CSP2021 考前复习（所有模板）
作者：互联网
一、\(\rm dp\)
1. \(01\) 背包
#include <iostream>
#include <cstdio>
using namespace std;
const int MAXM = 105;
const int MAXT = 1005;
int v[MAXM], w[MAXM];
int dp[MAXT];
int main()
{
int t, m;
scanf("%d%d", &t, &m);
for (int i = 1; i <= m; i++)
{
scanf("%d%d", v + i, w + i);
}
for (int i = 1; i <= m; i++)
{
for (int j = t; j >= v[i]; j--)
{
dp[j] = max(dp[j], dp[j - v[i]] + w[i]);
}
}
printf("%d\n", dp[t]);
return 0;
}
\(2.\) 完全背包
tips：开long long。
#include <iostream>
#include <cstdio>
#define int long long
using namespace std;
const int MAXM = 1e4 + 5;
const int MAXT = 1e7 + 5;
int v[MAXM], w[MAXM];
int dp[MAXT];
signed main()
{
int t, m;
scanf("%lld%lld", &t, &m);
for (int i = 1; i <= m; i++)
{
scanf("%lld%lld", v + i, w + i);
}
for (int i = 1; i <= m; i++)
{
for (int j = v[i]; j <= t; j++)
{
dp[j] = max(dp[j], dp[j - v[i]] + w[i]);
}
}
printf("%lld\n", dp[t]);
return 0;
}
\(3.\) 多重背包（二进制优化）
#include <iostream>
#include <cstdio>
using namespace std;
const int MAXN = 1e5 + 5;
const int MAXT = 4e4 + 5;
int v[MAXN], w[MAXN];
int dp[MAXT];
int main()
{
int n, t, tot = 0;
scanf("%d%d", &n, &t);
for (int i = 1; i <= n; i++)
{
int a, b, c;
scanf("%d%d%d", &a, &b, &c);
for (int j = 1; j <= c; j <<= 1)
{
c -= j;
w[++tot] = a * j;
v[tot] = b * j;
}
if (c)
{
w[++tot] = a * c;
v[tot] = b * c;
}
}
for (int i = 1; i <= tot; i++)
{
for (int j = t; j >= v[i]; j--)
{
dp[j] = max(dp[j], dp[j - v[i]] + w[i]);
}
}
printf("%d\n", dp[t]);
return 0;
}
\(4.\) 区间 \(\rm dp\)（环形类问题）
#include <iostream>
#include <cstdio>
#include <cstring>
using namespace std;
const int MAXN = 205;
int a[MAXN], sum[MAXN], dp1[MAXN][MAXN], dp2[MAXN][MAXN];
int main()
{
memset(dp1, 0x3f, sizeof(dp1));
int n;
scanf("%d", &n);
for (int i = 1; i <= n; i++)
{
scanf("%d", a + i);
a[i + n] = a[i];
}
for (int i = 1; i <= (n << 1); i++)
{
sum[i] = sum[i - 1] + a[i];
dp1[i][i] = 0;
}
for (int len = 2; len <= n; len++)
{
for (int i = 1; i + len - 1 <= (n << 1); i++)
{
int j = i + len - 1;
dp1[i][j] = 0x3f3f3f3f;
for (int k = i; k < j; k++)
{
dp1[i][j] = min(dp1[i][j], dp1[i][k] + dp1[k + 1][j] + sum[j] - sum[i - 1]);
dp2[i][j] = max(dp2[i][j], dp2[i][k] + dp2[k + 1][j] + sum[j] - sum[i - 1]);
}
}
}
int ans1 = 0x3f3f3f3f, ans2 = 0;
for (int i = 1; i <= n; i++)
{
ans1 = min(ans1, dp1[i][i + n - 1]);
ans2 = max(ans2, dp2[i][i + n - 1]);
}
printf("%d\n%d\n", ans1, ans2);
return 0;
}
二、图论
1. 并查集（路径压缩+按秩合并）
#include <iostream>
#include <cstdio>
using namespace std;
const int MAXN = 1e4 + 5;
int n, m;
int fa[MAXN], dep[MAXN];
void init()
{
for (int i = 1; i <= n; i++)
{
fa[i] = i;
}
}
int find(int x)
{
if (x == fa[x])
{
return x;
}
return fa[x] = find(fa[x]);
}
void merge(int x, int y)
{
x = find(x), y = find(y);
if (x != y)
{
if (dep[x] > dep[y])
{
fa[y] = x;
}
else
{
fa[x] = y;
if (dep[x] == dep[y])
{
dep[y]++;
}
}
}
}
int main()
{
scanf("%d%d", &n, &m);
init();
while (m--)
{
int z, x, y;
scanf("%d%d%d", &z, &x, &y);
if (z == 1)
{
merge(x, y);
}
else
{
if (find(x) == find(y))
{
puts("Y");
}
else
{
puts("N");
}
}
}
return 0;
}
2. 最短路
\(I.\rm dijktra\)
#include <iostream>
#include <cstdio>
#include <queue>
#include <cstring>
using namespace std;
const int MAXN = 1e5 + 5;
const int MAXM = 2e5 + 5;
int cnt;
int head[MAXN];
struct edge
{
int to, dis, nxt;
}e[MAXM];
void add(int u, int v, int w)
{
e[++cnt] = edge{v, w, head[u]};
head[u] = cnt;
}
struct que
{
int pos, dis;
bool operator <(const que &x)const
{
return x.dis < dis;
}
};
int dis[MAXN];
bool vis[MAXN];
priority_queue<que> pq;
void dijkstra(int s)
{
memset(dis, 0x3f, sizeof(dis));
dis[s] = 0;
pq.push(que{s, 0});
while (!pq.empty())
{
int u = pq.top().pos;
pq.pop();
if (vis[u])
{
continue;
}
vis[u] = true;
for (int i = head[u]; i; i = e[i].nxt)
{
int v = e[i].to, w = e[i].dis;
if (dis[v] > dis[u] + w)
{
dis[v] = dis[u] + w;
pq.push(que{v, dis[v]});
}
}
}
}
int main()
{
int n, m, s;
scanf("%d%d%d", &n, &m, &s);
for (int i = 1; i <= m; i++)
{
int u, v, w;
scanf("%d%d%d", &u, &v, &w);
add(u, v, w);
}
dijkstra(s);
for (int i = 1; i <= n; i++)
{
printf("%d ", dis[i]);
}
return 0;
}
\(II.\rm spfa\)
#include <iostream>
#include <cstdio>
#include <cstring>
#include <queue>
using namespace std;
const int MAXN = 1e4 + 5;
const int MAXM = 5e5 + 5;
int cnt;
int head[MAXN];
struct edge
{
int to, dis, nxt;
}e[MAXM];
void add(int u, int v, int w)
{
e[++cnt] = edge{v, w, head[u]};
head[u] = cnt;
}
int dis[MAXN];
bool vis[MAXN];
queue<int> q;
void spfa(int s)
{
memset(dis, 0x3f, sizeof(dis));
dis[s] = 0;
vis[s] = true;
q.push(s);
while (!q.empty())
{
int u = q.front();
q.pop();
vis[u] = false;
for (int i = head[u]; i; i = e[i].nxt)
{
int v = e[i].to, w = e[i].dis;
if (dis[v] > dis[u] + w)
{
dis[v] = dis[u] + w;
if (!vis[v])
{
vis[v] = true;
q.push(v);
}
}
}
}
}
int main()
{
int n, m, s;
scanf("%d%d%d", &n, &m, &s);
for (int i = 1; i <= m; i++)
{
int u, v, w;
scanf("%d%d%d", &u, &v, &w);
add(u, v, w);
}
spfa(s);
for (int i = 1; i <= n; i++)
{
if (dis[i] == 0x3f3f3f3f)
{
printf("2147483647 ");
}
else
{
printf("%d ", dis[i]);
}
}
return 0;
}
标签：const,考前,CSP2021,int,MAXN,dp,include,模板,dis 来源： https://www.cnblogs.com/mangoworld/p/CSP2021-muban.html