Trick
Posted ShadowAA
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Trick
Do we really need to visit all the states?
Sometimes, the naive dp solution to a problem might take too long and too much memory. However, sometimes it is worth noting that most of the states can be ignored because they will never be reached and this can reduce your time complexity and memory complexity.
有时，问题的简单dp解决方案可能会占用太长时间和太多内存。然而，有时值得注意的是，大多数状态可以被忽略，因为它们永远不会被到达，这可以降低您的时间复杂性和内存复杂性。
Problem - C - Codeforces
Change the object to dp
更改dp的对象，选择小的dp
Problem - C - Codeforces
Problem - E - Codeforces
Open and Close Interval Trick
Problem - F - Codeforces
Problem - D - Codeforces
Problem - E - Codeforces
"Connected Component" DP
P5999 [CEOI2016] kangaroo
#include<bits/stdc++.h>
using namespace std;
long long n, m, k, i, j, f[2005][2005];
const int mod = 1e9 + 7;
int main()
cin >> n >> m >> k;
f[0][0] = 1;
for (i = 1; i <= n; i++)
for (j = 0; j <= n; j++)
if ((i != m) and (i != k))
(f[i][j] += f[i - 1][j + 1] * j) %= mod;
if (j) (f[i][j] += f[i - 1][j - 1] * (j - (i > m) - (i > k))) %= mod;
else
(f[i][j] += f[i - 1][j]) %= mod;
if (j) (f[i][j] += f[i - 1][j - 1]) %= mod;
cout << f[n][1] << \'\\n\';
摩天大楼
#2743. 「JOI Open 2016」摩天大楼 - 题目 - LibreOJ (loj.ac)
#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
ll n, m, i, j, k, u, o, p, l, s, f[105][105][1005][3], a[105];
const int mod = 1e9 + 7;
int main()
cin >> n >> m;
for (i = 1; i <= n; i++)
cin >> a[i];
if (n == 1) puts("1"); return 0;
sort(a + 1, a + n + 1);
f[0][0][0][0] = 1;
for (i=0;i<n;i++)
for (j=0;j<=i;j++)
for (k=0;k<=m;k++)
for (u = 0; u <= 2; u++)
o = a[i+1] - a[i];
if (i == 0) o = 0;
p = f[i][j][k][u];
l = k + o * (j * 2 - u);
if ((l > m) or (!p))continue;
if (j > 1) (f[i + 1][j - 1][l][u] += p * (j - 1))%=mod;
(f[i + 1][j + 1][l][u] += p * (j + 1 - u))%=mod;
if (u == 0)
(f[i + 1][j + 1][l][u + 1] += p * 2)%=mod;
if (u == 1)
(f[i + 1][j + 1][l][u + 1] += p)%=mod;
(f[i + 1][j][l][u] += p * (j * 2 - u))%=mod;
if ((j)and(u == 0))
(f[i + 1][j][l][u+1] += p * 2)%=mod;
if ((j)and(u == 1))
(f[i + 1][j][l][u + 1] += p)%=mod;
for (i = 0; i <= m; i++)
s = (s + f[n][1][i][2])%mod;
cout << s << \'\\n\';
Ant Man
#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
ll n, s, e, i, j, k, q, a[5005], b[5005], c[5005], d[5005], x[5005], f[5005][5005];
int main()
cin >> n >> s >> e;
for (i = 1; i <= n; i++)
cin >> x[i];
for (i = 1; i <= n; i++)
cin >> a[i];
for (i = 1; i <= n; i++)
cin >> b[i];
for (i = 1; i <= n; i++)
cin >> c[i];
for (i = 1; i <= n; i++)
cin >> d[i];
for (i = 0; i <= n; i++)
for (j = 0; j <= n; j++)
f[i][j] = 1e18;
f[0][0] = 0;
for (i = 0; i < n; i++)
for (j = 0; j <= n; j++)
if ((j == 1) and (k == 2))
f[i][j] = 1e18;
if (f[i][j] < 1e18)
if (i + 1 == s)
if (j < n) f[i + 1][j + 1] = min(f[i + 1][j + 1], f[i][j] + d[i + 1] - x[i + 1]);
if (j) f[i + 1][j] = min(f[i + 1][j], f[i][j] + c[i + 1] + x[i + 1]);
else
if (i + 1 == e)
if (j < n) f[i + 1][j + 1] = min(f[i + 1][j + 1], f[i][j] + b[i + 1] - x[i + 1]);
if (j) f[i + 1][j] = min(f[i + 1][j], f[i][j] + a[i + 1] + x[i + 1]);
else
if (j < n) f[i + 1][j + 1] = min(f[i + 1][j + 1], f[i][j] + b[i + 1] + d[i + 1] - x[i + 1] - x[i + 1]);
if (j > 1) f[i + 1][j - 1] = min(f[i + 1][j - 1], f[i][j] + a[i + 1] + c[i + 1] + x[i + 1] + x[i + 1]);
if (j)
if ((e > i + 1) or (j > 1)) f[i + 1][j] = min(f[i + 1][j], f[i][j] + a[i + 1] + d[i + 1]);
if ((s > i + 1) or (j > 1)) f[i + 1][j] = min(f[i + 1][j], f[i][j] + c[i + 1] + b[i + 1]);
k = k + ((i + 1 == s) or (i + 1 == e));
cout << f[n][1] << \'\\n\';
嘻嘻
实用Trick
实用Trick
开栈(跑大样例必备)
-Wl,--stack=134217728手动开O2(能快一倍?)
#pragma GCC optimize(2)五十行优化(非必须)
#pragma GCC diagnostic error "-std=c++11"
#pragma GCC optimize(3)
#pragma GCC optimize("Ofast")
#pragma GCC optimize("inline")
#pragma GCC optimize("-fgcse")
#pragma GCC optimize("-fgcse-lm")
#pragma GCC optimize("-fipa-sra")
#pragma GCC optimize("-ftree-pre")
#pragma GCC optimize("-ftree-vrp")
#pragma GCC optimize("-fpeephole2")
#pragma GCC optimize("-ffast-math")
#pragma GCC optimize("-fsched-spec")
#pragma GCC optimize("unroll-loops")
#pragma GCC optimize("-falign-jumps")
#pragma GCC optimize("-falign-loops")
#pragma GCC optimize("-falign-labels")
#pragma GCC optimize("-fdevirtualize")
#pragma GCC optimize("-fcaller-saves")
#pragma GCC optimize("-fcrossjumping")
#pragma GCC optimize("-fthread-jumps")
#pragma GCC optimize("-funroll-loops")
#pragma GCC optimize("-fwhole-program")
#pragma GCC optimize("-freorder-blocks")
#pragma GCC optimize("-fschedule-insns")
#pragma GCC optimize("inline-functions")
#pragma GCC optimize("-ftree-tail-merge")
#pragma GCC optimize("-fschedule-insns2")
#pragma GCC optimize("-fstrict-aliasing")
#pragma GCC optimize("-fstrict-overflow")
#pragma GCC optimize("-falign-functions")
#pragma GCC optimize("-fcse-skip-blocks")
#pragma GCC optimize("-fcse-follow-jumps")
#pragma GCC optimize("-fsched-interblock")
#pragma GCC optimize("-fpartial-inlining")
#pragma GCC optimize("no-stack-protector")
#pragma GCC optimize("-freorder-functions")
#pragma GCC optimize("-findirect-inlining")
#pragma GCC optimize("-fhoist-adjacent-loads")
#pragma GCC optimize("-frerun-cse-after-loop")
#pragma GCC optimize("inline-small-functions")
#pragma GCC optimize("-finline-small-functions")
#pragma GCC optimize("-ftree-switch-conversion")
#pragma GCC optimize("-foptimize-sibling-calls")
#pragma GCC optimize("-fexpensive-optimizations")
#pragma GCC optimize("-funsafe-loop-optimizations")
#pragma GCC optimize("inline-functions-called-once")
#pragma GCC optimize("-fdelete-null-pointer-checks")快读快输(卡常必备)
template <typename T>
void read(T &x) {
x = 0; bool f = 0;
char c = getchar();
for (;!isdigit(c);c=getchar()) if (c=='-') f=1;
for (;isdigit(c);c=getchar()) x=x*10+(c^48);
if (f) x=-x;
}
template <typename T>
void write(T x) {
if (x < 0) putchar('-'), x = -x;
if (x >= 10) write(x / 10);
putchar('0' + x % 10);
}
超级快读快输
static char buf[100000],*p1=buf,*p2=buf;
#define gc p1==p2&&(p2=(p1=buf)+fread(buf,1,100000,stdin),p1==p2)?EOF:*p1++;
inline int read()
{
register int x=0;register char a=gc;
while(a<'0'||a>'9')a=gc;
while(a>='0'&&a<='9')x=(x<<1)+(x<<3)+(a^48),a=gc;
return x;
}
char pbuf[10000000] , *pp = pbuf;
inline void write(int x)
{
static int sta[35];
register int top = 0;
if(!x)sta[++top]=0;
while(x) sta[++top] = x % 10 , x /= 10;
while(top) *pp ++ = sta[top -- ] ^ '0';
}notice: 使用快速输出时应按照如下规则:
for (; t != ed; ++t) write(*t), *pp ++ = ' ';
fwrite(pbuf , 1 , pp - pbuf , stdout);以上是关于Trick的主要内容，如果未能解决你的问题，请参考以下文章
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