T1 网络连接

• 高深的sscanf$and$sprintf

sscanf是以字符串作为输入源

sscanf(str, "%d.%lld:%c||%lf", &a, &b, &c, &d);
这里的a是int, b是long long, c是char, d是double，他会返回所读取到的元素个数
同时自动忽略.:||

sprintf则是送回字符串中

#include <unordered_map>
#include <algorithm>
#include <iostream>
#include <cstring>
#include <stdio.h>
#include <cstdio>

using namespace std;

int n;
unordered_map<string, int> h;

bool check(string ip){
    int t[] = {0, -1, -1, -1, -1, -1};
    int cnt = sscanf(ip.c_str(), "%d.%d.%d.%d:%d", &t[1], &t[2], &t[3], &t[4], &t[5]);

    if (cnt != 5) return false;

    for (int i = 1; i <= 4; i ++ )
        if (t[i] < 0 || t[i] > 255)
            return false;
    if (t[5] < 0 || t[5] > 65535) return false;

    char now[100];
    sprintf(now, "%d.%d.%d.%d:%d", t[1], t[2], t[3], t[4], t[5]);

    return now == ip; 
}

int main(){

    scanf("%d", &n);
    for (int i = 1; i <= n; i ++ ){
        string op, ad;
        cin >> op >> ad;
        if (!check(ad)) puts("ERR");
        else {
            if (op == "Server"){
                if (!h[ad]) puts("OK"), h[ad] = i;
                else puts("FAIL");
            } else {
                if (!h[ad]) puts("FAIL");
                else printf("%d\n", h[ad]);
            }
        }
    }

    return 0;
}

T3 Beingawesomeism

一般像这种大模拟思路简单，但是代码量极大
luogu的CF

其实样例给的很良心，所以我们从样例入手

• 其实我们可以发现，最多的步数就是 4 ，因为上下左右肯定就把这个矩阵覆盖掉了

$步数为0$：全是A

AAA
AAA
AAA

$步数为1$：有一条完整的棱

PPP
PPP
AAA

$步数为2$：角上有A或者中间有完整的边

APPP
PPPP
APAP
PPPP

或者

PPPP
AAAA
PPPP
PPPP

$步数为3$：棱上有A

PPPP
APPP
PPPP
PPPP

$步数为4$：中间有点

PPPP
PPAP
PAPP
PPPP

$无解$：根本没有A

PPPP
PPPP
PPPP
PPPP

理清思路后，代码如下

#include <iostream>
#include <algorithm>
#include <cstring>

using namespace std;

const int N = 65;

int T, n, m;
char s[N][N];
bool allA;

bool find_len(){ // 有一条完整的棱
    allA = true;
    for (int j = 0; j < m; j ++ )
        if (s[0][j] != 'A')
            allA = false;
    if (allA) return true;
    allA = true;
    for (int j = 0; j < m; j ++ )
        if (s[n - 1][j] != 'A')
            allA = false;
    if (allA) return true;
    allA = true;
    for (int i = 0; i < n; i ++ )
        if (s[i][0] != 'A')
            allA = false;
    if (allA) return true;
    allA = true;
    for (int i = 0; i < n; i ++ )
        if (s[i][m - 1] != 'A')
            allA = false;
    if (allA) return true;
    return false;
}

// 角上是否有A
bool angle(){return s[0][0] == 'A' || s[0][m - 1] == 'A' || s[n - 1][0] == 'A' || s[n - 1][m - 1] == 'A';}

bool find_ct(){ // 棱上有A
    allA = false;
    for (int j = 1; j < m - 1; j ++ )
        if (s[0][j] == 'A')
            allA = true;
    if (allA) return 3;
    allA = false;
    for (int i = 1; i < n - 1; i ++ )
        if (s[i][0] == 'A')
            allA = true;
    if (allA) return 3;
    allA = false;
    for (int i = 1; i < n - 1; i ++ )
        if (s[i][m - 1] == 'A')
            allA = true;
    if (allA) return 3;
    allA = false;
    for (int j = 1; j < m - 1; j ++ )
        if (s[n - 1][j] == 'A')
            allA = true;
    if (allA) return 3;
    return false;
}

bool find_smlen(){ // 有完整的边
    for (int i = 1; i < n - 1; i ++ ){
        bool allA = true;
        for (int j = 0; j < m; j ++ )
            if (s[i][j] != 'A')
                allA = false;
        if (allA) return true;
    }
    for (int j = 1; j < m - 1; j ++ ){
        bool allA = true;
        for (int i = 0; i < n; i ++ )
            if (s[i][j] != 'A')
                allA = false;
        if (allA) return true;
    }
    return false;
}

inline bool allisA(){ // 全是A
    allA = true;
    for (int i = 0; i < n; i ++ )
        for (int j = 0; j < m; j ++ )
            if (s[i][j] != 'A')
                allA = false;
    return allA;
}

int step(){
    if (allisA()) return 0; 
    if (find_len()) return 1;
    if (find_smlen() || angle()) return 2;
    if (find_ct()) return 3;
    return 4; // 中间有点
}

int main(){

    scanf("%d", &T);
    while (T -- ){
        scanf("%d%d", &n, &m);
        bool havA = false;
        for (int i = 0; i < n; i ++ ){
            scanf("%s", s[i]);
            if (havA) continue;
            for (int j = 0; j < m; j ++ ) if (s[i][j] == 'A') havA = true;
        }

        if (!havA) puts("MORTAL");
        else printf("%d\n", step());
    }

    return 0;
}

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