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  • HDU 5583 Kingdom of Black and White 暴力

    Kingdom of Black and White

    Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
    Total Submission(s): 1890    Accepted Submission(s): 588


    Problem Description
    In the Kingdom of Black and White (KBW), there are two kinds of frogs: black frog and white frog.

    Now N frogs are standing in a line, some of them are black, the others are white. The total strength of those frogs are calculated by dividing the line into minimum parts, each part should still be continuous, and can only contain one kind of frog. Then the strength is the sum of the squared length for each part.

    However, an old, evil witch comes, and tells the frogs that she will change the color of at most one frog and thus the strength of those frogs might change.

    The frogs wonder the maximum possible strength after the witch finishes her job.
     
    Input
    First line contains an integer T, which indicates the number of test cases.

    Every test case only contains a string with length N, including only 0 (representing
    a black frog) and 1 (representing a white frog).

     1T50.

     for 60% data, 1N1000.

     for 100% data, 1N105.

     the string only contains 0 and 1.
     
    Output
    For every test case, you should output "Case #x: y",where x indicates the case number and counts from 1 and y is the answer.
     
    Sample Input
    2
    000011
    0101
     
    Sample Output
    Case #1: 26
    Case #2: 10
     
    Source
     
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    #include <iostream>
    #include <cstring>
    #include <cstdio>
    #include <algorithm>
    #include <queue>
    #include <vector>
    #include <iomanip>
    #include <math.h>
    #include <map>
    using namespace std;
    #define FIN     freopen("input.txt","r",stdin);
    #define FOUT    freopen("output.txt","w",stdout);
    #define INF     0x3f3f3f3f
    #define INFLL   0x3f3f3f3f3f3f3f
    #define lson    l,m,rt<<1
    #define rson    m+1,r,rt<<1|1
    typedef long long LL;
    typedef pair<int,int> PII;
    
    const int MAXN = 1e5 + 5;
    int T;
    char a[MAXN];
    LL dp[MAXN];
    LL b[MAXN];
    
    int main()
    {
        //FIN
        scanf("%d", &T);
        int cas = 1;
        while(T--) {
            scanf("%s", a);
            int lena = strlen(a);
            dp[0] = 1;
            LL sum = 0;
            int t = 0;
            LL mx = 0;
            memset(b, 0, sizeof(b));
            for(int i = 1; i < lena; i++) {
                if(a[i] == a[i - 1]) dp[i] = dp[i - 1] + 1;
                else {
                    mx = max(dp[i - 1], mx);
                    b[t++] = dp[i - 1];
                    sum += LL(dp[i - 1] * dp[i - 1]);
                    dp[i] = 1;
                }
            }
    
            b[t] = dp[lena - 1];
            mx = max(b[t], mx);
            sum += LL(b[t] * b[t]);
            LL ans = sum;
    
            if(b[0] == lena) {
                ans = sum;
                printf("Case #%d: %lld
    ", cas++, ans);
                continue;
            }
    
            LL tt;
    
            //cout << "mx=" << mx << endl;
    
            for(int i = 0; i <= t; i++) {
                if(b[i] == mx) {
                    //cout << "i=" << i << endl;
                    if(i == 0) tt = b[1];
                    else if(i == t) tt = b[t - 1];
                    else tt = min(b[i - 1], b[i + 1]);
                    LL now = sum + (mx + 1) * (mx + 1) - (mx * mx) + (tt - 1) * (tt -1) - tt * tt;
                    ans = max(ans, now);
                }
            }
    
            for(int i = 0; i <= t; i++) {
                if(b[i] == 1) {
                    LL now = sum - b[i - 1] * b[i - 1] - b[i + 1] * b[i + 1] - 1 + (b[i + 1] + 1 + b[i - 1]) * (b[i + 1] + 1 + b[i - 1]);
                    ans = max(ans, now);
                }
            }
    
            printf("Case #%d: %lld
    ", cas++, ans);
        }
        return 0;
    }
    

      

     
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  • 原文地址:https://www.cnblogs.com/Hyouka/p/5905675.html
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