Va 1444-knowledge for the masses (efficient)

Link to the Q & A 1444-knowledge for the masses

Given R and l, r indicates that there are R rows, and l indicates the maximum length of a row.
For each row, n is given, followed by the number of n. If arr [I] is 0, it indicates space and length is 1. Otherwise, it indicates the shelf and length is arr [I]. Now, I want to move from the top to the bottom and ask to move at least a few bookshelves and output the path coordinates that can be passed.

Solution: C [I] indicates how many lines can pass through the I coordinate. When C [I] = r, it indicates that a person can pass through the entire shelf from this position. G [I] indicates the cost of passing through from the I position. Then, for each row, POS [I] records the position of the I-th space. vis [I] records the Left shift cost, which is used to compare with the right shift cost and take the minimum value. Note that when the second row is output, each number must be followed by a space. Otherwise, PE is used.

#include <cstdio>#include <cstring>#include <algorithm>using namespace std;const int maxn = 1e6+5;const int INF = 0x3f3f3f3f;int R, L, c[maxn], g[maxn];int n, arr[maxn], pos[maxn], vis[maxn];void  input () {    scanf("%d", &n);    memset(vis, -1, sizeof(vis));    int cnt = 0, mv = 0;    for (int i = 0; i < n; i++) {        scanf("%d", &arr[i]);        if (arr[i] == 0) {            mv++;            pos[cnt++] = i;            c[mv-1]++;        } else {            int k = min(cnt, arr[i]);            mv = mv + arr[i];            for (int j = 1; j <= k; j++) {                int u = mv - j;                vis[u] = i - pos[cnt-j] - j + 1;                g[u] += vis[u];                c[u]++;            }        }    }    reverse(arr, arr+n);    cnt = 0;    for (int i = 0; i < n; i++) {        if (arr[i] == 0) {            mv--;            pos[cnt++] = i;        } else {            int k = min(cnt, arr[i]);            mv = mv - arr[i];            for (int j = 0; j < k; j++) {                int u = mv + j;                if (vis[u] == -1) {                    vis[u] = i - pos[cnt-j-1] - j;                    g[u] += vis[u];                    c[u]++;                } else {                    int tmp = i - pos[cnt-j-1] - j;                    g[u] += min(0, tmp - vis[u]);                }            }        }    }}void init () {    scanf("%d%d", &R, &L);    memset(c, 0, sizeof(c));    memset(g, 0, sizeof(g));    for (int i = 0; i < R; i++)        input();}void solve () {    int ans = INF, cnt = 0;    for (int i = 0; i < L; i++) {        if (c[i] == R) {            if (g[i] < ans) {                cnt = 0;                ans = g[i];            }            if (g[i] == ans)                vis[cnt++] = i;        }    }    /*    printf("%d\n%d", ans, vis[0]);    for (int i = 1; i < cnt; i++)        printf(" %d", vis[i]);    printf("\n");    */    printf("%d\n", ans);    for (int i = 0; i < cnt; i++)        printf("%d ", vis[i]);    printf("\n");}int main () {    int cas;    scanf("%d", &cas);    while (cas--) {        init();        solve();    }    return 0;}

Va 1444-knowledge for the masses (efficient)