ACM/ICPC 之 SPFA练习两道(ZOJ3088-ZOJ3103)

两道题都需要进行双向SPFA，比范例复杂，代码也较长，其中第二题应该可以用DFS或者BFS做，如果用DFS可能需要的剪枝较多。

//利用SPFA找出下降最长路径和上升最短路径，输出最大的比值和回路路径
//Time:0Ms	Memory:328K
#include<iostream>
#include<cstring>
#include<cstdio>
#include<algorithm>
#include<queue>
using namespace std;

#define MAX 1005
#define INF 0x3f3f3f3f

struct Edge {
	int u, w, next;
	Edge(){}
	Edge(int uu,int ww,int nn):u(uu),w(ww),next(nn){}
}eu[MAX], ed[MAX];	//up-down

int n, m, k;
int hu[MAX], hd[MAX];	//邻接表头位置
int pu[MAX], pd[MAX];	//路径
int du[MAX], dd[MAX];	//最短路长
int tu[MAX], td[MAX];	//临时路径
bool v[MAX];

//上升最短路径
void spfa_u(int x)
{
	memset(du, INF, sizeof(du));
	memset(pu, -1, sizeof(pu));
	memset(v, false, sizeof(v));
	du[x] = 0;
	queue<int> q;
	q.push(x);	pu[x] = x;
	while (!q.empty()) {
		int cur = q.front();
		q.pop();	v[cur] = false;
		for (int i = hu[cur]; i != -1; i = eu[i].next)
		{
			int u = eu[i].u, w = eu[i].w;
			if (du[u] > du[cur] + w)
			{
				du[u] = du[cur] + w;
				pu[u] = cur;
				if (!v[u]) {
					v[u] = true; q.push(u);
				}
			}
		}
	}
}

//SPFA-下降最长路径
void spfa_d(int x)
{
	memset(dd, -1, sizeof(dd));
	memset(pd, -1, sizeof(pd));
	memset(v, false, sizeof(v));
	dd[x] = 0;
	queue<int> q;
	q.push(x);	pd[x] = x;
	while (!q.empty()) {
		int cur = q.front();
		q.pop();	v[cur] = false;
		for (int i = hd[cur]; i != -1; i = ed[i].next)
		{
			int u = ed[i].u, w = ed[i].w;
			if (dd[u] < dd[cur] + w)
			{
				dd[u] = dd[cur] + w;
				pd[u] = cur;
				if (!v[u]) {
					v[u] = true; q.push(u);
				}
			}
		}
	}
}

void path_u(int x)
{
	if (tu[x] != x) path_u(tu[x]);
	printf("%d ", x);
}

void path_d(int x)
{
	int i;
	for (i = td[x]; i != td[i]; i = td[i])
		printf("%d ", i);
	printf("%d\n", i);
}

int main()
{
	int T;
	scanf("%d", &T);
	while (T--)
	{
		memset(hu, -1, sizeof(hu));
		memset(hd, -1, sizeof(hd));
		scanf("%d%d%d", &n, &m, &k);
		int a, b, w;
		for (int i = 0; i < m; i++)
		{
			scanf("%d%d%d", &a, &b, &w);
			ed[i] = Edge(a, w, hd[b]);	//反向建图
			hd[b] = i;
		}
		for (int i = 0; i < k; i++)
		{
			scanf("%d%d%d", &a, &b, &w);
			eu[i] = Edge(b, w, hu[a]);	//正向建图
			hu[a] = i;
		}

		double rate = 0;
		int tmp = -1;
		for (int i = 1; i <= n; i++)
		{
			spfa_u(i);	spfa_d(i);
			for (int j = 1; j <= n; j++)
			{
				if (i == j || du[j] == INF)	continue;
				if (rate < 1.0 * dd[j] / du[j]) {
					rate = 1.0 * dd[j] / du[j];
					tmp = j;
					memcpy(tu, pu, (n+1)*sizeof(int));
					memcpy(td, pd, (n+1)*sizeof(int));
				}
			}
		}
		path_u(tmp);	path_d(tmp);
		printf("%.3f\n", rate);
	}
	return 0;
}

//需要理清题意，比较复杂，建立双向邻接表，并须计算双向最短路
//第一次边表设大了MLE了...考虑最大边数 < n*18个，因此设为MAX*18
//Time:190Ms	Memory:1152K
#include<iostream>
#include<cstring>
#include<cstdio>
#include<algorithm>
#include<queue>
using namespace std;

#define MAXW 32
#define MAXH 62
#define MAX MAXW*MAXH
#define INF 0x3f3f3f3f
#define IN_RANGE(x,y) (x >= 0 && x < H && y >= 0 && y < W)

struct Edge {
	int u, w, next;
	Edge() {}
	Edge(int uu, int ww, int nn) :u(uu), w(ww), next(nn) {}
}e[2][MAX*18];	//0:左脚点	1::右脚点

int W, H, n;
int board[MAX];
int h[2][MAX], le[2];
int d[2][MAX];	//双向最短距离
bool v[MAX];

int mov0[9][2] = { { 0, 1 },{0, 2 },{0, 3 },{-1, 1 },{ -1, 2 },{ -2, 1 },{ 1, 1 },{ 1, 2 },{ 2, 1 } };	//左脚踩住，右脚移动位置
int mov1[9][2] = { { 0, -1},{0, -2},{0, -3},{-1, -1}, {-1, -2}, {-2, -1}, {1, -1}, {1, -2}, {2, -1} };	//右脚踩住，左脚移动位置

void spfa(int x)
{
	memset(v, false, sizeof(v));
	memset(d, INF, sizeof(d));
	queue<int> q;
	q.push(x);	d[0][x] = d[1][x] = 0;
	while (!q.empty()) {
		int cur = q.front();
		q.pop();	v[cur] = false;
		for (int k = 0; k < 2; k++)	//双向最短路
			for (int i = h[k][cur]; i != -1; i = e[k][i].next)
			{
				int u = e[k][i].u;
				int w = e[k][i].w;
				if (d[!k][u] > d[k][cur] + w)	//交叉影响
				{
					d[!k][u] = d[k][cur] + w;
					if (!v[u]) {
						v[u] = true; q.push(u);
					}
				}

			}
	}
}

int main()
{
	while (scanf("%d%d", &W, &H), W && H)
	{
		char s[3];
		n = W*H;
		memset(h, -1, sizeof(h));
		//一维序列表示各点
		for (int i = 0; i < n; i++)
		{
			scanf("%s", s);
			if (s[0] == ‘S‘ || s[0] == ‘T‘)	board[i] = 0;
			else if (s[0] == ‘X‘)	board[i] = INF;
			else board[i] = s[0] - ‘0‘;
		}

		//构建邻接表
		le[0] = le[1] = 0;
		for (int i = 0; i < n; i++)
		{
			if ((i < W && board[i] == 0) || board[i] == INF) continue;
			for (int j = 0; j < 9; j++)
			{
				int x = i / W, y = i % W;	//计算行与列
				int x0 = x + mov0[j][0], y0 = y + mov0[j][1];
				int x1 = x + mov1[j][0], y1 = y + mov1[j][1];
				int n0 = x0*W + y0, n1 = x1*W + y1;
				if (IN_RANGE(x0,y0) && board[n0] != INF) {
					e[0][le[0]] = Edge(n0, board[n0], h[0][i]);
					h[0][i] = le[0]++;
				}
				if (IN_RANGE(x1,y1) && board[n1] != INF) {
					e[1][le[1]] = Edge(n1, board[n1], h[1][i]);
					h[1][i] = le[1]++;
				}
			}
		}

		int Min = INF;
		for (int i = (H - 1) * W; i < n; i++)	//枚举最后一行‘S‘进行SPFA
			if (board[i] == 0)
			{
				spfa(i);
				for (int j = 0; j < W; j++)	//遍历第一行‘T‘的最短路长
					if(board[j] == 0)	Min = min(min(Min, d[0][j]), d[1][j]);
			}
		if (Min == INF)	Min = -1;
		printf("%d\n", Min);
	}
	return 0;
}