标签:struct void 贪心 .com 复杂度 push fread efi main
问题等价于选出$n / 3$个不相邻元素是权值和最大
这是一个经典贪心问题,同种树,拿堆维护即可,复杂度$O(n \log n)$
#include <queue> #include <cstdio> #include <cstring> using namespace std; extern inline char gc() { static char RR[23456], *S = RR + 23333, *T = RR + 23333; if(S == T) fread(RR, 1, 23333, stdin), S = RR; return *S ++; } inline int read() { int p = 0, w = 1; char c = gc(); while(c > ‘9‘ || c < ‘0‘) { if(c == ‘-‘) w = -1; c = gc(); } while(c >= ‘0‘ && c <= ‘9‘) p = p * 10 + c - ‘0‘, c = gc(); return p * w; } #define ll long long #define ri register int #define sid 200050 int n, k; bool gg[sid]; ll ans, v[sid], pre[sid], nxt[sid]; struct Man { ll id, val; Man() {} Man(ll _i, ll _v) : id(_i), val(_v) {} friend bool operator < (Man a, Man b) { return a.val < b.val; } } ; priority_queue <Man> q; inline void del(int o) { pre[nxt[o]] = pre[o]; nxt[pre[o]] = nxt[o]; gg[o] = 1; } int main() { n = read(); for(ri i = 1; i <= n; i ++) v[i] = read(); for(ri i = 1; i <= n; i ++) { pre[i] = i - 1; nxt[i] = i + 1; q.push(Man(i, v[i])); } pre[1] = n; nxt[n] = 1; k = n / 3; for(ri i = 1; i <= k; i ++) { Man o; int id; while(1) { o = q.top(); q.pop(); id = o.id; if(!gg[id]) break; } ans += v[id]; v[id] = v[pre[id]] + v[nxt[id]] - v[id]; q.push(Man(id, v[id])); del(pre[id]); del(nxt[id]); } printf("%lld\n", ans); return 0; }
标签:struct void 贪心 .com 复杂度 push fread efi main
原文地址:https://www.cnblogs.com/reverymoon/p/9558001.html