[SNOI 2017] 炸弹

题目描述: 给定炸弹和爆炸范围，求对于每个炸弹连锁爆炸的炸弹总和对\(1e9+7\)取膜

思路：

为啥都是线段树+TS+tarjan呢？

实在是搞不懂~~

线性\(O(n)\)递推即可.

#include<bits/stdc++.h>
using namespace std;
const int maxn = 1000010;
const int mod = 1e9+7ull;
#define debug(x) cout<<"x:"<<x<<endl;
#define ll long long
inline ll read() {
    ll q=0,f=1;char ch=getchar();
    while(!isdigit(ch)) {
        if(ch==‘-‘)f =-1;ch=getchar();
    }
    while(isdigit(ch)) {
        q=q*10+ch-‘0‘;ch=getchar();
    }
    return q*f;
}
struct node{
    ll num;
    ll pos;
}nod[maxn];

ll n;
ll L[maxn];
ll R[maxn];
ll ans;
int main() {
    n = read();
    for(ll i = 1;i <= n; ++i) {
        nod[i].pos = read(),nod[i].num = read();
    }
    for(ll i = 1;i <= n; ++i) {
        L[i] = i;
        while(nod[i].pos - nod[L[i] - 1].pos <= nod[i].num && L[i] > 1) {
            L[i] = L[L[i] - 1];
            nod[i].num = max(nod[i].num,nod[L[i]].num - (nod[i].pos - nod[L[i]].pos));
        }
    }
    for(ll i = n;i >= 1; --i) {
        R[i] = i;
        while(nod[R[i] + 1].pos - nod[i].pos <= nod[i].num && R[i] < n) {
            R[i] = R[R[i] + 1];
            L[i] = min(L[i],L[R[i]]);
        }
    }
    for(ll i = 1;i <= n; ++i) {
//      debug(i);
//      debug(L[i]);
//      debug(R[i]);
        ans = (ans + (R[i] - L[i] + 1) * i) % mod;
    }
    printf("%lld\n",ans);
    return 0;
}
/*
题目有毒orz
*/

[SNOI 2017] 炸弹

原文：https://www.cnblogs.com/akoasm/p/9408357.html