URAL 1650. Billionaires（线段树）

题意：求城市财富排行第一的天数。有两个城市的财富值都最多，则不计算。

首先输入n行，每行代表一个人。 名字，最初所在的城市，拥有的财富（不一定能被100000000整除）。

计算m天的城市财富排行。这些人有k次迁移。

k行，每一行分别是：在第几天晚上迁移的，迁移的人，迁入的城市。

用线段树记录每一个城市的财富值、

#include <iostream>#include <cmath>#include <cstdio>#include <cstring>#include <algorithm>#include <queue>#include <stack>#include <map>#include <set>#include <list>#include <deque>#include <string>#define LL long long#define DB double#define SI(a) scanf("%d",&a)#define SD(a) scanf("%lf",&a)#define SS(a) scanf("%s",a)#define SF scanf#define PF printf#define MM(a,b) memset(a,b,sizeof(a))#define REP(i,a,b) for(int (i)=(a);(i)<(b);(i)++)#define N 100009#define INF 0x3f3f3f3f#define EPS 1e-8#define bug puts("bug")using namespace std;map<string,int> city;map<string,int> peop;struct nod{    char ch[2][29];    LL k;} re1[N],re2[N];int n,m,d;LL money[N];int sum[N];char where[N][29];string recity[N];struct TR{    LL mx;int two,p;} tree[N<<2];#define lson rt<<1,l,m#define rson rt<<1|1,m+1,rvoid update(int rt){    int l=rt<<1,r=rt<<1|1;    if(tree[l].mx>tree[r].mx)    {        tree[rt] = tree[l];    }else if(tree[l].mx<tree[r].mx)    {        tree[rt] = tree[r];    }else    {        tree[rt].mx = tree[l].mx;        tree[rt].two = 1;        tree[rt].p = -1;    }}void build(int rt,int l,int r){    if(l==r)    {        tree[rt].mx = 0;        tree[rt].two = 0;        tree[rt].p = l;        return ;    }    int m = (l+r)>>1;    build(lson);    build(rson);    update(rt);}void add(int p,LL c,int rt,int l,int r){    if(l==r)    {        tree[rt].mx += c;        tree[rt].two = 0;        return ;    }    int m = (l+r)>>1;    if(p<=m) add(p,c,lson);    else add(p,c,rson);    update(rt);}struct AN{    string cit;    bool operator<(const AN t)const    {        return cit<t.cit;    }    int val;} ans[N];int main(){    #ifndef ONLINE_JUDGE    freopen("in.txt","r",stdin);    #endif    SI(n);    int cnt = 0;    REP(i,0,n)    {        SS(re1[i].ch[0]);//peop        SS(re1[i].ch[1]);//city        LL tm;        cin>>tm;        money[i] = re1[i].k = tm;        peop[string(re1[i].ch[0])]=i;        recity[cnt] = re1[i].ch[1];        city[string(re1[i].ch[1])]=cnt++;        memcpy(where[i],re1[i].ch[1],sizeof(re1[i].ch[1]));    }    SI(d);SI(m);    REP(i,0,m)    {        cin>>re2[i].k;        SS(re2[i].ch[0]);//name        SS(re2[i].ch[1]);//city        if(city.find(re2[i].ch[1])==city.end())        {            recity[cnt] = re2[i].ch[1];            city[string(re2[i].ch[1])] = cnt++;        }    }    build(1,0,cnt-1);    REP(i,0,n){        add(city[re1[i].ch[1]],money[i],1,0,cnt-1);    }    int dy = 0;    REP(i,0,m)    {        int tmp = re2[i].k - dy;        dy = re2[i].k;        if(tree[1].two==0)        {            sum[tree[1].p] += tmp;        }        int who = peop[string(re2[i].ch[0])];        add(city[string(where[who])],0-money[who],1,0,cnt-1);        add(city[string(re2[i].ch[1])],money[who],1,0,cnt-1);        strcpy(where[who],re2[i].ch[1]);    }    if(tree[1].two==0)    {        sum[tree[1].p] += d-dy;    }    int an = 0;    REP(i,0,cnt)    {        if(sum[i]>0)        {            ans[an].cit = recity[i];            ans[an].val = sum[i];            an++;        }    }    sort(ans,ans+an);    REP(i,0,an)    {        cout<<ans[i].cit<<" "<<ans[i].val<<endl;    }    return 0;}