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  • HDU2819(二分图匹配,记录过程)

    Swap

    Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
    Total Submission(s): 2236    Accepted Submission(s): 801
    Special Judge


    Problem Description
    Given an N*N matrix with each entry equal to 0 or 1. You can swap any two rows or any two columns. Can you find a way to make all the diagonal entries equal to 1?
     
    Input
    There are several test cases in the input. The first line of each test case is an integer N (1 <= N <= 100). Then N lines follow, each contains N numbers (0 or 1), separating by space, indicating the N*N matrix.
     
    Output
    For each test case, the first line contain the number of swaps M. Then M lines follow, whose format is “R a b” or “C a b”, indicating swapping the row a and row b, or swapping the column a and column b. (1 <= a, b <= N). Any correct answer will be accepted, but M should be more than 1000.

    If it is impossible to make all the diagonal entries equal to 1, output only one one containing “-1”. 
     
    Sample Input
    2
    0 1
    1 0
    2
    1 0
    1 0
     
    Sample Output
    1
    R 1 2
    -1
     
    Source
     
    很好的一道题,值得深思。
    题意:可交换任意两行或任意两列,最终是主对角线上全为1,输出交换过程
    如果可行的话,一定可以只交换列或只交换行得到,因为不管怎么交换,原先在同一行的始终在同一行,原先在同一列的始终在同一列。想到这,构图也就出来了,对行和列构图,如果mp[i][j]==1则在i和j之间加边
    /*
    ID: LinKArftc
    PROG: 2819.cpp
    LANG: C++
    */
    
    #include <map>
    #include <set>
    #include <cmath>
    #include <stack>
    #include <queue>
    #include <vector>
    #include <cstdio>
    #include <string>
    #include <utility>
    #include <cstdlib>
    #include <cstring>
    #include <iostream>
    #include <algorithm>
    using namespace std;
    #define eps 1e-8
    #define randin srand((unsigned int)time(NULL))
    #define input freopen("input.txt","r",stdin)
    #define debug(s) cout << "s = " << s << endl;
    #define outstars cout << "*************" << endl;
    const double PI = acos(-1.0);
    const double e = exp(1.0);
    const int inf = 0x3f3f3f3f;
    const int INF = 0x7fffffff;
    typedef long long ll;
    
    const int maxn = 110;
    
    int mp[maxn][maxn];
    int linker[maxn];
    bool vis[maxn];
    int n;
    
    bool dfs(int u) {
        for (int v = 1; v <= n; v ++) {
            if (!vis[v] && mp[u][v]) {
                vis[v] = true;
                if (linker[v] == -1 || dfs(linker[v])) {
                    linker[v] = u;
                    return true;
                }
            }
        }
        return false;
    }
    
    int hungry() {
        memset(linker, -1, sizeof(linker));
        int ret = 0;
        for (int i = 1; i <= n; i ++) {
            memset(vis, 0, sizeof(vis));
            if (dfs(i)) ret ++;
        }
        return ret;
    }
    
    int a[maxn], b[maxn];
    
    int main() {
        //input;
        while (~scanf("%d", &n)) {
            int tmp;
            memset(mp, 0, sizeof(mp));
            for (int i = 1; i <= n; i ++) {
                for (int j = 1; j <= n; j ++) {
                    scanf("%d", &tmp);
                    if (tmp) mp[i][j] = 1;
                }
            }
            int ans = hungry();
            if (ans < n) printf("-1
    ");
            else {
                int cnt = 0;
                for (int i = 1; i <= n; i ++) {
                    int j;
                    for (j = i; j <= n; j ++) {
                        if (linker[j] == i) break;
                    }
                    if (j != i) {
                        cnt ++;
                        a[cnt] = i; b[cnt] = j;
                        swap(linker[i], linker[j]);
                    }
                }
                printf("%d
    ", cnt);
                for (int i = 1; i <= cnt; i ++) printf("C %d %d
    ", a[i], b[i]);
            }
        }
    
        return 0;
    }
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  • 原文地址:https://www.cnblogs.com/LinKArftc/p/4908965.html
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