cf #254 (Div. 2)

a题

#include<stdio.h>
#include<string.h>
char c[101][101];
int main()
{
    long n,m,i,j;
    scanf("%ld%ld",&n,&m);
    gets(c[0]);
    for(i=1;i<=n;i++)
        gets(c[i]);

    for(i=1;i<=n;i++)
        {
            for(j=0;j<m;j++)
                if(c[i][j]==‘-‘)
                    printf("-");
                else
                {
                    if((i%2)^(j%2))
                        printf("W");
                    else
                        printf("B");
                }
            printf("\n");
        }

    return 0;

}

b题 其实就是每个联通快的节点数减1，dfs，并查集都可以。

#include <iostream>
#include <stdio.h>
#include <math.h>
#include <string.h>

using namespace std;

int p[100];
int n, m;

int find(int x)
{
    int s;
    for(s = x; p[s] >= 0; s = p[s]);
    while(s != x)
    {
        int tmp = p[x];
        p[x] = s;
        x = tmp;
    }
    return s;
}

int unionSet(int a, int b)
{
    int t1 = find(a);
    int t2 = find(b);
    int tmp = p[t1] + p[t2];
    if(p[t1] > p[t2])
    {
        p[t1] = t2;
        p[t2] = tmp;
    }
    else
    {
        p[t2] = t1;
        p[t1] = tmp;
    }
}

int main()
{
    while(scanf("%d%d", &n, &m) != EOF)
    {
        int a, b;
        for(int i = 0; i <= n; i++)
        {
            p[i] = -1;
        }
        for(int i = 0; i < m; i++)
        {
            scanf("%d%d", &a, &b);
            int r1 = find(a);
            int r2 = find(b);
            if(r1 != r2)
            {
                unionSet(a, b);
            }
        }
        int mul = 0;
        for(int i = 1; i <= n; i++)
        {
            if(p[i] < 0)
            {
                mul -= (p[i]+1);
            }
        }
        __int64 ans = pow(2.0, mul);

        printf("%I64d\n", ans);
    }
    return 0;
}

c题 求最小密度子图，这个在当时学最大密度子图的时候想过，其实就是一条边两个点，因为再添加一条边之后的密度要小于这两部分中最大的那个。

#include <stdio.h>
int n,m;
double vval[550],eval[500000];
int main()
{
    double ans=0;
    int i,j,u,v;
    scanf("%d%d",&n,&m);
    for(i=1;i<=n;i++) scanf("%lf",&vval[i]);
    for(i=1;i<=m;i++)
    {
        scanf("%d%d%lf",&u,&v,&eval[i]);
        if((vval[u]+vval[v])/eval[i]>ans)
            ans=(vval[u]+vval[v])/eval[i];
    }
    printf("%.9lf\n",ans);
    return 0;
}

cf #254 (Div. 2)