目錄
雖然三道題上了200，但還是有點菜
一.優雅的序列
1.題目
2.題解
這道題難就難在重複的數字上，如果沒有重複的數字，直接n-1。
不難發現，越小的數越要往前面放，但是又爲了儘量避免重複，我們就輪換着放，比如：
1 1 1 2 3 4 4 4 5
就應該變成：
1 2 3 4 5 1 4 1 4
這樣就能保證有最多對a[i] < a[i + 1]；
具體實踐如下：
step1：把每個數字排序後的起點和重複數量存在一個pair數組內；
step2：用len表示最後排好的數組長度，每次枚舉每一個數字，直到枚舉完每一個數字爲止。
step3：統計滿足條件的i的個數。
3.Code
#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
using namespace std;
#define LL long long
int n, k, len;
LL a[100005], ans, a1[100005];
pair <int, int> p[100005];
int main (){
//freopen ("grace.in", "r", stdin);
//freopen ("grace.out", "w", stdout);
scanf ("%d", &n);
for (register int i = 1; i <= n; i ++)
scanf ("%lld", &a[i]);
sort (a + 1, a + 1 + n);
for (register int i = 2; i <= n; i ++){
if (a[i] == a[i - 1]){
int j;
for (j = i; j <= n; j ++)
if (a[j] != a[j - 1])
break;
p[++ k].first = i - 1;
p[k].second = j - i + 1;
i = j;
}
else{
p[++ k].first = i - 1;
p[k].second = 1;
}
}
if (a[n] != a[n - 1]){
p[++ k].first = n;
p[k].second = 1;
}
while (len < n){
for (register int i = 1; i <= k; i ++){
if (p[i].second){
a1[++ len] = a[p[i].first];
p[i].first ++;
p[i].second --;
}
}
}
for (register int i = 1; i <= len; i ++)
if (a1[i] < a1[i + 1])
ans ++;
printf ("%lld\n", ans);
return 0;
}
二.甲蟲入侵
1.題目
2.題解
這道題是整個最難的題目，要用離散化+暴力搜索。
首先輸入時可以把每次移動看做一個點，起點是（0,0）例如：樣例第一個就是：（8,0）
每次移動要存入兩個點，這是爲後面的離散化做準備的；
然後，對於我們存入的每一個點，我們要進行去重操作（裏面要排序），橫座標和縱座標分別進行。
接着就進行離散化，我存入的每一個點都對應他們自己的一個下標，就可以用這些下標表示他們的值，一共只有1000個點，離散化之後整個的空間就變成了2000。
然後再來標記農夫走過的田地，這裏我們用upper_bound找到農夫每次走之後的座標離散化之後的下標，把灑過農藥的田地標爲1。
接着暴力搜索一波，把甲蟲能到的位置標爲-1；
最後就是統計面積，我們先來看一看離散化之後農場的樣子：（標黃的是我們要求的不被甲蟲侵犯的面積）
相當於每一個點都被我擴散成了兩個點，那麼面積的求法就是1 * 4 + (x2 - x1 - 1) * (y1 - y2 + 1) + (y1 - y2 - 1) * (x2 - x1 + 1);
就這樣求面積就完了。
有點懵，上代碼：
3.Code
#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
using namespace std;
#define LL long long
#define M 2005
int n, len[M], cntx, cnty, curx, cury, realx[M], realy[M], vis[M][M];
char dir[M];
int d[4][2] = {{1, 0}, {-1, 0}, {0, -1}, {0, 1}};
LL ans;
void unique (int *x, int &cnt){
sort (x, x + cnt);
int cnt1 = 1;
for (int i = 1; i < cnt; i ++)
if (x[i] != x[i - 1])
x[cnt1 ++] = x[i];
cnt = cnt1;
}
bool pd (int tox, int toy){
if (tox < 0 || tox >= cntx || toy < 0 || toy >= cnty || vis[tox][toy])
return 0;
return 1;
}
void DFS (int x, int y){
vis[x][y] = -1;
for (int i = 0; i < 4; i ++){
int tox = x + d[i][0];
int toy = y + d[i][1];
if (pd (tox, toy))
DFS (tox, toy);
}
}
int main (){
scanf ("%d", &n);
for (int i = 1; i <= n; i ++){
scanf ("\n%c %d", &dir[i], &len[i]);
if (dir[i] == 'L'){
realx[cntx ++] = curx + 1;
realy[cnty ++] = cury;
realx[cntx ++] = curx -= len[i];
realy[cnty ++] = cury + 1;
}
if (dir[i] == 'R'){
realx[cntx ++] = curx;
realy[cnty ++] = cury;
realx[cntx ++] = (curx += len[i]) + 1;
realy[cnty ++] = cury + 1;
}
if (dir[i] == 'U'){
realx[cntx ++] = curx;
realy[cnty ++] = cury;
realx[cntx ++] = curx + 1;
realy[cnty ++] = (cury += len[i]) + 1;
}
if (dir[i] == 'D'){
realx[cntx ++] = curx;
realy[cnty ++] = cury + 1;
realx[cntx ++] = curx + 1;
realy[cnty ++] = cury -= len[i];
}
}
unique (realx, cntx);
unique (realy, cnty);
int Indexx = lower_bound (realx, realx + cntx, curx = 0) - realx,
Indexy = lower_bound (realy, realy + cnty, cury = 0) - realy,
now;
for (int i = 1; i <= n; i ++){
if (dir[i] == 'L'){
now = lower_bound (realx, realx + cntx, curx -= len[i]) - realx;
for ( ; Indexx > now; Indexx --)
vis[Indexx][Indexy] = 1;
}
if (dir[i] == 'R'){
now = lower_bound (realx, realx + cntx, curx += len[i]) - realx;
for ( ; Indexx < now; Indexx ++)
vis[Indexx][Indexy] = 1;
}
if (dir[i] == 'U'){
now = lower_bound (realy, realy + cnty, cury += len[i]) - realy;
for ( ; Indexy < now; Indexy ++)
vis[Indexx][Indexy] = 1;
}
if (dir[i] == 'D'){
now = lower_bound (realy, realy + cnty, cury -= len[i]) - realy;
for ( ; Indexy > now; Indexy --)
vis[Indexx][Indexy] = 1;
}
}
vis[Indexx][Indexy] = 1;
for (int i = 0; i <= cntx; i ++){
if (! vis[i][0]) DFS (i, 0);
if (! vis[i][cnty]) DFS (i, cnty);
}
for (int i = 1; i < cnty; i ++){
if (! vis[0][i]) DFS (0, i);
if (! vis[cntx][i]) DFS (cntx, i);
}
for (int i = 1; i <= cntx; i ++){
for (int j = 1; j <= cnty; j ++){
if (~vis[i][j])
ans += (LL) (realx[i] - realx[i - 1]) * (realy[j] - realy[j - 1]);
}
}
printf ("%lld\n", ans);
return 0;
}
三.大逃亡
1.題目
2.題解
很簡單，因爲答案具有單調性，直接二分答案，然後暴力搜索檢查答案，注意有細節，特別是二分和距離有可能是0
3.Code
#include <cstdio>
#include <iostream>
#include <cstring>
#include <queue>
using namespace std;
#define M 10005
struct node {
int xx, yy, step;
node (){};
node (int XX, int YY, int STEP){
xx = XX;
yy = YY;
step = STEP;
}
};
int n, X, Y, sx, sy, tx, ty, dx[M], dy[M], ans1, ans2, dis[1005][1005];
int dir[4][2] = {{1, 0}, {0, -1}, {-1, 0}, {0, 1}};
bool flag[1005][1005], vis[1005][1005];
int fabs (int x){
if (x < 0)
return -x;
return x;
}
bool pd1 (int tox, int toy){
if (tox < 0 || tox >= X || toy < 0 || toy >= Y || vis[tox][toy])
return 0;
return 1;
}
void prepare (){
queue <node> Q;
for (int i = 1; i <= n; i ++)
Q.push (node (dx[i], dy[i], 0));
while (! Q.empty ()){
node f = Q.front ();
if (! dis[f.xx][f.yy])
dis[f.xx][f.yy] = f.step;
Q.pop ();
for (int i = 0; i < 4; i ++){
int tox = f.xx + dir[i][0];
int toy = f.yy + dir[i][1];
if (pd1 (tox, toy)){
vis[tox][toy] = 1;
Q.push (node (tox, toy, f.step + 1));
}
}
}
}
bool pd (int tox, int toy, int now){
if (tox < 0 || tox >= X || toy < 0 || toy >= Y || vis[tox][toy] || dis[tox][toy] < now)
return 0;
return 1;
}
inline int BFS (int now){
memset (vis, 0, sizeof vis);
queue <node> Q;
vis[sx][sy] = 1;
Q.push (node (sx, sy, 0));
while (! Q.empty ()){
node f = Q.front ();
Q.pop ();
if (f.xx == tx && f.yy == ty)
return f.step;
for (register int i = 0; i < 4; i ++){
int tox = f.xx + dir[i][0];
int toy = f.yy + dir[i][1];
if (pd (tox, toy, now)){
vis[tox][toy] = 1;
Q.push (node (tox, toy, f.step + 1));
}
}
}
return -1;
}
inline int check (int now){
int tmp = BFS (now);
return tmp;
}
int main (){
//freopen ("escape.in", "r", stdin);
//freopen ("escape.out", "w", stdout);
int l = 0, r = 0x3f3f3f3f, mid;
scanf ("%d %d %d %d %d %d %d", &n, &X, &Y, &sx, &sy, &tx, &ty);
for (register int i = 1; i <= n; i ++){
scanf ("%d %d", &dx[i], &dy[i]);
r = min (r, fabs (dx[i] - sx) + fabs (dy[i] - sy));
r = min (r, fabs (dx[i] - tx) + fabs (dy[i] - ty));
}
prepare ();
int tmp;
while (l + 1 < r){
mid = (l + r) / 2;
tmp = check (mid);
if (tmp != -1){
l = mid;
}
else
r = mid - 1;
}
ans1 = l, ans2 = tmp;
int tp = check (r);
if (tp != -1)
ans1 = r, ans2 = tmp;
printf ("%d %d\n", ans1, ans2);
return 0;
}
