C++博客-purplest-随笔分类-图

hdu 4034 Graph

purplest — Mon, 12 Sep 2011 04:29:00 GMT

昨天成都网赛的题，我后面接队友手读的，看到数据规模N<=100，瞬间想到了floyd，想了下，果然是floyd的变种，敲完后，WA了俩次，然后AC。。。

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

int ncase, n;
int map[110][110];
int sign[110][110];

void floyd()
{
    int i, j, k;
    for ( k = 0 ; k < n ; k++ )
        for ( i = 0 ; i < n ; i++ )
            for ( j = 0 ; j < n ; j++ )
                if ( i != j && j!=k && i!=k )
                    if ( map[i][j] == map[i][k]+map[k][j] ) {
                        sign[i][j] = 1;
                    }
}

int judge()
{
    int i, j, k;
    for ( k = 0 ; k < n ; k++ )
        for ( i = 0 ; i < n ; i++ )
            for ( j = 0 ; j < n ; j++ )
                if ( map[i][j] > map[i][k]+map[k][j] )
                    return 1;
    return 0;
}

int main()
{
    scanf("%d", &ncase);
    int i, j, k;
    for ( k = 1 ; k <= ncase ; k++ )
    {
        scanf("%d", &n);
        for ( i = 0 ; i < n ; i++ )
            for ( j = 0 ; j < n ; j++ )
                scanf("%d", map[i]+j);
        printf("Case %d: ", k);
        memset(sign, 0, sizeof(sign));
        if (judge())
            printf("impossible\n");
        else
        {
            floyd();
            int num= 0;
            for ( i = 0 ; i < n ; i++ ) {
                for ( j = 0 ; j < n ; j++ ) {
                    if (!sign[i][j] && map[i][j]) {
                        num++;
                    }
                }
            }
            printf("%d\n", num);
        }
    }
    return 0;
}

purplest 2011-09-12 12:29 发表评论

poj 1778 All Discs Considered

purplest — Tue, 23 Aug 2011 16:47:00 GMT

拓扑排序

#include <cstdio>
#include <iostream>
#include <queue>
#include <vector>
#include <algorithm>

using namespace std;

const int MAX= 100100;

vector<int> num[MAX];
queue<int> que[2];
int n1, n2, d;
int dat[MAX], bak[MAX];

int line(int k)
{
    return k>n1?1:0;
}

int bfs( int k )
{
    int i;
    for ( i = 0 ; i < 2 ; i++ )
    {
        while ( !que[i].empty() )
            que[i].pop();
    }
    for ( i = 1 ; i <= n1+n2 ; i++ )
    {
        if ( !dat[i] )
            que[line(i)].push(i);
    }
    int count = 1 ;
    while ( !que[0].empty() || !que[1].empty() )
    {
        while(!que[k].empty())
        {
            int t= que[k].front();
            que[k].pop();
            for ( i = 0 ; i < num[t].size() ; i++ )
            {
                if ( !(--dat[num[t][i]]) )
                {
                    que[line(num[t][i])].push(num[t][i]);
                }
            }
        }
        k^=1;
        count++;
    }
    return count;
}

int main()
{
    while ( scanf("%d%d%d", &n1, &n2, &d) , n1 || n2 || d )
    {
        int i, x, y;
        for ( i = 1 ; i <= n1+n2 ; i++ )
        {
            num[i].clear();
            bak[i]= 0;
        }
        for ( i = 0 ; i < d ; i++ )
        {
            scanf("%d%d", &x, &y);
            num[y].push_back(x);
            bak[x]++;
        }
        for ( i = 1 ; i <= n1+n2 ; i++ )
            dat[i]=bak[i];
        x=bfs(0);
        for ( i = 1 ; i <= n1+n2 ; i++ )
            dat[i]=bak[i];
        y=bfs(1);
        printf("%d\n", x<y?x:y);
    }
    return 0;
}

purplest 2011-08-24 00:47 发表评论

poj 2253 Frogger

purplest — Tue, 16 Aug 2011 16:56:00 GMT

floyd的变种

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

int n, times= 1;
int data[300][3];
int num[300][300];

int dist(int x1, int y1, int x2, int y2)
{
    return (x1-x2)*(x1-x2)+(y1-y2)*(y1-y2);
}

int max(int a, int b)
{
    return a>b?a:b;
}

void floyd()
{
    for ( int k = 0 ; k < n ; k++ )
        for ( int i = 0 ; i < n ; i++ )
            for ( int j = 0 ; j < n ; j++ )
                if ( num[i][k] < num[i][j] && num[k][j] < num[i][j] )
                    num[i][j]= max(num[i][k], num[k][j]);
}

int main()
{
    while ( scanf("%d", &n), n )
    {
        int i, j;
        for ( i = 0 ; i < n ; i++ )
            scanf("%d%d", data[i], data[i]+1);
        for ( i = 0 ; i < n ; i++ )
            for ( j = 0 ; j < n ; j++ )
                num[i][j]= num[j][i]= dist(data[i][0], data[i][1], data[j][0], data[j][1]);
        floyd();
        printf("Scenario #%d\nFrog Distance = %.3f\n\n", times++, sqrt(num[0][1]));
    }
    return 0;
}

purplest 2011-08-17 00:56 发表评论