題解:
中間可能存在的最大值是於兩邊值+1有關,所以我們從左和從右開始維護該點可以最多的路即可。
#include <algorithm>
#include <iostream>
#include <cstdlib>
#include <cstring>
#include <cstdio>
#include <string>
#include <vector>
#include <bitset>
#include <stack>
#include <cmath>
#include <deque>
#include <queue>
#include <list>
#include <set>
#include <map>
#define line printf("---------------------------\n")
#define mem(a, b) memset(a, b, sizeof(a))
#define pi acos(-1)
using namespace std;
typedef long long ll;
const int inf = 0x3f3f3f3f;
const int mod = 1e9+7;
const int maxn = 200000+10;
ll s[maxn], g[maxn];
int main(){
int n;
scanf("%lld", &n);
for(int i = 0; i < n; i++){
scanf("%lld %lld", &s[i], &g[i]);
g[i] += s[i];
}
for(int i = 1; i < n; i++){
g[i] = min(g[i], g[i-1]+1);
}
for(int i = n-2; i >= 0; i--){
g[i] = min(g[i], g[i+1]+1);
}
ll ok = 1, ans = 0;
for(int i = 0; i < n; i++){
if(s[i] <= g[i]){
ans += g[i]-s[i];
}
else{
ok = 0;
break;
}
}
if(!ok){
puts("-1");
}
else{
printf("%lld\n", ans);
printf("%lld", g[0]);
for(int i = 1; i < n; i++){
printf(" %lld", g[i]);
}
puts("");
}
}