SRM 451 DIV 1 总结

250p:这次是有史以来做的最快的一次250p。。。看题花了两分钟,敲代码最多一分钟。。。太明显了题意~


500p:这题水了。。。每次都这样。。。很显然用DP来做,不过前面状态表示有问题了。。。搞了好久还是错的。。


题意:给你一系列点,每个点都有一个硬币,刚开始从点(0, 0)开始走,每次只能按照以下规律走,第一次从(x, y)走到(x+k1, y+1),第二次从当前点(x, y)走到(x+k2, y+2),要保证k1 < k2 < ... < kn,问最多能收集几个硬币


解题思路:用dp[i][j]表示到了当前i这个点收集到j个硬币的最小步数,最小步数表示y-1走到当前y最少需要的步数,注意而不是前一个点走到当前点的步数。


250p:

class MagicalSource {
    public:
    long long calculate(long long x) ;

};

long long MagicalSource::calculate(long long x) {
    LL cur = 1;
    while(x >= cur*10+1) {
        cur = cur*10+1;
    }
    while(x%cur != 0) {
        cur /= 10;
    }
    return x/cur;
}

 

 

500p:

class BaronsAndCoins {
    public:
    int getMaximum(vector <int> x, vector <int> y)  ;


};

struct PP {
    int x, y;
    bool operator < (const PP &a) const {
        return y < a.y;
    }
}a[55];

int dp[55][55];

int cal(int n) {
    return  n*(n+1)/2;
}

int max(int a, int b) {
    return a > b ? a : b;
}

int BaronsAndCoins::getMaximum(vector <int> xx, vector <int> yy) {
    int len = xx.size();
    int i, j, k;
    for(i = 1;i <= len; i++) {
        a[i].x = xx[i-1];
        a[i].y = yy[i-1];
    }
    sort(a+1, a + len+1);
    memset(dp, -1, sizeof(dp));
    dp[0][0] = 0;
    a[0].x = 0, a[0].y = 0;
    for(i = 0;i <= len; i++) {
        for(j = i+1;j <= len; j++) {
            if(a[j].y == a[i].y || a[j].x <= a[i].x)    continue;
            for(k = 0;k <= len; k++) if(dp[i][k] != -1){
                int dis = a[j].x - a[i].x;
                int step = a[j].y - a[i].y;
                int now = dp[i][k]*step + cal(step);
                if(now > dis)   continue;
                int curk = step+dp[i][k];
                while(now < dis) {
                    now += step;
                    curk++;
                }
                if(dp[j][k+1] == -1)    dp[j][k+1] = curk;
                else if(dp[j][k+1] > curk)  dp[j][k+1] = curk;
            }
        }
    }
    int ans = 0;
    for(i = 0;i <= len; i++)
    for(j = 0;j <= len; j++) if(dp[i][j] != -1) {
        ans = max(ans, j);
    }
    return ans;
}

 

 

 

posted @ 2013-08-05 19:19  javawebsoa  Views(312)  Comments(0Edit  收藏  举报