BestCoder Round #42

标签：acm   bestcoder   

题目传送：BestCoder Round #42

1、Shaking hands

思路：简单题，ans = n * 2 + 矩阵上的1的个数

AC代码：

#include <cstdio>
#include <cstring>
#include <string>
#include <cstdlib>
#include <iostream>
#include <algorithm>
#include <cmath>
#include <queue>
#include <stack>
#include <vector>
#include <map>
#include <set>
#include <deque>
#include <cctype>
#define LL long long
#define INF 0x7fffffff
using namespace std;

int main() {
    int n;
    while(scanf("%d", &n) != EOF) {
        int ans = n * 2;
        int tmp = 0;
        for(int i = 0; i < n; i++) {
            for(int j = 0; j < n; j ++) {
                int t;
                scanf("%d", &t);
                tmp += t;
            }
        }
        printf("%d\n", ans + tmp);
    }
    return 0;
}

思路：map+queue模拟过去，存下鸟的位置，然而我比赛的代码TLE了，然后同样的代码在OJ上交过啦，也是醉了，这真的是太不爽了，希望以后BC能有所改进吧，毕竟给中国选手提供了一个比赛环境。

AC代码：

#include <cstdio>
#include <cstring>
#include <string>
#include <cstdlib>
#include <iostream>
#include <algorithm>
#include <cmath>
#include <queue>
#include <stack>
#include <vector>
#include <map>
#include <set>
#include <deque>
#include <cctype>
#define LL long long
#define INF 0x7fffffff
using namespace std;

inline int readint()
{
    char c = getchar();
    while(!isdigit(c)) c = getchar();
    
    int x = 0;
    while(isdigit(c))
    {
        x = x * 10 + c - '0';
        c = getchar();
    }
    return x;
}

int buf[10];    
inline void writeint(int i)
{
    int p = 0;
    if(i == 0) p++;    
    else while(i)
    {
        buf[p++] = i % 10;
        i /= 10;
    }
    for(int j = p-1; j >= 0; j--) putchar('0' + buf[j]);
}

int main() {
    int n, m;
    while(scanf("%d %d", &n, &m) != EOF) {
        map<int, queue<int> > mp;
        for(int i = 1; i <= n; i ++) {
            int t;
            t = readint();
            mp[t].push(i);
        }
        for(int i = 1; i <= m; i ++) {
            int t;
            t = readint();
            if(mp.find(t) == mp.end() || mp[t].empty()) {
                puts("-1");
            }
            else {
                writeint(mp[t].front());
                printf("\n");
                mp[t].pop();
            }
        }
    }
    return 0;
}

3、Happy birthday

思路：dp,dp[i][j][k]表示走到i和j这个位置容量是否可以为k

AC代码：

#include <cstdio>
#include <cstring>
#include <string>
#include <cstdlib>
#include <iostream>
#include <algorithm>
#include <cmath>
#include <queue>
#include <stack>
#include <vector>
#include <map>
#include <set>
#include <deque>
#include <cctype>
#define LL long long
#define INF 0x7fffffff
using namespace std;

int dp[105][105][105];

int a[105][105];

int main() {
    int n, m, k;
    while(scanf("%d %d %d", &n, &m, &k) != EOF) {
        memset(dp, 0, sizeof(dp));
        dp[0][0][0] = 1;
        for(int i = 1; i <= n; i ++) {
            dp[i][0][0] = 1;
        }
        for(int i = 1; i <= m; i ++) {
            dp[0][i][0] = 1;
        }
        for(int i = 1; i <= n; i ++) {
            for(int j = 1; j <= m; j ++) {
                scanf("%d", &a[i][j]);
                dp[i][j][0] = 1;
            }
        }
        
        for(int i = 1; i <= n; i ++) {
            for(int j = 1; j <= m; j ++) {
                for(int l = 0; l <= k; l ++) {
                    if(dp[i-1][j][l] == 1 || dp[i][j-1][l] == 1 
                    || (l >= a[i][j] && dp[i-1][j][l-a[i][j]] == 1)
                    || (l >= a[i][j] && dp[i][j-1][l-a[i][j]] == 1)) dp[i][j][l] = 1;
                }
            }
        }
        for(int t = k; t >= 0; t --) {
            if(dp[n][m][t] == 1) {
                printf("%d\n", t);
                break;
            }
        }
    }
    return 0;
}

BestCoder Round #42

标签：acm   bestcoder   

原文地址：http://blog.csdn.net/u014355480/article/details/45938187