# 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
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; }

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By |2017-04-07T22:11:50+00:00November 1st, 2016|Categories: C/C++, Programming||0 Comments