计蒜客 跳跃游戏 贪心

           思路：假设当前位于pos，这个点能跳跃的最大距离是len[pos]，那么下一个点应该是next_pos = j满足max{j + len[j]}，即下一个点要能到达最远的距离。例如:2 0 2 0 1，当pos=0，它能到达1 和 2，2 + len[2] > 1 + len[1]，所以选择下一个点是2而不是1.

AC代码

#include <cstdio>#include <cmath>#include <cctype>#include <bitset>#include <algorithm>#include <cstring>#include <utility>#include <string>#include <iostream>#include <map>#include <set>#include <vector>#include <queue>#include <stack>using namespace std;#pragma comment(linker, "/STACK:1024000000,1024000000") #define eps 1e-10#define inf 0x3f3f3f3f#define PI pair<int, int> typedef long long LL;const int maxn = 500 + 5;int a[maxn];int main() {int n;while(scanf("%d", &n) == 1) {for(int i = 0; i < n; ++i) {scanf("%d", &a[i]);}int pos = 0;while(a[pos]) {if(pos == n-1) {break;}int jump = a[pos];int next_pos = pos+1;for(int i = pos+1; i <= pos+jump && i < n; ++i) {if(a[i] + i >= next_pos + a[next_pos]) next_pos = i;}pos = next_pos;}if(pos == n-1) printf("true\n");else printf("false\n");}return 0;}

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