第三周 1.31-2.6

1.31

HDU 5617 Jam's maze

不会dp。

#include <iostream>
#include <cstdio>
#include <cstring>
using namespace std;
const int mod = 5201314;
int dp[555][555], cpy[555][555];
char G[555][555];

int main(void)
{
    int T;
    scanf("%d", &T);
    while(T--)
    {
        int N;
        scanf("%d", &N);
        for(int i = 0; i < N; i++) scanf("%s", G[i]);
        memset(dp, 0, sizeof(dp));
        dp[0][N-1] = G[0][0] == G[N-1][N-1];
        for(int k = 1; k < N; k++)
        {
            memcpy(cpy, dp, sizeof(cpy));
            memset(dp, 0, sizeof(dp));
            for(int x1 = 0; x1 <= k; x1++)
            {
                for(int x2 = N - 1; x2 >= N - k - 1; x2--)
                {
                    int y1 = k - x1, y2 = 2 * N - 2 - k - x2;
                    if(G[x1][y1] == G[x2][y2])
                    {
                        if(x1 != k && x2 != N - 1 - k) dp[x1][x2] += cpy[x1][x2];
                        if(x1 != k && x2 != N - 1) dp[x1][x2] += cpy[x1][x2+1];
                        if(x1 != 0 && x2 != N - 1 - k) dp[x1][x2] += cpy[x1-1][x2];
                        if(x1 != 0 && x2 != N - 1) dp[x1][x2] += cpy[x1-1][x2+1];
                        dp[x1][x2] %= mod;
                    }
                }
            }
        }
        int ans = 0;
        for(int i = 0; i < N; i++) ans = ( ans + dp[i][i] ) % mod;
        printf("%d
", ans);
    }
    return 0;
}

2.1

什么都没干。

2.2

CF 621 D Rat Kwesh and Cheese

精度之迷。

#include <iostream>
#include <cstdio>
#include <cmath>
using namespace std;
long double x, y, z;
char * s[] = {
    "",
    "x^y^z", "x^z^y", "(x^y)^z", "(x^z)^y",
    "y^x^z", "y^z^x", "(y^x)^z", "(y^z)^x",
    "z^x^y", "z^y^x", "(z^x)^y", "(z^y)^x"
};

long double f(int op)
{
    if(x < 1.001 && y < 1.001 && z < 1.001)
    {
        if(op == 1) return pow(y, z) * log(x);
        if(op == 2) return pow(z, y) * log(x);
        if(op == 3) return z * y * log(x);

        if(op == 5) return pow(x, z) * log(y);
        if(op == 6) return pow(z, x) * log(y);
        if(op == 7) return x * z * log(y);

        if(op == 9) return pow(x, y) * log(z);
        if(op == 10) return pow(y, x) * log(z);
        if(op == 11) return x * y * log(z);
    }

    if(op == 1) return log(log(x)) + z * log(y);
    if(op == 2) return log(log(x)) + y * log(z);
    if(op == 3) return log(z * y * log(x));

    if(op == 5) return log(log(y)) + z * log(x);
    if(op == 6) return log(log(y)) + x * log(z);
    if(op == 7) return log(z * x * log(y));

    if(op == 9) return log(log(z)) + y * log(x);
    if(op == 10) return log(log(z)) + x * log(y);
    if(op == 11) return log(x * y * log(z));

}

int main(void)
{
    cin >> x >> y >> z;
    long double M = -1e300;
    int pos;
    if(x < 1.001 && y < 1.001 && z < 1.001 || x > 0.999)
        for(int i = 1; i < 4; i++)
            if(f(i) > M) M = f(i), pos = i;
    if(x < 1.001 && y < 1.001 && z < 1.001 || y > 0.999)
        for(int i = 5; i < 8; i++)
            if(f(i) > M) M = f(i), pos = i;
    if(x < 1.001 && y < 1.001 && z < 1.001 || z > 0.999)
        for(int i = 9; i < 12; i++)
            if(f(i) > M) M = f(i), pos = i;
    printf("%s
", s[pos]);
    return 0;
}

2.3

CF 618 D Hamiltonian Spanning Tree

不是很懂你们贪心。

#include <iostream>
#include <cstdio>
using namespace std;
typedef long long LL;
const int maxn = 2e5 + 10;
int cnt, h[maxn], deg[maxn];
int num;

struct edge
{
    int to, pre;
} e[maxn<<1];

void add(int from, int to)
{
    cnt++;
    e[cnt].pre = h[from];
    e[cnt].to = to;
    h[from] = cnt;
}

int dfs(int p, int f)
{
    int tmp = 0;
    for(int i = h[p]; i; i = e[i].pre)
    {
        int to = e[i].to;
        if(to == f) continue;
        tmp += dfs(to, p);
    }
    if(!tmp) return 1;
    else if(tmp == 1){ num++; return 1;}
    else{ num += 2; return 0; }
}

int main(void)
{
    int n, leaf = 0;
    LL x, y;
    scanf("%d %I64d %I64d", &n, &x, &y);
    for(int i = 1; i < n; i++)
    {
        int u, v;
        scanf("%d %d", &u, &v);
        add(u, v), add(v, u);
        deg[u]++, deg[v]++;
    }
    for(int i = 1; i <= n; i++) if(deg[i] == 1) leaf++;
    if(x >= y) printf("%I64d
", y * (n - 2) + (leaf >= n - 1 ? x : y));
    else dfs(1, 0), printf("%I64d
", x * num + y * (n - 1 - num));
    return 0;
}

 2.4-2.5

什么都没干。

2.6

HDU 5623 KK's Number

这个出题人有点萌。

#include <iostream>
#include <cstdio>
#include <algorithm>
using namespace std;
typedef long long LL;
const int maxn = 5e4 + 10;
bool cmp(LL x, LL y){return x > y;}
LL a[maxn], dp[maxn];

int main(void)
{
    int T;
    scanf("%d", &T);
    while(T--)
    {
        int N;
        scanf("%d", &N);
        for(int i = 0; i < N; i++) scanf("%I64d", a + i);
        sort(a, a + N, cmp);
        LL M = a[N] = 0LL;
        for(int i = N - 1; i >= 0; i--)
        {
            dp[i] = a[i] - M;
            M = max(M, dp[i]);
        }
        printf("%I64d
", M);
    }
    return 0;
}