ACM/ICPC 之 SPFA范例两道(POJ3268-POJ3259)

两道以SPFA算法求解的最短路问题，比较水，第二题需要掌握如何判断负权值回路。

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POJ3268-Silver Cow Party

//计算正逆最短路径之和的最大值

//Time:32Ms	Memory:360K

#include<iostream>

#include<cstring>

#include<cstdio>

#include<queue>

#include<algorithm>

using namespace std;

#define MAX 1005

#define MAXE 100005

#define INF 0x3f3f3f3f

struct Edge {

	int v, w, next;

	Edge() {}

	Edge(int vv, int ww, int nn):v(vv), w(ww), next(nn) {}

}e[2][MAXE];

int n, m, x;

int h[2][MAX], d[2][MAX];

bool v[MAX];

void spfa(int x, int r)

{

	memset(v, false, sizeof(v));

	queue<int> q;

	q.push(x);	d[r][x] = 0;

	while (!q.empty()){

		int cur = q.front();

		q.pop();	v[cur] = false;

		for (int i = h[r][cur]; i != -1; i = e[r][i].next)

		{

			int tv = e[r][i].v;

			int tw = e[r][i].w;

			if (d[r][tv] > d[r][cur] + tw) {

				d[r][tv] = d[r][cur] + tw;

				if (!v[tv]) {

					q.push(tv);	v[tv] = true;

				}

			}

		}

	}

}

int main()

{

	memset(d, INF, sizeof(d));

	memset(h, -1, sizeof(h));

	scanf("%d%d%d", &n, &m, &x);

	for (int i = 0; i < m; i++)

	{

		int a, b, t;

		scanf("%d%d%d", &a, &b, &t);

		if (a == b) {

			m--; i--; continue;

		}

		e[0][i] = Edge(b, t, h[0][a]);	//正向

		e[1][i] = Edge(a, t, h[1][b]);	//反向

		h[0][a] = h[1][b] = i;

	}

	spfa(x, 0);	//正向路径找返回最短路

	spfa(x, 1);	//反向路径找出发最短路

	int Max = 0;

	for (int i = 1; i <= n; i++)

		Max = max(d[0][i] + d[1][i], Max);

	printf("%d\n", Max);

	return 0;

}

---

POJ3259-Wormholes

//SPFA或Bellman Ford算法

//判断是否有负权值回路

//Time:172Ms	Memory:252K

#include<iostream>

#include<cstring>

#include<cstdio>

#include<queue>

using namespace std;

#define MAX 505

#define MAXN 5500

#define INF 0x3f3f3f3f

struct Edge {

	int u, w, next;

	Edge() {}

	Edge(int uu, int ww, int nn) :u(uu), w(ww), next(nn) {}

}e[MAXN];

int n, m, w;

int le;

int h[MAX], d[MAX];

int cnt[MAX];

bool v[MAX];

bool spfa(int x)

{

	memset(cnt, 0, sizeof(cnt));

	memset(v, false, sizeof(v));

	memset(d, INF, sizeof(d));

	d[x] = 0;

	queue<int> q;

	q.push(x); cnt[x] = 1;

	while (!q.empty()) {

		int cur = q.front();

		q.pop();	v[cur] = false;

		for (int i = h[cur]; i != -1; i = e[i].next)

		{

			int u = e[i].u, w = e[i].w;

			if (d[u] > d[cur] + w)

			{

				d[u] = d[cur] + w;

				if (!v[u]) {

					v[u] = true;  q.push(u);

					//如果某点入队列达到n次，则一定存在负权值回路

					if (++cnt[u] == n)	return true;

				}

			}

		}

	}

	return false;

}

int main()

{

	int T;

	scanf("%d", &T);

	while (T--)

	{

		le = 0;

		memset(h, -1, sizeof(h));

		scanf("%d%d%d", &n, &m, &w);

		int a, b, t;

		while (m--) {

			scanf("%d%d%d", &a, &b, &t);

			e[le] = Edge(b, t, h[a]);	h[a] = le++;

			e[le] = Edge(a, t, h[b]);	h[b] = le++;

		}

		while (w--) {

			scanf("%d%d%d", &a, &b, &t);

			e[le] = Edge(b, -t, h[a]);	h[a] = le++;

		}

		/*如果不是连通图则需要遍历所有点	//Time:1750Ms

		int flag = false;

		for (int i = 1; i <= n; i++)

			if (spfa(i)) {

				flag = true; break;

			}

		题目未说清是否为连通图

		*/

		if (spfa(1)) printf("YES\n");

		else printf("NO\n");

	}

	return 0;

}