.h>
#define ls x<<1
#define rs x<<1|1
#define fi first
#define se second
#define ll long long
#define pb push_back
#define mp make_pair
#define fun function
#define sz(x) (x).size()
#define lowbit(x) (x)&(-x)
#define all(x) (x).begin(),(x).end()
#define mem(a,b) memset(a,b,sizeof(a))

namespace FastIO {
#define BUF_SIZE 100000
#define OUT_SIZE 100000
    bool IOerror=0;
    inline char nc() {
        static char buf[BUF_SIZE],*p1=buf+BUF_SIZE,*pend=buf+BUF_SIZE;
        if(p1==pend) {
            p1=buf;
            pend=buf+fread(buf,1,BUF_SIZE,stdin);
            if(pend==p1) {
                IOerror=1;
                return -1;
            }
        }
        return *p1++;
    }
    inline bool blank(char ch) {
        return ch==' '||ch=='\n'||ch=='\r'||ch=='\t';
    }
    template<class T> inline bool read(T &x) {
        bool sign=0;
        char ch=nc();
        x=0;
        for(; blank(ch); ch=nc());
        if(IOerror)return false;
        if(ch=='-')sign=1,ch=nc();
        for(; ch>='0'&&ch<='9'; ch=nc())x=x*10+ch-'0';
        if(sign)x=-x;
        return true;
    }
    template<class T,class... U>bool read(T& h,U&... t) {
        return read(h)&&read(t...);
    }
#undef OUT_SIZE
#undef BUF_SIZE
};
using namespace std;
using namespace FastIO;
using namespace __gnu_pbds;

mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
typedef tree<int, null_type, less<int>, rb_tree_tag, tree_order_statistics_node_update> rbtree;

const int INF = 0x3f3f3f3f;
const int N = 2e5+10;

int mx[N],a[N]; 

int main() {

#ifdef xiaofan
    freopen("1.in","r",stdin);
    freopen("1.out","w",stdout);
#endif

    int t;
    read(t);
    while(t--){
        int n;
        read(n);
        for(int i=1;i<=n;i++) read(a[i]),mx[i]=0;
        int m;
        read(m);
        for(int i=1;i<=m;i++){
            int p,s;
            read(p,s);
            mx[s]=max(mx[s],p);
        }
        for(int i=n-1;i>=0;i--) mx[i]=max(mx[i],mx[i+1]);
        int ok=1;
        int pos=1,ans=0;
        while(pos<=n){
            ans++;
            int temp=pos;
            int ma=0;
            while(1){
                ma=max(ma,a[temp]);
                if(ma>mx[temp-pos+1]) break;
                ++temp;
            }
            if(temp==pos) {
                ok=0;
                break;
            }
            pos=temp;
        }
        printf("%d\n",ok?ans:-1);
    }




    return 0;
}