607A Chain Reaction(DP)

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.

Input
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.

Output
Print a single integer — the minimum number of beacons that could be destroyed if exactly one beacon is added.

Sample test(s)
input
4
1 9
3 1
6 1
7 4
output
1
input
7
1 1
2 1
3 1
4 1
5 1
6 1
7 1
output
3
Note
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 9 with 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 position 1337 with power level 42.

一堆培根，每个培根在数轴上一个位置a，拥有能量b，当激活这个培根的时候，它会将在它左面距离他b以内（包括b）的培根破坏，一旦培根遭到破坏，就无法激活，现在你要放第n+1个培根在给的n个培根的右面，能量自定，位置自定，但一定是最右面的一个，然后从右向左依次激活培根，问最少破坏几个培根。

想了一下就是一个递推，每个培根只有破坏和不破坏两个状态，然后最右面那个培根的作用就是控制右面连续几个培根被破坏。
也就是说，这个培根坏了，左面转移过来可能是好的也可能是坏的培根，但是这个培根如果是好的，那么一定是从好的培根转移过来。

dp[0][i]代表这个培根坏了，dp[1][i]代表这个培根是好的。
dp[0][i] = max(dp[0][i - 1],dp[1][i - 1]);
dp[1][i] = dp[1][i - b[i]] + 1;

#include<cstdio>#include<cstring>#include<iostream>#include<queue>#include<vector>#include<algorithm>#include<string>#include<cmath>#include<set>#include<map>#include<vector>#include<stack>#include<utility>#include<sstream>using namespace std;typedef long long ll;const int inf = 0x3f3f3f3f;const int maxn = 1005;int dp[2][1000005],leap[1000005];int main(){    #ifdef LOCAL    freopen("C:\\Users\\巍巍\\Desktop\\in.txt","r",stdin);    //freopen("C:\\Users\\巍巍\\Desktop\\out.txt","w",stdout);    #endif // LOCAL    int n;    scanf("%d",&n);    memset(leap,0,sizeof(leap));    memset(dp,0,sizeof(dp));    for(int i = 1;i <= n;i++)    {        int a,b;        scanf("%d%d",&a,&b);        leap[a] = b;    }    if(leap[0])dp[0][0] = dp[1][0] = 1;    else dp[0][0] = dp[1][0] = 0;    for(int i = 1;i <= 1000000;i++)    {        if(leap[i])        {            if(i > leap[i])            {                dp[0][i] = max(dp[0][i - 1],dp[1][i - 1]);                dp[1][i] = dp[1][i - leap[i] - 1] + 1;            }            else            {                dp[0][i] = max(dp[0][i - 1],dp[1][i - 1]);                dp[1][i] = 1;            }        }        else        {            dp[0][i] = max(dp[0][i - 1],dp[1][i - 1]);            dp[1][i] = dp[1][i - 1];        }    }    int ans = 0;    for(int i = 0;i <= 1000000;i++)        ans = max(ans,max(dp[0][i],dp[1][i]));    printf("%d",n - ans);    return 0;}