HDU 5037 Frog (efficient)

Link: HDU 5037 frog

Given N, M, L, indicates that there is a river with a width of M. The frog must jump from 0 to m, and the maximum jump capability of the frog is L, now there are n stones, and now we need to put any number of stones, so that frogs need to jump to m with the maximum number of steps.

Solution: maintain two pointers. Pos indicates the current position of the frog, and MV indicates the farthest position that the frog can jump to in the previous position. Then the length of the period L + 1 allows the frog to jump twice. However, note that the last cycle of each segment must be determined, because it may only take one step.

#include <cstdio>#include <cstring>#include <vector>#include <algorithm>using namespace std;int N, M, L;vector<int> g;void init () {    int x;    g.clear();    scanf("%d%d%d", &N, &M, &L);    for (int i = 0; i < N; i++) {        scanf("%d", &x);        g.push_back(x);    }    g.push_back(0);    g.push_back(M);    sort(g.begin(), g.end());}int solve () {    int ret = 0;    int pos = 0, mv = 0;    while (true) {        int idx = upper_bound(g.begin(), g.end(), pos + L) - g.begin() - 1;        if (g[idx] <= pos) {            int k = (g[idx+1] - pos) / (L + 1);            if (k == 1) {                ret++;                int tmp = mv + 1;                mv = pos + L;                pos = tmp;            } else {                k--;                ret += k * 2;                pos = pos + k * (L + 1);                mv = mv + k * (L + 1);            }        } else {            mv = pos + L;            pos = g[idx];            ret++;        }        if (pos == M)            break;    }    return ret;}int main () {    int cas;    scanf("%d", &cas);    for (int kcas = 1; kcas <= cas; kcas++) {        init();        printf("Case #%d: %d\n", kcas, solve());    }    return 0;}

HDU 5037 Frog (efficient)