题意：给定 n 个工作的最好开始时间，和持续时间，现在有两种方法，第一种，如果当前的工作能够恰好在最好时间开始，那么就开始，第二种，如果不能，那么就从前找最小的时间点，来完成。
析：直接暴力，每次都先去看看能不能在最好时间完成，如果不能，就去找最小的时间点。
代码如下：
#pragma comment(linker, "/STACK:1024000000,1024000000")
#include <cstdio>
#include <string>
#include <cstdlib>
#include <cmath>
#include <iostream>
#include <cstring>
#include <set>
#include <queue>
#include <algorithm>
#include <vector>
#include <map>
#include <cctype>
#include <cmath>
#include <stack>
#define freopenr freopen("in.txt", "r", stdin)
#define freopenw freopen("out.txt", "w", stdout)
using namespace std;
typedef long long LL;
typedef pair<int, int> P;
const int INF = 0x3f3f3f3f;
const double inf = 0x3f3f3f3f3f3f;
const LL LNF = 0x3f3f3f3f3f3f;
const double PI = acos(-1.0);
const double eps = 1e-8;
const int maxn = 1e2 + 100;
const int mod = 1e9 + 7;
const int dr[] = {-1, 0, 1, 0};
const int dc[] = {0, 1, 0, -1};
const char *Hex[] = {"0000", "0001", "0010", "0011", "0100", "0101", "0110", "0111", "1000", "1001", "1010", "1011", "1100", "1101", "1110", "1111"};
int n, m;
const int mon[] = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
const int monn[] = {0, 31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
inline int Min(int a, int b){ return a < b ? a : b; }
inline int Max(int a, int b){ return a > b ? a : b; }
inline LL Min(LL a, LL b){ return a < b ? a : b; }
inline LL Max(LL a, LL b){ return a > b ? a : b; }
inline int gcd(int a, int b){ return b ? gcd(b, a%b) : a; }
inline int lcm(int a, int b){ return a * b / gcd(a, b); }
inline bool is_in(int r, int c){
return r >= 0 && r < n && c >= 0 && c < m;
}
struct Node{
int l, r;
Node() { }
Node(int ll, int rr) : l(ll), r(rr) { }
bool operator < (const Node &p) const{
return l < p.l;
}
};
Node a[205];
int main(){
while(scanf("%d", &n) == 1){
int s, d;
int cnt = 1;
a[0] = Node(0, 0);
cin >> s >> d;
printf("%d %d
", s, s+d-1);
a[cnt++] = Node(s, s+d-1);
for(int i = 1; i < n; ++i){
cin >> s >> d;
--d;
sort(a, a+cnt);
if(a[cnt-1].r < s){
printf("%d %d
", s, s+d);
a[cnt++] = Node(s, s+d);
continue;
}
bool ok = false;
for(int j = 0; j < cnt-1; ++j){
if(s > a[j].r && s+d < a[j+1].l){
ok = true;
printf("%d %d
", s, s+d);
a[cnt++] = Node(s, s+d);
break;
}
}
if(!ok){
bool ok1 = false;
for(int j = 1; j < cnt; ++j){
if(a[j].l - a[j-1].r > d+1){
printf("%d %d
", a[j-1].r+1, a[j-1].r+d+1);
a[cnt++] = Node(a[j-1].r+1, a[j-1].r+d+1);
ok1 = true;
break;
}
}
if(!ok1){
printf("%d %d
", a[cnt-1].r+1, a[cnt-1].r+1+d);
a[cnt] = Node(a[cnt-1].r+1, a[cnt-1].r+d+1);
++cnt;
}
}
}
}
return 0;
}