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;
}

 

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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;
}

 

转载于:https://www.cnblogs.com/Inkblots/p/5475879.html