hdu 4831 Scenic Popularity(模拟)
题目链接:hdu 4831 Scenic Popularity
题目大意:略。
解题思路:对于休闲区g[i][0]和g[i][1]记录的是近期的两个景点的id(仅仅有一个近期的话g[i][1]为0),对于景点来说。g[i][0]为-1(表示该id相应的是景点),g[i][1]为该景点的热度值.主要就是模拟,注意一些细节就能够了。
#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <algorithm>
using namespace std;
const int N = 100005;
const int INF = 0x3f3f3f3f;
int n, pos[N], g[N][2];
inline int dis (int a, int b) {
return abs(pos[a]-pos[b]);
}
inline void cat (int x, int mv) {
if (mv == n+1)
return;
int& tmp = g[x][0];
if (tmp == 0 || dis(x, mv) < dis(x, tmp)) {
g[x][0] = mv;
} else if (dis(x, mv) == dis(x, tmp)) {
g[x][1] = mv;
}
}
void init () {
scanf("%d", &n);
memset(g, 0, sizeof(g));
memset(pos, 0, sizeof(pos));
int mv = 0, val;
for (int i = 1; i <= n; i++) {
scanf("%d%d", &pos[i], &val);
if (val) {
mv = i;
g[i][0] = -1;
g[i][1] = val;
} else {
g[i][0] = mv;
}
}
mv = n+1;
for (int i = n; i; i--) {
if (g[i][0] < 0) {
mv = i;
} else {
cat(i, mv);
}
}
}
int find (int x) {
if (x == 0)
return 0;
if (g[x][0] < 0)
return g[x][1];
else
return max(find(g[x][0]), find(g[x][1]));
}
int query (int k) {
int ans = 0;
for (int i = 1; i <= n; i++) {
if (find(i) <= k)
ans++;
}
return ans;
}
void solve () {
int m, a, b;
char str[5];
scanf("%d", &m);
for (int i = 0; i < m; i++) {
scanf("%s", str);
if (str[0] == 'Q') {
scanf("%d", &a);
printf("%d\n", query(a));
} else {
scanf("%d%d", &a, &b);
g[a+1][1] = b;
}
}
}
int main () {
int cas;
scanf("%d", &cas);
for (int i = 1; i <= cas; i++) {
init();
printf("Case #%d:\n", i);
solve();
}
return 0;
}
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