刷题笔记day4

Dikjstra算法

//
#include<limits.h>
const int INF = INT_MAX;
const int maxv = 1000;

int n,g[maxv][maxv];
int d[maxv];
bool vis[maxv] = {false};


void dij(int s) //s为起点
{
    int i;
    for(i=0;i<maxv;i++)
        d[i] = INF;
    d[s] = 0;
    for(i=0;i<n;i++)
    {
        int u = -1;
        int MIN = INF;

        for(int j=0;j<n;j++)
        {
            if(vis[j] == false && d[j] < MIN)
            {
                u = j;
                MIN = d[j];
            }
        }

        if(u == -1) return;
        vis[u] = true;
        for(int v = 0;v<n;v++)
        {
            if(vis[v] == false && g[u][v] != INF && d[u] + g[u][v] < d[u])
            {
                d[v] = d[u] + g[u][v];
            }
        }
    }
}

【PAT A1003 】Emergency （要复习）

使用多标尺的Dikjstra算法，求最短路和最大医疗资源。

增加了双向路、带权路、求最小路径数的情况。

#include<stdio.h>
#include<limits.h>
int rescue[505];
int r[505] = {0};
int g[505][505];
int d[505];
int vis[505] = {0};
int num[505] = {0};

int main()
{
    int n,m,now,save;
    scanf("%d%d%d%d",&n,&m,&now,&save);
    
    int i;
    for(i=0;i<n;i++)
    {
        scanf("%d",&rescue[i]);
    }
    
    for(i=0;i<n;i++)
    {
        for(int k=0;k<n;k++)
            g[i][k] = INT_MAX;
    }
    int c1,c2;
    for(i=0;i<m;i++)
    {
        scanf("%d%d",&c1,&c2);
        scanf("%d",&g[c1][c2]);
        g[c2][c1] = g[c1][c2];
    }

    for(i=0;i<n;i++)
    {
        d[i] = INT_MAX;
    }
    
    d[now] = 0;
    r[now] = rescue[now];
    num[now] = 1;


    for(int q=0;q<n;q++)
    {
        int u = -1;
        int min = INT_MAX;
        for(i=0;i<n;i++)
        {
            if(d[i] < min && vis[i] == 0)
            {
                u = i;
                min = d[i];
            }
        }

        vis[u] = 1;

        for(i=0;i<n;i++)
        {
            if(vis[i] == 0 && g[u][i] != INT_MAX)
            {
                if(d[u]+g[u][i] < d[i])
                {
                    d[i] = d[u] + g[u][i];
                    r[i] = r[u] + rescue[i]; 
                    num[i] = num[u];
                }
                else if(d[u]+g[u][i] == d[i])
                {
                    if(r[u] + rescue[i] > r[i])
                    {
                        r[i] = r[u] + rescue[i];
                    }
                    num[i] += num[u];
                }
            }
        }
    }

    printf("%d %d\n",num[save],r[save]);
    return 0;
}

【PAT A1005】 Spell It Right

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

char str[105];
char str2[105];
char num[10][10] = {"zero","one","two","three","four","five","six","seven","eight","nine"};
int main()
{
    scanf("%s",str);
    int len = strlen(str);
    int ans = 0;
    for(int i=0;i<len;i++)
    {
        ans  = ans + str[i] - '0';
    }

    sprintf(str2,"%d",ans);

    int len2 = strlen(str2);

    for(int j=0;j<len2;j++)
    {
        if(j!=0)printf(" ");
        
        printf("%s",num[str2[j]-'0']);
    }
}

【PAT A1035 】 Password 

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

int isp[1005] = {0};
char name[1005][20];
char pwd[1005][20];

int main()
{

    int n;
    scanf("%d",&n);
    for(int i=0;i<n;i++)
    {
        scanf("%s %s",name[i],pwd[i]);
        //printf("%s %s",name[i],pwd[i]);
    }
    
    if(n == 1 ||n == 0)
    {
        printf("There is 1 account and no account is modified\n");
        return 0;
    }
    
    int mod = 0;
    int ccount = 0;
    for(int j=0;j<n;j++)
    {
        int jlen = strlen(pwd[j]);

        for(int k=0;k<jlen;k++)
        {
            char c = pwd[j][k];
            if(c == '1' || c == '0'||c == 'l' ||c == 'O')
            {
                mod = 1;
                isp[j] = 1;
            }

            if(c == '1')
            {
                pwd[j][k] = '@';
            }
            else if(c == '0')
            {
                pwd[j][k] = '%';
            }
            else if(c == 'l')
            {
                pwd[j][k] = 'L';
            }
            else if(c == 'O')
            {
                pwd[j][k] = 'o';
            }
        }
        if(isp[j] == 1) ccount++;
    }

    if(mod == 0)
    {
        printf("There are %d accounts and no account is modified\n",n);
    }
    else
    {
        printf("%d\n",ccount);
        int first = 1;
        for(int q=0;q<n;q++)
        {
            if(isp[q] == 1)
            {
                if(first == 1) first = 0;
                    else printf("\n");
                printf("%s %s",name[q],pwd[q]);
            }
                
        }
    }
}

【PAT B1001】 害死人不偿命的(3n+1)猜想

水一题。。

#include<stdio.h>

int main()
{
    int n;
    scanf("%d",&n);
    int ccount = 0;
    while(n!=1)
    {
        if(n%2==0) n /= 2;
        else n = (3*n+1)/2;
        ccount++;
    }
    printf("%d\n",ccount);
}

【PAT B1022】 D进制的A+B 

要注意当a+b等于0时，因为使用了另外的数组进行存储结果，需要特别讨论。

#include<stdio.h>

int main()
{
    int a,b,d;
    scanf("%d%d%d",&a,&b,&d);
    int c = a+b;
    int ans[100];
    int ccount = 0;
    
    if(c == 0) {printf("0\n");return 0;}
    while(c)
    {
        ans[ccount++] = c % d;
        c /= d;
    }

    for(int i=ccount-1;i>=0;i--)
        printf("%d",ans[i]);
    printf("\n");
}