[Offer收割]编程练习赛57

A.1-偏差排列

dp？

#include <bits/stdc++.h>
using namespace std;
typedef long long LL;
LL f[55][3];

int main(){
    int N;
    scanf("%d", &N);
    f[1][0] = f[1][2] = 1;
    for(int i = 1; i < N; ++i){
        f[i+1][0] = f[i][0] + f[i][1];
        f[i+1][1] = f[i][2];
        f[i+1][2] = f[i][0] + f[i][1];
    }
    printf("%lld\n", f[N][0] + f[N][1]);
    return 0;
}

 

B.递增N元组

dp？

#include <bits/stdc++.h>
using namespace std;
typedef long long LL;
const LL mod = 1e9 + 7;
LL f[105][1000005];

int main(){
    int N, M, x;
    scanf("%d %d", &N, &M);
    for(int i = 0; i <= 100001; ++i) f[0][i] = 1;
    for(int i = 1; i <= N; ++i){
        for(int j = 1; j <= M; ++j) {
            scanf("%d", &x), x++;
            f[i][x] = (f[i][x] + f[i-1][x-1]) % mod;
        }
        for(int j = 1; j <= 100001; ++j) f[i][j] = (f[i][j] + f[i][j-1]) % mod;
    }
    printf("%lld\n", f[N][100001]);
    return 0;
}

 

C.逃离迷宫5

bfs？

#include <bits/stdc++.h>
using namespace std;
typedef pair<int, int> pii;
typedef pair<pii, int> piii;
const int INF = 1e9;
int G[1111][1111], f[1111][1111][2];
int dx[] = {-1, 0, 1, 0};
int dy[] = {0, 1, 0, -1};
char s[1111];
queue<piii> q;

int main(){
    int N;
    scanf("%d", &N);
    for(int i = 1; i <= N; ++i){
        scanf("%s", s + 1);
        for(int j = 1; j <= N; ++j) G[i][j] = s[j] == '#' ? 1 : 0, f[i][j][0] = f[i][j][1] = INF;
    }
    if(G[1][1]) q.push(piii(pii(1, 1), 1)), f[1][1][1] = 0;
    else q.push(piii(pii(1, 1), 0)), f[1][1][0] = 0;
    while(!q.empty()){
        piii now = q.front(); q.pop();
        int nx = now.first.first, ny = now.first.second, o = now.second;
        for(int i = 0; i < 4; ++i){
            int xx = nx + dx[i], yy = ny + dy[i];
            if(xx < 1 || xx > N || yy < 1 || yy > N) continue;
            if(G[xx][yy] && o) continue;
            int oo = o || G[xx][yy];
            if(f[xx][yy][oo] != INF) continue;
            f[xx][yy][oo] = f[nx][ny][o] + 1;
            q.push(piii(pii(xx, yy), oo));
        }
    }
    int ans = min(f[N][N][0], f[N][N][1]);
    printf("%d\n", ans == INF ? -1 : ans);
    return 0;
}

 

D.最大割集

……无语了

#include <bits/stdc++.h>
using namespace std;
typedef long long LL;
const int maxn = 1e5 + 10;
LL W[maxn], V[maxn], b[66];
int id[maxn];

bool cmp(int i, int j){
    return W[i] > W[j];
}

// Template
//清除前缀0，如果结果是空字符串则设为0
inline void clear(string& a){
    while(a.length()>0 && a[0]=='0')
        a.erase(0, 1);
    if(a == "")
        a = "0";
}

//如果a>=b则返回真（如果包含前缀零会被消除）
bool isBigger(string a, string b){
    clear(a);
    clear(b);
    if(a.length() > b.length())
        return true;
    if(a.length()==b.length() && a>=b)
        return true;
    return false;
}


//两个高精度正整数加法 a+b
string Add(string a, string b){
    //1、对位，将两个数补零直到其具有相同长度
    while(a.length() < b.length())
        a = '0' + a;
    while(a.length() > b.length())
        b = '0' + b;
    //2、补零，在开头再加一个0以便进位
    a = '0' + a;
    b = '0' + b;
    //3、从低位开始相加，注意进位
    for(int i=a.length()-1; i>=0; i--){
        a[i] = a[i] + b[i] - '0';
        if(a[i] > '9'){
            a[i] = a[i] - 10;
            a[i-1] += 1;
        }
    }
    clear(a);
    return a;
}

//两个高精度正整数减法 a-b
string Minus(string a, string b){
    bool aBigger = true;
    //1、对位，将两个数补零直到其具有相同长度
    while(a.length() < b.length())
        a = '0' + a;
    while(a.length() > b.length())
        b = '0' + b;
    //2、推测结果正负值，调整为前大后小
    if(a < b)
    {
        aBigger = false;
        string buf = b;
        b = a;
        a = buf;
    }
    //3、从低位开始相减，注意借位
    for(int i=a.length()-1; i>=0; i--){
        if(a[i] >= b[i]){
            a[i] = a[i] - (b[i] - '0');
        }else{
            a[i] = a[i] + 10;
            a[i-1] -= 1;
            a[i] = a[i] - (b[i] - '0');
        }
    }
    clear(a);
    if(!aBigger)
        a = '-' + a;
    return a;
}

template<typename T> string to(const T& t){
    ostringstream oss;  //创建一个格式化输出流
    oss<<t;             //把值传递如流中
    return oss.str();
}

int main(){
    int N, M;
    scanf("%d %d", &N, &M);
    string sum("0"), ans("0");
    for(int i = 1; i <= N; ++i) scanf("%lld", W + i), sum = Add(sum, to(W[i])), id[i] = i;
    sort(id + 1, id + 1 + N, cmp);
    for(int i = 1; i <= M; ++i){
        int u, v;
        LL w;
        scanf("%d %d %lld", &u, &v, &w);
        V[u] ^= w, V[v] ^= w;
    }
    for(int i = 1; i <= N; ++i){
        int x = id[i];
        for(int j = 0; j <= 60; ++j) {
            if(V[x] & (1LL << j)){
                if(b[j]) V[x] ^= b[j];
                else {b[j] = V[x], ans = Add(ans, to(W[x])); break;}
            }
        }
    }
    cout << Minus(Add(ans, ans), sum) << endl;
    return 0;
}

学习了py

N, M = list(map(int, raw_input().strip().split()))
W = list(map(int, raw_input().strip().split()))
sum = sum(W)
id = sorted([(W[i], i) for i in range(N)], reverse=True)
V = [0 for i in range(N)]
for i in range(M):
    u, v, w = list(map(int, raw_input().strip().split()))
    V[u-1] ^= w
    V[v-1] ^= w
b = [0 for i in range(61)]
ans = 0
for i in range(N):
    x = id[i][1]
    for j in range(61):
        if V[x] & (1 << j):
            if b[j]:
                V[x] ^= b[j]
            else:
                b[j] = V[x]
                ans += W[x]
                break
print(ans + ans - sum)