Codeforces Gym 100340C ePig 模拟

题目大意：

题目很长.....不过实现并不难....当年华科校赛遇到这个题没做...现在补一下

就是现在P2P协议进行文件传输

现在又n个客户端, 一个包被分成k个部分进行发送, 发送到所有客户端需要的时间

发送规则：

首先初始状态只有Client 1拥有完整文件(所有的包)进行提供

每一轮 开始时所有Client机器决定需要的包, 每个Client需要的包是其没有的所有包中供源最少的包, 如果有多个选择序号最小的包

然后每个Client决定要从哪里获得包, 获得包的原则是在所有能提供这个包的Client中选择, 并且选择从拥有包的数量最少的Client那里获得, 如果依旧有多个选择就选择序号最小的Client

然后每个Client都会有得到的请求, 这个时候所有Client决定要给谁包, 每个Client只能满足一个请求, 每个Client会选择曾经给它包的数量最多的Client, 如果有多个, 那么选择拥有包数量最少的Client, 如果依旧有多个, 选择序号最小的

于是每轮都这样重复, 到某一轮之后所有Client有所有包, 传输结束

输出除了Client 1之外, 其他Client得到完整的全部包的轮数

大致思路：

上面的过程看懂了很容易过的....

模拟题还是要耐心...

代码如下：

Result  :  Accepted     Memory  :  460 KB     Time  :  560 ms

/* * Author: Gatevin * Created Time:  2015/9/1 16:41:44 * File Name: C.cpp */#include<iostream>#include<sstream>#include<fstream>#include<vector>#include<list>#include<deque>#include<queue>#include<stack>#include<map>#include<set>#include<bitset>#include<algorithm>#include<cstdio>#include<cstdlib>#include<cstring>#include<cctype>#include<cmath>#include<ctime>#include<iomanip>using namespace std;const double eps(1e-8);typedef long long lint;int n, k;struct Client{    int own[210];    int pro[210];    int has;    int giv;    int complete;    int need;    int req;    int gei;    Client()    {        memset(own, 0, sizeof(own));        memset(pro, 0, sizeof(pro));        has = giv = 0;        complete = -1;        need = req = -1;        gei = -1;    }};Client client[110];int all = 0;int provide[210];vector<int> V[110];int g[110][110];void start(){    memset(g, 0, sizeof(g));    int time = 1;    client[1].complete = 1;    fill(client[1].own, client[1].own + k + 1, 1);    fill(client[1].pro, client[1].pro + k + 1, 1);    client[1].has = client[1].giv = k;    client[1].gei = -1;    all = 1;    for(int i = 1; i <= k; i++)        provide[i] = 1;        while(all < n)    {        for(int i = 1; i <= n; i++)            client[i].need = client[i].req = -1;        for(int i = 1; i <= n; i++)//decide which chunk need        {            if(client[i].has == k)            {                client[i].need = -1;                continue;            }            int get = -1;            for(int j = 1; j <= k; j++)            {                if(client[i].own[j] == 1) continue;                if(get == -1 || provide[j] < provide[get]) get = j;            }            client[i].need = get;        }        for(int i = 1; i <= n; i++)//decide which client to request        {            int need = client[i].need;            if(need == -1)            {                client[i].req = -1;                continue;            }            int req = -1;            int number = -1;            for(int j = 1; j <= n; j++)            {                if(client[j].own[need])                {                    if(req == -1)                    {                        req = j;                        number = client[j].giv;                        continue;                    }                    if(number > client[j].giv)                    {                        req = j;                        number = client[j].giv;                    }                }            }            client[i].req = req;        }                for(int i = 1; i <= n; i++) V[i].clear();        for(int i = 1; i <= n; i++)            if(client[i].req != -1)            {                V[client[i].req].push_back(i);            }                for(int i = 1; i <= n; i++)        {            int now = -1;            for(int j = 0, sz = V[i].size(); j < sz; j++)            {                int nex = V[i][j];                if(now == -1)                {                    now = V[i][j];                    continue;                }                if(g[i][nex] > g[i][now])                    now = nex;                else if(g[i][nex] == g[i][now])                {                    if(client[nex].has < client[now].has)                        now = nex;                }            }            client[i].gei = now;        }                for(int i = 1; i <= n; i++)        {            if(client[i].gei != -1)            {                int nex = client[i].gei;                int thing = client[nex].need;                client[nex].own[thing] = 1;                client[nex].has++;                if(client[nex].has == k) client[nex].complete = time, all++;                client[nex].pro[thing] = 1;                provide[thing]++;                client[nex].giv++;                g[nex][i]++;            }        }        time++;    }    for(int i = 2; i <= n; i++)        printf("%d%c", client[i].complete, i == n ? '\n' : ' ');}int main(){    freopen("epig.in", "r", stdin);    freopen("epig.out", "w", stdout);    scanf("%d %d", &n, &k);    start();    return 0;}