1003. Emergency (25)

1003. Emergency (25)

时间限制

400 ms

内存限制

65536 kB

代码长度限制

16000 B

判题程序

Standard

作者

CHEN, Yue

As an emergency rescue team leader of a city, you are given a special map of your country. The map shows several scattered cities connected by some roads. Amount of rescue teams in each city and the length of each road between any pair of cities are marked on the map. When there is an emergency call to you from some other city, your job is to lead your men to the place as quickly as possible, and at the mean time, call up as many hands on the way as possible.

Input

Each input file contains one test case. For each test case, the first line contains 4 positive integers: N (<= 500) - the number of cities (and the cities are numbered from 0 to N-1), M - the number of roads, C1 and C2 - the cities that you are currently in and that you must save, respectively. The next line contains N integers, where the i-th integer is the number of rescue teams in the i-th city. Then M lines follow, each describes a road with three integers c1, c2 and L, which are the pair of cities connected by a road and the length of that road, respectively. It is guaranteed that there exists at least one path from C1 to C2.

Output

For each test case, print in one line two numbers: the number of different shortest paths between C1 and C2, and the maximum amount of rescue teams you can possibly gather.
All the numbers in a line must be separated by exactly one space, and there is no extra space allowed at the end of a line.

Sample Input

5 6 0 2
1 2 1 5 3
0 1 1
0 2 2
0 3 1
1 2 1
2 4 1
3 4 1

Sample Output

2 4

---

提交代码

题解： 题意是给定N个点, M条边, 给定起点和终点, 每个点都有权值, 每条边也有权值, 问起点到终点最短的路径有多少条, 并且在最短的路径当中, 求点权值的和最大是多少.

我们可以先用Dijkstra算法求得起点到各点的最短路径(我用的是链式前向星存图, 每个测试样例都是1ms), 然后用DFS统计到达每一个点的路径的条数以及到达该点的最大权值是多少.

#include <cstdio>
#include <cstring>

const int N = 510;
const int INF = 0x3f3f3f3f;
bool vis[N];
int num[N], cnt[N];
int head[N], dis[N], cnt_num[N];
int total;

struct node{
	int v, val, next;
}Node[N * 5];

void add_edge(int u, int v, int val) {
	Node[total].v = v;
	Node[total].val = val;
	Node[total].next = head[u];
	head[u] = total++;
}

void dijkstra(int vertex, int root) {
	dis[root] = 0;
	vis[root] = true;

	for (int i = head[root]; ~i; i = Node[i].next) 
		if (dis[Node[i].v] > Node[i].val)
			dis[Node[i].v] = Node[i].val;

	for (int i = 1; i < vertex; ++i) {
		int tmp = INF;
		int k = root;

		for (int j = 0; j < vertex; ++j)
			if (!vis[j] && dis[j] < tmp)
				tmp = dis[k = j];
		vis[k] = true;

		for (int j = head[k]; ~j; j = Node[j].next) {
			if (dis[k] + Node[j].val < dis[Node[j].v])
				dis[Node[j].v] = dis[k] + Node[j].val;
		}
	}
}

void dfs(int pre, int u) {

	for (int i = head[u]; ~i; i = Node[i].next) {
		int v = Node[i].v;
		if (v != pre && dis[u] + Node[i].val == dis[v]) {
			cnt[v]++;
			if (cnt_num[v] < cnt_num[u] + num[v])
				cnt_num[v] = cnt_num[u] + num[v];
			dfs(u, v);
		}
	}
}

void Init() {
	total = 0;
	memset(head, -1, sizeof(head));
	memset(dis, INF, sizeof(dis));
}

int main() {
	int vertex, edge, start, end;
	scanf("%d%d%d%d", &vertex, &edge, &start, &end);

	for (int i = 0; i < vertex; ++i)
		scanf("%d", num + i);

	Init();
	int u, v, val;
	for (int i = 0; i < edge; ++i) {
		scanf("%d%d%d", &u, &v, &val);
		add_edge(u, v, val);
		add_edge(v, u, val);
	}

	dijkstra(vertex, start);
	cnt[start] = 1;
	cnt_num[start] = num[start];
	dfs(-1, start);

	printf("%d %d\n", cnt[end], cnt_num[end]);

	return 0;
}