Billionaires

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题意：给出n个人所在的城市以及个人财富，接下来有k起事件，表示某天某人到达某个城市，问每个城市拥有的财富值最大的天数（比其他所有城市都多）。
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解法：用stl直接模拟就行了
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队友代码

#include <cstdio>#include <map>#include <string>#include <algorithm>#include <set>using namespace std;char c[200], d[200];struct peo {    string name;    long long wealth;} a[60010];struct event {    int day;    string who;    string place;} b[60010];struct ss {    int day;    string p;} aaa[60010];struct pl {    string name;    long long num;    bool operator <(const pl& other) const {        if (num == other.num) return name < other.name;        return num < other.num;    }};bool cmp(ss a, ss b){    return a.p < b.p;}map<string,long long> peosw;map<string,long long> howmuch;map<string,string> whichplace;map<string,int> ans;set<pl> sett;long long s;int main(){    int n, m, k;    scanf("%d", &n);    for(int i = 1; i <= n; i++)    {        scanf("%s %s %I64d", c, d, &s);        string aa(c), bb(d); whichplace[aa] = bb;        a[i].name = aa; a[i].wealth = s; peosw[aa] = s;        if (!howmuch[bb]) howmuch[bb] = s;        else howmuch[bb] = howmuch[bb] + s;    }    map<string,long long>::iterator it;    for (it = howmuch.begin(); it != howmuch.end(); it++)    {        pl newx;        newx.name = it -> first; newx.num = it -> second;        sett.insert(newx);    }    scanf("%d %d", &m, &k);    for (int i = 1; i <= k; i++)    {        scanf("%d %s %s", &b[i].day, c, d);        string aa(c), bb(d);        b[i].who = aa; b[i].place = bb;    }    int pos = 1;    for (int i = 1; i <= m; i++)    {        set<pl>::iterator it1, it2;        if (sett.size() > 1)        {            it1 = sett.end();            it1--; it2 = it1; it2--;            if (it1 -> num > it2 -> num)            {                if (!ans[it1 -> name]) ans[it1 -> name] = 1;                else ans[it1 -> name] = ans[it1 -> name] + 1;            }        }        else if (sett.size() == 1)        {            it1 = sett.end();            it1--;            if (!ans[it1 -> name]) ans[it1 -> name] = 1;            else ans[it1 -> name] = ans[it1 -> name] + 1;        }        while (i == b[pos].day && pos <= k)        {            pl oldp, newp1, newp2, oldp1;            oldp.name = whichplace[b[pos].who];            oldp.num = peosw[b[pos].who];            newp1.name = b[pos].place;            whichplace[b[pos].who] = b[pos].place;            if (!howmuch[newp1.name])            {                newp1.num = oldp.num;                howmuch[newp1.name] = oldp.num;                sett.insert(newp1);            }            else            {                newp1.num = howmuch[newp1.name] + oldp.num;                oldp1.name = newp1.name; oldp1.num = howmuch[newp1.name];                howmuch[newp1.name] = howmuch[newp1.name] + oldp.num;                sett.insert(newp1); sett.erase(oldp1);            }            if (howmuch[oldp.name] > oldp.num)            {                newp2.name = oldp.name; newp2.num = howmuch[oldp.name] - oldp.num;                sett.insert(newp2);            }            long long tmp = oldp.num;            oldp.num = howmuch[oldp.name];            howmuch[oldp.name] = howmuch[oldp.name] - tmp;            sett.erase(oldp);            pos++;        }    }    map<string, int>::iterator itt;    int cnt = 0;    for (itt = ans.begin(); itt != ans.end(); itt++)    {        ss news;        news.day = itt -> second; news.p = itt -> first;        aaa[++cnt] = news;    }    sort(aaa + 1, aaa + cnt + 1, cmp);    for (int i = 1; i <= cnt; i++)    {        printf("%s %d\n", aaa[i].p.c_str(), aaa[i].day);    }}