ACM/ICPC 之 Floyd范例两道(POJ2570-POJ2263)

两道以Floyd算法为解法的范例，第二题如果数据量较大，须采用其他解法

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POJ2570-Fiber Network

//经典的传递闭包问题，由于只有26个公司可以采用二进制存储
//Time:141Ms	Memory:328K
#include<iostream>
#include<cstring>
#include<cstdio>
using namespace std;

#define MAX 205
#define MAXS 28

int n;
int d[MAX][MAX];

void floyd()
{
	for (int k = 1; k <= n; k++)
		for (int i = 1; i <= n; i++)
			for (int j = 1; j <= n; j++)
				d[i][j] |= d[i][k] & d[k][j];	//找出两条线路的相同公司
}

int main()
{
	while (scanf("%d", &n), n)
	{
		memset(d, 0, sizeof(d));
		int a, b;
		char s[MAXS];
		while (scanf("%d%d", &a, &b), a && b)
		{
			scanf("%s", s);
			int len = strlen(s);
			for (int i = 0; i < len; i++)
				d[a][b] |= 1 << (s[i] - 'a');
		}

		floyd();
		while (scanf("%d%d", &a, &b), a && b)
		{
			if (d[a][b] == 0)	printf("-\n");
			else {
				int tmp = d[a][b];
				for (int i = 0; i < 26; i++, tmp >>= 1)
					if (tmp & 1) putchar('a' + i);	//单个字符输出putchar较快
				printf("\n");
			}
		}
		printf("\n");
	}
	return 0;
}

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POJ2263-Heavy Cargo

//求图中起点到终点的公路最大承载量
//Time:32Ms	Memory:500K
#include<iostream>
#include<cstring>
#include<cstdio>
#include<algorithm>
using namespace std;

#define MAX 205
#define MAXS 32

struct City {
	char s[MAXS];
}c[MAX];

int n, m;
int lc;
int board[MAX][MAX];
int d[MAX][MAX];

int find(char s[MAXS])
{
	for (int i = 0; i < lc; i++)
		if (!strcmp(s, c[i].s))	return i;
	return -1;
}

void floyd()
{
	memcpy(d, board, sizeof(board));

	for (int k = 0; k < lc; k++)
		for (int i = 0; i < lc; i++)
			for (int j = 0; j < lc; j++)
				d[i][j] = max(d[i][j], min(d[i][k], d[k][j]));
}

int main()
{
	int cas = 0;
	while (scanf("%d%d", &n, &m), n && m)
	{
		int dis; lc = 0;
		memset(board, -1, sizeof(board));
		for (int i = 0; i < m; i++)
		{
			scanf("%s%s%d", c[lc].s, c[lc+1].s, &dis);
			int n1 = find(c[lc].s);
			int n2 = find(c[lc + 1].s);
			if (n1 == -1) {
				n1 = lc++;
				if (n2 == -1) n2 = lc++;
			}
			else if (n2 == -1)
				c[n2 = lc++] = c[lc + 1];
			board[n1][n2] = board[n2][n1] = dis;
		}

		char s1[MAXS], s2[MAXS];
		scanf("%s%s", s1, s2);
		int n1 = find(s1);
		int n2 = find(s2);

		floyd();

		printf("Scenario #%d\n", ++cas);
		printf("%d tons\n\n", d[n1][n2]);

	}

	return 0;
}