Ural 1650 Billionaires（线段树）

http://acm.timus.ru/problem.aspx?space=1&num=1650

　　用线段树统计的模拟题，开始的时候数组开小了，以为城市只有50000个，所以wa了两次，后来开大了就过了！～ ^_^

这题用STL解决代码量比较小，据说可以用set来充当我的线段树。代码如下：

View Code

#include <cstdio>
#include <cstring>
#include <algorithm>
#include <vector>
#include <set>
#include <string>
#include <map>

using namespace std;

typedef long long ll;
typedef vector<ll> vll;
typedef vector<int> vi;
typedef vector<string> vstr;
set<string> cityName, billName;
map<string, int> cityId, billId;

const int max1 = 10005;
const int max2 = 60005;
char city[max2][25], bill[max1][25];
int curPos[max1], cntDay[max2];
ll money[max1];

vstr buf1[2], buf2[2];
vll buff1;
vi buff2;

void initAll() {
cityName.clear();
cityId.clear();
billName.clear();
billId.clear();
buff1.clear();
buff2.clear();
memset(city, 0, sizeof(city));
memset(bill, 0, sizeof(bill));
memset(money, 0, sizeof(money));
memset(curPos, 0, sizeof(curPos));
memset(cntDay, 0, sizeof(cntDay));
for (int i = 0; i < 2; i++) {
buf1[i].clear();
buf2[i].clear();
}
}

void input1(int n) {
char tmp1[25], tmp2[25];
ll tmp;

while (n--) {
scanf("%s%s%lld", tmp1, tmp2, &tmp);
buf1[0].push_back(tmp1);
buf1[1].push_back(tmp2);
buff1.push_back(tmp);
billName.insert(tmp1);
cityName.insert(tmp2);
}
}

void input2(int k) {
char tmp1[25], tmp2[25];
int tmp;

while (k--) {
scanf("%d%s%s", &tmp, tmp1, tmp2);
buff2.push_back(tmp);
buf2[0].push_back(tmp1);
buf2[1].push_back(tmp2);
//        billName.insert(tmp1);
cityName.insert(tmp2);
}
}

void makeMap() {
set<string>::iterator ii;
int pos;

for (pos = 0, ii = cityName.begin(); ii != cityName.end(); ii++) {
strcpy(city[pos], (*ii).c_str());
cityId[*ii] = pos++;
}
for (pos = 0, ii = billName.begin(); ii != billName.end(); ii++) {
strcpy(bill[pos], (*ii).c_str());
billId[*ii] = pos++;
}
}

#define lson l, m, rt << 1
#define rson m + 1, r, rt << 1 | 1

struct segTree {
ll mx[max2 << 2];
int id[max2 << 2];

void up(int rt) {
int ls = rt << 1, rs = rt << 1 | 1;

mx[rt] = max(mx[ls], mx[rs]);
if (mx[ls] == mx[rs]) {
id[rt] = -1;
} else {
id[rt] = mx[ls] > mx[rs] ? id[ls] : id[rs];
}
}

void build(int l, int r, int rt) {
mx[rt] = 0;
if (l == r) {
id[rt] = l;
return ;
}
int m = (l + r) >> 1;

build(lson);
build(rson);
up(rt);
}

void update(int p, ll k, int l, int r, int rt) {
if (l == r) {
mx[rt] += k;
return ;
}
int m = (l + r) >> 1;

if (p <= m) {
update(p, k, lson);
} else {
update(p, k, rson);
}
up(rt);
}
} segT;

void initStat(int n) {
int size = buff1.size();

segT.build(0, n - 1, 1);
for (int i = 0; i < size; i++) {
int bId = billId[buf1[0][i]];
int cId = cityId[buf1[1][i]];

segT.update(cId, buff1[i], 0, n - 1, 1);
curPos[bId] = cId;
money[bId] = buff1[i];
}
//    for (int i = 0; i < size; i++) {
//        printf("%s : %s\n", bill[i], city[curPos[i]]);
//    }
//    puts("~~");
}

void work(int n, int t) {
int size = buff2.size();
int curTime = 0;

for (int i = 0; i < size; i++) {
//        if (~segT.id[1]) printf("highest %s\n", city[segT.id[1]]);
if (~segT.id[1]) cntDay[segT.id[1]] += buff2[i] - curTime;
curTime = buff2[i];

int bId = billId[buf2[0][i]];
int cId = cityId[buf2[1][i]];

segT.update(curPos[bId], -money[bId], 0, n - 1, 1);
segT.update(cId, money[bId], 0, n - 1, 1);
curPos[bId] = cId;

//        for (int j = 0; j < cityName.size(); j++) {
//            printf("%s : %d\n", city[j], cntDay[j]);
//        }
//        puts("!!!");
//        for (int j = 0; j < billName.size(); j++) {
//            printf("%s : %s\n", bill[j], city[curPos[j]]);
//        }
//        puts("~~");
}
if (~segT.id[1]) cntDay[segT.id[1]] += t - curTime;
for (int j = 0, endj = cityName.size(); j < endj; j++) {
if (cntDay[j]) printf("%s %d\n", city[j], cntDay[j]);
}
//    puts("!!!");

}

int main() {
int n, m, k;

//    freopen("in", "r", stdin);
//    freopen("out", "w", stdout);

while (~scanf("%d", &n)) {
initAll();
input1(n);
scanf("%d%d", &m, &k);
input2(k);
makeMap();

n = cityName.size();
initStat(n);
work(n, m);
}

return 0;
}

——written by Lyon