洛谷P1133 教主的花园
作者:互联网
恢复手感的第二天,一道简单dp
需要注意的是,1和n是相邻的,这个是环形
刚交的时候就WA了三个点,后来想到了一个结论,就是如果1和n相邻,那么这个限制比单纯的链肯定是次优的,所以环形从2 到 n + 1再做一遍就好了
可以滚动数组优化
#include <bits/stdc++.h> using namespace std; #define limit (200000 + 5)//防止溢出 #define INF 0x3f3f3f3f #define inf 0x3f3f3f3f3f #define lowbit(i) i&(-i)//一步两步 #define EPS 1e-9 #define FASTIO ios::sync_with_stdio(false);cin.tie(0),cout.tie(0); #define ff(a) printf("%d\n",a ); #define pi(a,b) pair<a,b> #define rep(i, a, b) for(ll i = a; i <= b ; ++i) #define per(i, a, b) for(ll i = b ; i >= a ; --i) #define MOD 998244353 #define traverse(u) for(int i = head[u]; ~i ; i = edge[i].next) #define FOPEN freopen("C:\\Users\\tiany\\CLionProjects\\akioi\\data.txt", "rt", stdin) #define FOUT freopen("C:\\Users\\tiany\\CLionProjects\\akioi\\dabiao.txt", "wt", stdout) typedef long long ll; typedef unsigned long long ull; char buf[1<<23],*p1=buf,*p2=buf,obuf[1<<23],*O=obuf; inline ll read(){ #define getchar() (p1==p2&&(p2=(p1=buf)+fread(buf,1,1<<21,stdin),p1==p2)?EOF:*p1++) ll sign = 1, x = 0;char s = getchar(); while(s > '9' || s < '0' ){if(s == '-')sign = -1;s = getchar();} while(s >= '0' && s <= '9'){x = (x << 3) + (x << 1) + s - '0';s = getchar();} return x * sign; #undef getchar }//快读 void print(ll x) { if(x/ 10) print(x / 10); *O++=x % 10+'0'; } void write(ll x, char c = 't') { if(x < 0)putchar('-'),x = -x; print(x); if(!isalpha(c))*O++ = c; fwrite(obuf,O-obuf,1,stdout); O = obuf; } int kase; int n, m,k; int a[limit][4]; int dp[2][4][4]; //状态转移方程: // 30可以从20和 void solve(){ cin>>n; int ans = 0; int cur = 1, prev = 0; swap(cur, prev); rep(i,1,n){ cin>>a[i][1]>>a[i][2]>>a[i][3]; rep(f,1,3){ a[i + n][f] = a[i][f]; } rep(j, 1, 3){ rep(h, 1, 3){ rep(l, 1, 3){ if(h > l and h > j){ dp[cur][l][h] = max(dp[cur][l][h], dp[prev][h][j] + a[i][l]); } if(h < l and h < j){ dp[cur][l][h] = max(dp[cur][l][h], dp[prev][h][j] + a[i][l]); } } } } if(i != n)swap(cur, prev); } int ans1 = 0; rep(i,1,3){ rep(j, 1, 3){ ans = max(ans, dp[cur][i][j]); } } memset(dp, 0, sizeof(dp)); rep(i,2,n + 1){ rep(j, 1, 3){ rep(h, 1, 3){ rep(l, 1, 3){ if(h > l and h > j){ dp[cur][l][h] = max(dp[cur][l][h], dp[prev][h][j] + a[i][l]); } if(h < l and h < j){ dp[cur][l][h] = max(dp[cur][l][h], dp[prev][h][j] + a[i][l]); } } } } if(i != n + 1)swap(cur, prev); } rep(i,1,3){ rep(j, 1, 3){ ans1 = max(ans1, dp[cur][i][j]); } } cout<<min(ans1, ans)<<endl; } int32_t main() { #ifdef LOCAL FOPEN; // FOUT; #endif FASTIO // cin>>kase; // while (kase--) solve(); cerr << "Time elapsed: " << 1.0*clock()/CLOCKS_PER_SEC << "s\n"; return 0; }
标签:洛谷,cur,max,rep,教主,P1133,prev,dp,define 来源: https://www.cnblogs.com/tiany7/p/15732539.html