There are n beacons located at distinct positions on a number line. The i-th beacon has position ai and power level bi. When the i-th beacon is activated, it destroys all beacons to its left (direction of decreasing coordinates) within distance bi inclusive. The beacon itself is not destroyed however. Saitama will activate the beacons one at a time from right to left. If a beacon is destroyed, it cannot be activated.
Saitama wants Genos to add a beacon strictly to the right of all the existing beacons, with any position and any power level, such that the least possible number of beacons are destroyed. Note that Genos‘s placement of the beacon means it will be the first beacon activated. Help Genos by finding the minimum number of beacons that could be destroyed.
The first line of input contains a single integer n (1 ≤ n ≤ 100 000) — the initial number of beacons.
The i-th of next n lines contains two integers ai and bi (0 ≤ ai ≤ 1 000 000, 1 ≤ bi ≤ 1 000 000) — the position and power level of the i-th beacon respectively. No two beacons will have the same position, so ai ≠ aj if i ≠ j.
Print a single integer — the minimum number of beacons that could be destroyed if exactly one beacon is added.
4
1 9
3 1
6 1
7 4
1
7
1 1
2 1
3 1
4 1
5 1
6 1
7 1
3
For the first sample case, the minimum number of beacons destroyed is 1. One way to achieve this is to place a beacon at position 9with power level 2.
For the second sample case, the minimum number of beacons destroyed is 3. One way to achieve this is to place a beacon at position1337 with power level 42.
题意：给你 n个灯，每个灯在ai有一个威力值bi， 开启它，使得左边与其距离小于等于 bi 的会损坏
现在给你一个机会 可以在最右端 放一个 任意位置任意威力值的灯，并且从最右端的灯开始开启 问你最少的损坏的灯个数是多少
题解：设DP[i] 表示从1到i的这段距离内 不会损坏的 灯个数是多少
因为我们是开启灯 灯就不会损坏
dp[i] = dp[a[cnt]-b[cnt]-1] + 1; (1<=cnt<=n);
//meek///#include<bits/stdc++.h> #include <cstdio> #include <cmath> #include <cstring> #include <algorithm> #include<iostream> #include<bitset> using namespace std ; #define mem(a) memset(a,0,sizeof(a)) #define pb push_back #define fi first #define se second #define MP make_pair typedef long long ll; const int N = 201000; const int M = 1000001; const int inf = 0x3f3f3f3f; const int MOD = 100003; const double eps = 0.000001; struct ss{ int p,s,S; }a[N]; int n,b[N],H[N],t[M],sum[N],dp[M+5]; int cmp(ss s1,ss s2) {return s1.p<s2.p;} int main () { scanf("%d",&n); for(int i=1;i<=n;i++) { scanf("%d%d",&a[i].p,&a[i].s); a[i].S=a[i].p-a[i].s-1; } int cnt=0; int tmp; sort(a+1,a+n+1,cmp); int ans=0,L=0; int cc=1; for(int i=0;i<=M;i++) { if(i==a[cc].p) { if(a[cc].S>=0) { dp[i] = dp[a[cc].S] +1; } else dp[i] = 1; cc++; } else dp[i] = dp[i-1]; ans=max(dp[i],ans); // cout<<dp[i]<<endl; } cout<<n-ans<<endl; return 0; }
Codeforces Round #336 (Div. 2)C. Chain Reaction DP
原文：http://www.cnblogs.com/zxhl/p/5072699.html
