題目描述

The good folks in Texas are having a heatwave this summer. Their Texas Longhorn cows make for good eating but are not so adept at creating creamy delicious dairy products. Farmer John is leading the charge to deliver plenty of ice cold nutritious milk to Texas so the Texans will not suffer the heat too much.

FJ has studied the routes that can be used to move milk from Wisconsin to Texas. These routes have a total of T (1 <= T <= 2,500) towns conveniently numbered 1..T along the way (including the starting and ending towns). Each town (except the source and destination towns) is connected to at least two other towns by bidirectional roads that have some cost of traversal (owing to gasoline consumption, tolls, etc.). Consider this map of seven towns; town 5 is the

source of the milk and town 4 is its destination (bracketed integers represent costs to traverse the route):

                              [1]----1---[3]-
                             /               \
                      [3]---6---[4]---3--[3]--4 / / /| 5 --[3]-- --[2]- | \ / / | [5]---7---[2]--2---[3]--- | / [1]------
 

Traversing 5-6-3-4 requires spending 3 (5->6) + 4 (6->3) + 3 (3->4) = 10 total expenses.

Given a map of all the C (1 <= C <= 6,200) connections (described as two endpoints R1i and R2i (1 <= R1i <= T; 1 <= R2i <= T) and costs (1 <= Ci <= 1,000), find the smallest total expense to traverse from the starting town Ts (1 <= Ts <= T) to the destination town Te (1 <= Te <= T).

德克薩斯純樸的民眾們這個夏天正在遭受巨大的熱浪！！！他們的德克薩斯長角牛吃起來不錯，可是他們並不是很擅長生產富含奶油的乳製品。Farmer John此時以先天下之憂而憂，後天下之樂而樂的精神，身先士卒地承擔起向德克薩斯運送大量的營養冰涼的牛奶的重任，以減輕德克薩斯人忍受酷暑的痛苦。

FJ已經研究過可以把牛奶從威斯康星運送到德克薩斯州的路線。這些路線包括起始點和終點先一共經過T (1 <= T <= 2,500)個城鎮，方便地標號為1到T。除了起點和終點外地每個城鎮由兩條雙向道路連向至少兩個其它地城鎮。每條道路有一個通過費用（包括油費，過路費等等）。

給定一個地圖，包含C (1 <= C <= 6,200)條直接連線2個城鎮的道路。每條道路由道路的起點Rs，終點Re (1 <= Rs <= T; 1 <= Re <= T)，和花費(1 <= Ci <= 1,000)組成。求從起始的城鎮Ts (1 <= Ts <= T)到終點的城鎮Te(1 <= Te <= T)最小的總費用。

輸入輸出格式

第一行: 4個由空格隔開的整數: T, C, Ts, Te

第2到第C+1行: 第i+1行描述第i條道路。有3個由空格隔開的整數: Rs, Re和Ci

一個單獨的整數表示從Ts到Te的最小總費用。資料保證至少存在一條道路。

輸入輸出樣例

7 11 5 4
2 4 2
1 4 3
7 2 2
3 4 3
5 7 5
7 3 3
6 1 1
6 3 4
2 4 3
5 6 3
7 2 1

7

說明

【樣例說明】

5->6->1->4 (3 + 1 + 3)

#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cstdlib>
#include<cstring>
#include<string>
#include<cmath>
#include<map>
#include<set>
#include<vector>
#include<queue>
#include<bitset>
#include<ctime>
#include<deque>
#include<stack>
#include<functional>
#include<sstream>
//#include<cctype>
//#pragma GCC optimize(2)
using namespace std;
#define maxn 200005
#define inf 0x7fffffff
//#define INF 1e18
#define rdint(x) scanf("%d",&x)
#define rdllt(x) scanf("%lld",&x)
#define rdult(x) scanf("%lu",&x)
#define rdlf(x) scanf("%lf",&x)
#define rdstr(x) scanf("%s",x)
typedef long long  ll;
typedef unsigned long long ull;
typedef unsigned int U;
#define ms(x) memset((x),0,sizeof(x))
const long long int mod = 1e9 + 7;
#define Mod 1000000000
#define sq(x) (x)*(x)
#define eps 1e-3
typedef pair<int, int> pii;
#define pi acos(-1.0)
//const int N = 1005;
#define REP(i,n) for(int i=0;i<(n);i++)
typedef pair<int, int> pii;
inline ll rd() {
	ll x = 0;
	char c = getchar();
	bool f = false;
	while (!isdigit(c)) {
		if (c == '-') f = true;
		c = getchar();
	}
	while (isdigit(c)) {
		x = (x << 1) + (x << 3) + (c ^ 48);
		c = getchar();
	}
	return f ? -x : x;
}

ll gcd(ll a, ll b) {
	return b == 0 ? a : gcd(b, a%b);
}
int sqr(int x) { return x * x; }


/*ll ans;
ll exgcd(ll a, ll b, ll &x, ll &y) {
	if (!b) {
		x = 1; y = 0; return a;
	}
	ans = exgcd(b, a%b, x, y);
	ll t = x; x = y; y = t - a / b * y;
	return ans;
}
*/

struct node {
	int u, v, w, nxt;
	bool operator<(const node&rhs)const {
		return w > rhs.w;
	}
}edge[maxn];
int n, m;
int s;
int dis[maxn], head[maxn], vis[maxn];
int tot;
void init() {
	memset(head, -1, sizeof(head)); 
}
void addedge(int u, int v, int w) {
	edge[tot].v = v; edge[tot].w = w;
	edge[tot].nxt = head[u]; head[u] = tot++;
}
void dijkstra(int s) {
	memset(dis, 0x3f, sizeof(dis)); ms(vis);
	priority_queue<node>q;
	node tmp1, tmp2;
	tmp1.v = s; dis[s] = 0;
	q.push(tmp1);
	while (!q.empty()) {
		tmp1 = q.top(); q.pop();
		int u = tmp1.v;
		if (vis[u])continue;
		vis[u] = 1;
		for (int i = head[u]; i != -1; i = edge[i].nxt) {
			int v = edge[i].v; int w = edge[i].w;
			if (dis[v] > dis[u] + w && !vis[v]) {
				dis[v] = dis[u] + w; tmp2.v = v; tmp2.w = dis[v];
				q.push(tmp2);
			}
		}
	}
}

int main() {
	//ios::sync_with_stdio(0);
	init(); rdint(n); rdint(m);
	rdint(s); int ed; rdint(ed);
	while (m--) {
		int u, v, w; rdint(u); rdint(v); rdint(w);
		addedge(u, v, w); addedge(v, u, w);
	}
	dijkstra(s);
	cout << dis[ed] << endl;
	return 0;
}