Gym

来源：互联网发布：s7总决赛现场数据编辑：程序博客网时间：2024/05/18 01:01

题意：

告诉你n 个操作过程， + 代表在t 时刻来了k 个车， - 代表在t 时刻来了k 个人，每个人都要开走车，没车的话一直在队列里等，告诉你每一天开始有多少辆车，求所有人的等待时间总和，如果有人拿不到车就输出 inf。

思路：

数形结合的思想。

n 个操作可以分成n-1 段，记录每一段的人和等待时间。

那么一开始有x 辆车的话，那么只需要将大于x 的段减掉x 乘以等待时间即可。

那么这样与每一段的顺序无关，直接按照人排序，然后枚举x，二分查找左边界，做个前缀和差分即可。

#include <cstdio>#include <cstring>#include <algorithm>#include <queue>#include <string>#include <iostream>#include <cctype>#include <sstream>#include <vector>using namespace std;const int maxn = 100005;struct node1{    char cmd[4];    int t, k;}p[maxn];struct node{    int num, len;}nod[maxn];bool cmp(const node& lhs, const node& rhs){    return lhs.num < rhs.num;}long long pre[maxn];long long pre1[maxn];int main(){    freopen("expect.in","r",stdin);    freopen("expect.out","w", stdout);    int n,q;    while(~scanf("%d %d",&n, &q)){        int x, y;        int sum = 0;        long long peo = 0LL;        for (int i = 0; i < n; ++i){            scanf("%s%d%d", p[i].cmd, &p[i].t, &p[i].k);            if (p[i].cmd[0] == '-'){                peo += p[i].k;            }            else {                peo -= p[i].k;            }        }        if (p[0].cmd[0] == '-')            sum = p[0].k;        else            sum = -p[0].k;        int cnt = 0;        for (int i = 1; i < n; ++i){            nod[++cnt].num = sum;            nod[cnt].len = p[i].t - p[i-1].t;            if (p[i].cmd[0] == '-'){                sum += p[i].k;            }            else {                sum -= p[i].k;//                sum = max(sum, 0);            }        }        sort(nod + 1, nod + cnt + 1, cmp);        for (int i = 1; i <= cnt; ++i){            pre[i] = pre[i-1] + (long long)nod[i].num * nod[i].len;            pre1[i] = pre1[i-1] + nod[i].len;        }////        for (int i = 1; i <= cnt; ++i){//            printf("%d %d\n", nod[i].num, nod[i].len);////        }//        printf("peo = %lld\n", peo);        for (int i = 0; i < q; ++i){            long long ans = 0LL;            int x;            scanf("%d",&x);            if (x < peo) {                printf("INFINITY\n");                continue;            }            int l = 1, r = cnt;            while (l <= r){                int m = l + r >> 1;                if (nod[m].num > x){                    r = m - 1;                }                else {                    l = m + 1;                }            }//            printf("r = %d\n", r);            ans = (pre[cnt] - pre[r]) - x * (pre1[cnt] - pre1[r]);            printf("%I64d\n", ans);        }    }    return 0;}/**This is ACM North Eastern European Regional Contest,sponsored by International Business Machines.The. Best. Contest. Ever.A Great Opportunity for all contestants.ab Ab A Abc AB Abcd ABc Abcde AbCOh  No  Extra Spaces.And,Punctuation Ruin Everything10 4 1 21 2 3 4 5 6 7 8 9 1010 9 8 7 6 5 4 3 2 1EEESEEEESS686  100- 2 3- 4 3+ 5 2+ 7 4- 10 3+ 13 36  100- 2 3- 4 3+ 5 2+ 7 4- 10 3+ 13 35 100+ 2 1+ 4 2+ 5 3+ 9 10+ 10 11**/

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