【 2013 Multi-University Training Contest 5 】

HDU 4647 Another Graph Game

如果没有边的作用，显然轮流拿当前的最大值即可。

加上边的作用，将边权平均分给两个点，如果一个人选走一条边的两个点，就获得了边的权值；如果分别被两个人拿走，两人的差值不变。

将边权平均分配给点，对点的权值排序轮流选择。

#include<cstdio>
#include<algorithm>
#include<iostream>
#define MAXN 100010
#define EPS 1e-8
typedef long long LL;
using namespace std;
double arr[MAXN];
int main() {
    int n, m;
    int i;
    int x, y, val;
    double a, b;
    while (~scanf("%d%d", &n, &m)) {
        for (i = 1; i <= n; i++) {
            scanf("%lf", &arr[i]);
        }
        for (i = 0; i < m; i++) {
            scanf("%d%d%d", &x, &y, &val);
            arr[x] += val * 0.5 + EPS;
            arr[y] += val * 0.5 + EPS;
        }
        sort(arr + 1, arr + 1 + n);
        a = b = 0;
        for (i = n; i > 0; i -= 2) {
            a += arr[i];
            b += arr[i - 1];
        }
        cout << (LL) (a - b + EPS) << endl;
    }
    return 0;
}

---

HDU 4648 Magic Pen 6

求前缀和，枚举擦除的起点，求出最右端的位置，更新答案。

#include<cstdio>
#include<cstring>
#include<algorithm>
#define MAXN 100010
using namespace std;
int arr[MAXN];
int sum[MAXN];
int pos[MAXN];
int main() {
    int n, m;
    int i;
    int ans;
    while (~scanf("%d%d", &n, &m)) {
        sum[0] = 0;
        memset(pos, -1, sizeof(pos));
        for (i = 1; i <= n; i++) {
            scanf("%d", &arr[i]);
            arr[i] = (arr[i] % m + m) % m;
            sum[i] = (sum[i - 1] + arr[i]) % m;
            pos[sum[i]] = i;
        }
        ans = 0;
        for (i = 1; i <= n; i++) {
            ans = max(ans, pos[sum[i - 1]] - i + 1);
        }
        printf("%d
", ans);
    }
    return 0;
}

---

HDU 4649 Professor Tian

dp[i][j][k]表示第i位，处理到第j个数，该位的结果为k的概率。

答案就是每一位的dp[i][n][1]*(1<<i)相加。

#include<cstdio>
#include<cstring>
#define MAXN 220
#define MAXM 20
double dp[MAXM][MAXN][2];
int arr[MAXN];
char str[MAXN];
double p[MAXN];
int main() {
    int n;
    int ca = 1;
    int i, j, k, l;
    int tmp;
    double ans;
    while (~scanf("%d", &n)) {
        for (i = 0; i <= n; i++) {
            scanf("%d", &arr[i]);
        }
        for (i = 1; i <= n; i++) {
            scanf(" %c", &str[i]);
        }
        for (i = 1; i <= n; i++) {
            scanf("%lf", &p[i]);
        }
        memset(dp, 0, sizeof(dp));
        for (i = 0; i < MAXM; i++) {
            if (arr[0] & (1 << i)) {
                dp[i][0][1] = 1;
            } else {
                dp[i][0][0] = 1;
            }
            for (j = 1; j <= n; j++) {
                if (arr[j] & (1 << i)) {
                    l = 1;
                } else {
                    l = 0;
                }
                for (k = 0; k < 2; k++) {
                    if (str[j] == '&') {
                        tmp = k & l;
                    } else if (str[j] == '|') {
                        tmp = k | l;
                    } else {
                        tmp = k ^ l;
                    }
                    dp[i][j][tmp] += dp[i][j - 1][k] * (1 - p[j]);
                    dp[i][j][k] += dp[i][j - 1][k] * p[j];
                }
            }
        }
        ans = 0;
        for (i = 0; i < MAXM; i++) {
            ans += dp[i][n][1] * (1 << i);
        }
        printf("Case %d:
%lf
", ca++, ans);
    }
    return 0;
}