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

How old is a 80 pound blue catfish?

Unveiling the Secrets Behind the Age of an 80-Pound Blue Catfish Catching a massive river monster is the ultimate dream for many freshwater anglers across...

Stop Buying Cheap Alternatives This Viral Kitchen Gadgets Is the Real Deal

📢 As an Amazon Associate, I earn from qualifying purchases.After putting the Viral Kitchen Gadgets through a proper 30-day real-world test, I can tell...

Extra support arrives as Ohakune in race to restore water supply

Efforts to restore the main municipal water network in Ohakune have escalated significantly as specialist repair crews from across New Zealand arrive to assist...

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!