Codeforces Round #378 (Div. 2), Problem: (C) Epidemic in Monstropolis Solution in C

Hi guys , i just tried the Epidemic in Monstropolis problem , hope you might like it .

#include
#include

#define MAXN 500
int a[MAXN];
int b[MAXN];
int partial_sums[MAXN];
unsigned char d[MAXN + 1][MAXN + 1];
unsigned char equals[MAXN + 1][MAXN + 1];
int ans[MAXN + 1][MAXN + 1];
int n, k;
int i, j, r, t;

int dif[2];
int order[2];
char symbol[2];

int main(int argc, char *argv[]) {

#ifndef ONLINE_JUDGE
freopen(“input.txt”, “r”, stdin);
//freopen(“output.txt”, “w”, stdout);
#endif
// read and init data
scanf(“%d”, &n);
for (i = 0; i < n; ++i) { scanf("%d", &a[i]); } scanf("%d", &k); for (i = 0; i < k; ++i) { scanf("%d", &b[i]); } partial_sums[0] = a[0]; for (i = 1; i < n; ++i) { partial_sums[i] = partial_sums[i - 1] + a[i]; } d[0][0] = 1; for (i = 0; i < n; ++i) { equals[i][i] = 1; } for (i = 2; i <= n; ++i) { for (j = 0; j <= n - i; ++j) { equals[j][j + i - 1] = (equals[j][j + i - 2] && (a[j + i - 1] == a[j])); } } // dynamic programming for (i = 1; i <= n; ++i) { int border = k; if (i < k) { border = i; } for (j = 1; j <= border; ++j) { // int border2 = i - j + 1; for (r = 1; r <= i; ++r) { int sub_sum = partial_sums[i - 1] - partial_sums[i - r] + a[i - r]; unsigned char can_be_used = ((r == 1) || !equals[i - r][i - 1]); unsigned char check = (can_be_used && (sub_sum == b[j - 1])); if (d[i - r][j - 1] && check) { d[i][j] = 1; ans[i][j] = r; break; } } } } if (!d[n][k]) { printf("NO\n"); } else { printf("YES\n"); int curn = n; int curk = k; while (curn > 0) {
// print result
int maxval = -1;
int maxindx = curn;

for (i = curn – ans[curn][curk] + 1; i <= curn; ++i) { if (a[i - 1] > maxval) {
maxval = a[i – 1];
maxindx = i;
}
}

int left_cnt = maxindx – curn + ans[curn][curk] – 1;
int right_cnt = curn – maxindx;

dif[0] = 0; dif[1] = -1;
order[0] = right_cnt; order[1] = left_cnt;
symbol[0] = ‘R’; symbol[1] = ‘L’;

if (left_cnt == 0) {
while (maxindx < curn && a[maxindx] == maxval) { maxindx++; left_cnt++; right_cnt--; } dif[0] = 0; dif[1] = -1; order[0] = right_cnt; order[1] = left_cnt; symbol[0] = 'R'; symbol[1] = 'L'; } else if (right_cnt == 0 || a[maxindx] == a[maxindx - 1]) { dif[0] = -1; dif[1] = 0; order[0] = left_cnt; order[1] = right_cnt; symbol[0] = 'L'; symbol[1] = 'R'; } for (j = 0; j < 2; ++j) { for (i = 0; i < order[j]; ++i) { printf("%d %c\n", maxindx, symbol[j]); maxindx += dif[j]; } } curn -= ans[curn][curk]; curk -= 1; } } return 0; }

Hi guys , i just tried the Epidemic in Monstropolis problem , hope you might like it .

#include
#include

#define MAXN 500
int a[MAXN];
int b[MAXN];
int partial_sums[MAXN];
unsigned char d[MAXN + 1][MAXN + 1];
unsigned char equals[MAXN + 1][MAXN + 1];
int ans[MAXN + 1][MAXN + 1];
int n, k;
int i, j, r, t;

int dif[2];
int order[2];
char symbol[2];

int main(int argc, char *argv[]) {

#ifndef ONLINE_JUDGE
freopen(“input.txt”, “r”, stdin);
//freopen(“output.txt”, “w”, stdout);
#endif
// read and init data
scanf(“%d”, &n);
for (i = 0; i < n; ++i) { scanf("%d", &a[i]); } scanf("%d", &k); for (i = 0; i < k; ++i) { scanf("%d", &b[i]); } partial_sums[0] = a[0]; for (i = 1; i < n; ++i) { partial_sums[i] = partial_sums[i - 1] + a[i]; } d[0][0] = 1; for (i = 0; i < n; ++i) { equals[i][i] = 1; } for (i = 2; i <= n; ++i) { for (j = 0; j <= n - i; ++j) { equals[j][j + i - 1] = (equals[j][j + i - 2] && (a[j + i - 1] == a[j])); } } // dynamic programming for (i = 1; i <= n; ++i) { int border = k; if (i < k) { border = i; } for (j = 1; j <= border; ++j) { // int border2 = i - j + 1; for (r = 1; r <= i; ++r) { int sub_sum = partial_sums[i - 1] - partial_sums[i - r] + a[i - r]; unsigned char can_be_used = ((r == 1) || !equals[i - r][i - 1]); unsigned char check = (can_be_used && (sub_sum == b[j - 1])); if (d[i - r][j - 1] && check) { d[i][j] = 1; ans[i][j] = r; break; } } } } if (!d[n][k]) { printf("NO\n"); } else { printf("YES\n"); int curn = n; int curk = k; while (curn > 0) {
// print result
int maxval = -1;
int maxindx = curn;

for (i = curn – ans[curn][curk] + 1; i <= curn; ++i) { if (a[i - 1] > maxval) {
maxval = a[i – 1];
maxindx = i;
}
}

int left_cnt = maxindx – curn + ans[curn][curk] – 1;
int right_cnt = curn – maxindx;

dif[0] = 0; dif[1] = -1;
order[0] = right_cnt; order[1] = left_cnt;
symbol[0] = ‘R’; symbol[1] = ‘L’;

if (left_cnt == 0) {
while (maxindx < curn && a[maxindx] == maxval) { maxindx++; left_cnt++; right_cnt--; } dif[0] = 0; dif[1] = -1; order[0] = right_cnt; order[1] = left_cnt; symbol[0] = 'R'; symbol[1] = 'L'; } else if (right_cnt == 0 || a[maxindx] == a[maxindx - 1]) { dif[0] = -1; dif[1] = 0; order[0] = left_cnt; order[1] = right_cnt; symbol[0] = 'L'; symbol[1] = 'R'; } for (j = 0; j < 2; ++j) { for (i = 0; i < order[j]; ++i) { printf("%d %c\n", maxindx, symbol[j]); maxindx += dif[j]; } } curn -= ans[curn][curk]; curk -= 1; } } return 0; }

More from author

LEAVE A REPLY

Please enter your comment!
Please enter your name here

Related posts

Advertismentspot_img

Latest posts

Stop Buying Cheap Alternatives This Wireless Earbuds Accessories Is the Real Deal

📢 As an Amazon Associate, I earn from qualifying purchases.I stumbled across this while desperately trying to fix a recurring problem in my routine....

Where the OKC Thunder Rank Among the Top 10 NBA Teams Entering August

As the offseason progresses into August, the NBA landscape continues to undergo dramatic transformations. Major headline-grabbing moves, including LeBron James making his historic transition...

Stop Buying Cheap Alternatives This Portable Blender Bottle Is the Real Deal

📢 As an Amazon Associate, I earn from qualifying purchases.There comes a point when you realise that tolerating a frustrating situation is costing you...

Want to stay up to date with the latest news?

We would love to hear from you! Please fill in your details and we will stay in touch. It's that simple!