树上对抗搜索 - 树形dp

Alice and Bob are going on a trip. Alice is a lazy girl who wants to minimize the total travelling distance, while Bob as an active boy wants to maximize it. At the same time, they cannot let the value to be less than a given integer L since that will make them miss too much pleasure, and they cannot let the value to be greater than a given integer R since they don't want to get too exhausted.
The city they are visiting has n spots and the spots are connected by directed edges. The spots are connected in such a way that they form a tree and the root will always be at spot 0. They take turns to select which edge to go. Both of them choose optimally. Bob will go first.

InputThere are multiple test cases. For every test case, the first line has three integers, n, L and R (1<=n<=500000, 0<=L, R<=1000000000). The next n-1 lines each has three integers a, b and c, indicating that there is an edge going from spot a to spot b with length c (1<=c<=1000). The spots are labeled from 0 to n-1.
There is a blank line after each test case.
Proceed to the end of file.
OutputIf the total distance is not within the range [L, R], print "Oh, my god!" on a single line. Otherwise, print the most value Bob can get.Sample Input

Sample Output

Oh, my god!
2
6
2

题意：给你一棵树，以及树上的边权，两人轮流取走，先走的总是希望总路程最大，后走的人则希望总路程最小，但要求总的范围在（l, r）

思路分析：

　　思路比较好想，类似数字三角形，只不过这是在树上，从叶子向上去递推，明确在每层是谁去走，去怎么更新当前的结点，但是我的代码超时了...

待更新

代码示例：

const int maxn = 5e5+5;

void in(int &a)
{
    char c;
    while ((c = getchar()) == ' ' || c == '\n');
    a = c - 48;
    while ((c = getchar()) >= '0' && c <= '9')
        a = (a << 3) + (a << 1) + c - 48;
}
int n, l, r;
struct node
{
    int to, cost;
    
    node(int _to=0, int _cost=0):to(_to), cost(_cost){}
};
vector<node>ve[maxn];
int dis[maxn];
int dp[maxn];

void dfs(int x, int fa, int pt){
    if (dis[x] > r) {dp[x]=-1; return;}
    if (ve[x].size() == 1) dp[x] = 0;
    else dp[x] = pt?-1:inf;
    
    for(int i = 0; i < ve[x].size(); i++){
        int to = ve[x][i].to;
        int cost = ve[x][i].cost;
        if (to == fa) continue;
        dis[to] = dis[x]+cost;
        
        dfs(to, x, !pt);
        if (dp[to] == -1 || dp[to] == inf) continue;    
        if (dp[to]+dis[x]+cost >= l && dp[to]+dis[x]+cost <= r){ 
            if (pt == 1) dp[x] = max(dp[x], dp[to]+cost);
            else dp[x] = min(dp[x], dp[to]+cost);
        }
    }
    //printf("++++ x = %d ,pt = %d , dis = %d,  %d\n", x, pt, dis[x], dp[x]);
}

int main() {
    //freopen("in.txt", "r", stdin);
    //freopen("out.txt", "w", stdout);
    int a, b, c;
    
   while(~scanf("%d%d%d", &n, &l, &r)){
        for(int i = 0; i <= n; i++) ve[i].clear();
        for(int i = 1; i < n; i++){
            in(a), in(b), in(c);  
            ve[a].push_back(node(b, c));
            ve[b].push_back(node(a, c));            
        }
        //memset(dis, 0, sizeof(dis)); 
        //memset(dp, 0, sizeof(dp));
        dfs(0, -1, 1);
        //printf("+++ %d \n", dp[0]);
        if (dp[0] >= l && dp[0] <= r) printf("%d\n", dp[0]);
        else printf("Oh, my god!\n");
    }
    return 0;
}

徐州一道网络赛题

In a world where ordinary people cannot reach, a boy named "Koutarou" and a girl named "Sena" are playing a video game. The game system of this video game is quite unique: in the process of playing this game, you need to constantly face the choice, each time you choose the game will provide

That is, if there are three options in a selection, the score will be increased by

After all the choices have been made, the score will affect the ending of the game. If the score is greater than or equal to a certain value

Koutarou and Sena are playing very happy, but I believe you have seen through the final ending. Now give you the initial score, the

Input

The first line contains four integers $n,m,k,ln,m,k,ln,m,k,l（1≤n≤10001\le n \le 10001≤n≤1000, −100≤m≤100-100 \le m \le 100−100≤m≤100 , −100≤l<k≤100-100 \le l < k \le 100−100≤l<k≤100 ）, represents the number of choices, the initial score, the minimum score required to enter a good ending, and the highest score required to enter a bad ending, respectively.$

Each of the next $a,b,ca,b,ca,b,c（a≥0a\ge 0a≥0 , b≥0b\ge0b≥0 ,c=0c=0c=0 or c=1c=1c=1）,indicates the options that appear in this selection,in which a=0a=0a=0 means there is no option to increase the score in this selection, a>0a>0a>0 means there is an option in this selection to increase the score by aaa ; b=0b=0b=0 means there is no option to decrease the score in this selection, b>0b>0b>0 means there is an option in this selection to decrease the score by bbb; c=0c=0c=0 means there is no option to multiply the score by −1-1−1 in this selection , c=1c=1c=1 means there is exactly an option in this selection to multiply the score by −1-1−1. It is guaranteed that a,b,ca,b,ca,b,c are not equal to 000 at the same time.$

Output

One line contains the final ending of the game. If it will enter a good ending,print "Good Ending"(without quotes); if it will enter a bad ending,print "Bad Ending"(without quotes);otherwise print "Normal Ending"(without quotes).

样例输入1

样例输出1

Good Ending

样例输入2

样例输出2

Bad Ending

题意 ： 给你 n 组操作，一个初始分数 m ， 以及一个区间，最终分数大于区间右端点输出 good ，小于区间左端点输出 bad ，否则输出 normal ， 其实就是两个人在做游戏，每个人每次有三种操作可选，+a , -b, *-1 ,
　　先走的人希望最后的分数最大，后走的人希望最后的分数越小，问你最终的情况？
思路分析 ： 其实和上面的题目很类似，就是两个人轮流操作，对抗博弈，倒着推一遍就可以了，因为分数有负的，我们可以给其均加上 100 即可
　　　　定义 dp[i][j] 表示到第 i 行时,当前分数为 j 的最优解，
　　　　if (i&1) dp[i][j] = max(dp[i][j], dp[i+1][j+cost]);
　　　　else  dp[i][j] = min(dp[i][j], dp[i+1][j+cost]);
代码示例 ：

int n, m, k, l;
int arr[1005][3];
int dp[1005][205];

int main() {
    //freopen("in.txt", "r", stdin);
    //freopen("out.txt", "w", stdout);
    
    scanf("%d%d%d%d", &n, &m, &k, &l);
    for(int i = 1; i <= n; i++) scanf("%d%d%d", &arr[i][0], &arr[i][1], &arr[i][2]);
        
    for(int i = -100; i <= 200; i++) dp[n+1][i+100] = i;
    for(int i = n; i >= 1; i--){
        if (i&1) {
            for(int j = -100; j <= 100; j++){
                dp[i][j+100] = -100;
                if (arr[i][0]) dp[i][j+100] = max(dp[i][j+100], dp[i+1][min(200, j+arr[i][0]+100)]);
                if (arr[i][1]) dp[i][j+100] = max(dp[i][j+100], dp[i+1][max(0, j-arr[i][1]+100)]);
                if (arr[i][2]) dp[i][j+100] = max(dp[i][j+100], dp[i+1][-j+100]);
            }
        }
        else {
            for(int j = -100; j <= 100; j++){
                dp[i][j+100] = 100;
                if (arr[i][0]) dp[i][j+100] = min(dp[i][j+100], dp[i+1][min(200, j+arr[i][0]+100)]);
                if (arr[i][1]) dp[i][j+100] = min(dp[i][j+100], dp[i+1][max(0, j-arr[i][1]+100)]);
                if (arr[i][2]) dp[i][j+100] = min(dp[i][j+100], dp[i+1][-j+100]);
            }
        }
    }
    if (dp[1][min(200, m+100)] >= k) printf("Good Ending\n");
    else if (dp[1][min(200, m+100)] <= l) printf("Bad Ending\n");
    else printf("Normal Ending\n");
    return 0;
}

posted @ 2018-06-17 22:44 楼主好菜啊阅读(181) 评论(0) 编辑收藏举报

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