CodeForces-1253E Antenna Coverage

传送门：CodeForces-1253E

思路分析

设$dp[i]$表示前$1-i$已经覆盖，要覆盖$i+1$到$m$的最小花费,边界$dp[m]=0$
从后往前枚举，如果$i+1$已经被覆盖，$dp[i]=dp[i+1]$，可以直接继承
如果一条i小于某条线段的左端点，那么可以转移这条线段，覆盖$i-l$的代价为$cost$
那么转移就是$dp[i]=$$min$$(dp[i],dp[r+cost]+cost)$

样例输入

3 595
43 2
300 4
554 10

样例输出

281

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 = 1e5+10;

int dp[N];
pair<int,int>a[100];

int main() {

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

    int n,m;
    read(n,m);
    for(int i=1;i<=n;i++){
        int x,s;
        read(x,s);
        a[i].fi=max(0,x-s);
        a[i].se=min(m,x+s);
    }
    dp[m]=0;
    for(int i=m-1;i>=0;i--){
        dp[i]=m-i;
        for(int j=1;j<=n;j++){
            int l=a[j].fi,r=a[j].se;
            if(i+1>=l && i+1<=r) {
                dp[i]=dp[i+1];
                break;
            }
            if(i<l){
                int cost=(l-i-1);
                int rr=min(m,r+cost);
                dp[i]=min(dp[i],dp[rr]+cost);
            }
            
        }
    }
    cout<<dp[0]<<endl;




    return 0;
}