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  • HDU 5821 Ball （贪心）

    Ball

    题目链接：

    http://acm.hdu.edu.cn/showproblem.php?pid=5821

    Description

    ZZX has a sequence of boxes numbered 1,2,...,n. Each box can contain at most one ball. You are given the initial configuration of the balls. For 1≤i≤n, if the i-th box is empty then a[i]=0, otherwise the i-th box contains exactly one ball, the color of which is a[i], a positive integer. Balls with the same color cannot be distinguished. He will perform m operations in order. At the i-th operation, he collects all the balls from boxes l[i],l[i]+1,...,r[i]-1,r[i], and then arbitrarily put them back to these boxes. (Note that each box should always contain at most one ball) He wants to change the configuration of the balls from a[1..n] to b[1..n] (given in the same format as a[1..n]), using these operations. Please tell him whether it is possible to achieve his goal.

    Input

    First line contains an integer t. Then t testcases follow. In each testcase: First line contains two integers n and m. Second line contains a[1],a[2],...,a[n]. Third line contains b[1],b[2],...,b[n]. Each of the next m lines contains two integers l[i],r[i]. 1<=n<=1000,0<=m<=1000, sum of n over all testcases <=2000, sum of m over all testcases <=2000. 0<=a[i],b[i]<=n. 1<=l[i]<=r[i]<=n.

    Output

    For each testcase, print "Yes" or "No" in a line.

    Sample Input

    5 4 1 0 0 1 1 0 1 1 1 1 4 4 1 0 0 1 1 0 0 2 2 1 4 4 2 1 0 0 0 0 0 0 1 1 3 3 4 4 2 1 0 0 0 0 0 0 1 3 4 1 3 5 2 1 1 2 2 0 2 2 1 1 0 1 3 2 4

    Sample Output

    No No Yes No Yes

    Source

    2016 Multi-University Training Contest 8
    ##题意： 有N个盒子，每个盒子最多装一个球. 球的颜色不一定相同. 现在要进行m次区间操作： 每次操作 [l, r] 后可以随意将区间内的球重新分配回去. 问经过上述操作后是否有可能达到给定的状态.
    ##题解： 贪心. 为每个球标记它在最终结果中的序号. 对于颜色相同的球：左边的尽量分配小的序号. 对于m次区间操作，就将区间[l,r]中的球按最终序号排序. 每次排序都相当于让区间中的球向它们的最终位置更近一步. 最终再比较是否每个球都到位即可.
    官方题解： 假设有4个红球，初始时从左到右标为1,2,3,4。那么肯定存在一种方案，使得最后结束时红球的顺序没有改变，也是1,2,3,4。 那么就可以把同色球都写成若干个不同色球了。所以现在共有n个颜色互异的球。按照最终情况标上1,2，。。，n的序号，那么贪心的来每次操作就是把一个区间排序就行了。
    ##代码： ``` cpp #include #include #include #include #include #include #include #include #include #include #include #define LL long long #define eps 1e-8 #define maxn 1010 #define mod 100000007 #define inf 0x3f3f3f3f #define mid(a,b) ((a+b)>>1) #define IN freopen("in.txt","r",stdin); using namespace std;

    typedef pair<int,int> pii;
    pii ball[maxn];

    int main(int argc, char const *argv[])
    {
    //IN;

    int t; cin >> t;
while(t--)
{
    int n, m;
    scanf("%d %d", &n,&m);
    for(int i=1; i<=n; i++) {
        int x; scanf("%d", &x);
        ball[i] = make_pair(0, x);
    }
    for(int i=1; i<=n; i++) {
        int color; scanf("%d", &color);
        for(int j=1; j<=n; j++) {
            if(ball[j].second == color && !ball[j].first) {
                ball[j].first = i;
                break;
            }
        }
    }

    while(m--) {
        int l,r; scanf("%d %d", &l, &r);
        sort(ball+l, ball+r+1);
    }

    bool flag = 1;
    for(int i=1; i<=n; i++) {
        if(ball[i].first != i) {
            flag = 0; break;
        }
    }

    if(flag) puts("Yes");
    else puts("No");
}

return 0;

    }
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  • 原文地址：https://www.cnblogs.com/Sunshine-tcf/p/5762027.html
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