题目来源:http://codeforces.com/group/aUVPeyEnI2/contest/229509
时间限制：2s
空间限制：512MB
题目大意：
在一个车站中有若干人在队列中等待车辆，求所有人等待时间的期望值
首先给定n和q，随后是n行操作:
"+ t k":在t时刻有k个人加入队列等待车辆
"- t k":在t时刻有k个人乘车离开队列
然后是q个数字代表在初始时刻车站中有多少个车在等待
求出每个询问对应的所有人的等待时间，如果有人始终等不到车则输出"INFINITY"
样例：
代码：
#include <algorithm>
#include <iostream>
#include <cstring>
#include <vector>
#include <cstdio>
#include <string>
#include <cmath>
#include <queue>
#include <set>
#include <map>
#include <complex>
using namespace std;
typedef long long ll;
typedef long double db;
typedef pair<int,int> pii;
typedef vector<int> vi;
#define de(x) cout << #x << "=" << x << endl
#define rep(i,a,b) for(int i=(a);i<(b);i++)
#define all(x) (x).begin(),(x).end()
#define sz(x) (int)(x).size()
#define mp make_pair
#define pb push_back
#define fi first
#define se second
#define pi acos(-1.0)
#define mem0(a) memset(a,0,sizeof(a))
#define memf(b) memset(b,false,sizeof(b))
#define ll long long
#define eps 1e-10
#define inf 1e17
#define maxn 101010
int n, q;
int a[maxn], t[maxn];
int s[maxn];
int b[maxn], cnt;
long long ans[maxn];
struct node{
int x, id;
bool operator < (const node &rhs) const{
return x < rhs.x;
}
}c[maxn];
bool cmp(int i, int j){
return s[i] > s[j];
}
int main()
{
freopen("expect.in", "r", stdin);
freopen("expect.out", "w", stdout);
scanf("%d%d", &n, &q);
for(int i = 1; i <= n; i++){
char op[5];
scanf("%s%d%d", op, &t[i], &a[i]);
if(op[0] == '-') a[i] = -a[i];
}
for(int i = 1; i <= n; i++){
s[i] = s[i-1] + a[i];
if(i < n) t[i] = t[i+1] - t[i];
}
// for(int i = 1; i <= n; i++){
// printf("s[%d] = %d, t[%d] = %d\n", i, s[i], i, t[i]);
// }
long long sum1 = 0, sum2 = 0;
for(int i = 1; i <= n; i++){
if(s[i] < 0){
b[++cnt] = i;
sum1 += 1LL*(-s[i])*t[i];
sum2 += t[i];
}
}
sort(b + 1, b + cnt + 1, cmp);
int j = 1;
long long k = 0;
for(int i = 1; i <= q; i++){
scanf("%d", &c[i].x);
c[i].id = i;
}
sort(c + 1, c + q + 1);
for(int i = 1; i <= q; i++){
int x = c[i].x;
while(j <= cnt && (-s[b[j]]) <= x){
sum1 -= 1LL*(-s[b[j]]) * t[b[j]];
sum2 -= t[b[j]];
++j;
}
if(s[n] + x < 0){
ans[c[i].id] = -1;
}
else{
ans[c[i].id] = sum1 - x * sum2;
}
}
for(int i = 1; i <= q; i++){
if(ans[i] == -1) printf("INFINITY\n");
else printf("%lld\n", ans[i]);
}
return 0;
}