做了两天..终于AC了..

Invitation Cards

Description

In the age of television, not many people attend theater performances. Antique Comedians of Malidinesia are aware of this fact. They want to propagate theater and, most of all, Antique Comedies. They have printed invitation cards with all the necessary information and with the programme. A lot of students were hired to distribute these invitations among the people. Each student volunteer has assigned exactly one bus stop and he or she stays there the whole day and gives invitation to people travelling by bus. A special course was taken where students learned how to influence people and what is the difference between influencing and robbery.

The transport system is very special: all lines are unidirectional and connect exactly two stops. Buses leave the originating stop with passangers each half an hour. After reaching the destination stop they return empty to the originating stop, where they wait until the next full half an hour, e.g. X:00 or X:30, where 'X' denotes the hour. The fee for transport between two stops is given by special tables and is payable on the spot. The lines are planned in such a way, that each round trip (i.e. a journey starting and finishing at the same stop) passes through a Central Checkpoint Stop (CCS) where each passenger has to pass a thorough check including body scan.

All the ACM student members leave the CCS each morning. Each volunteer is to move to one predetermined stop to invite passengers. There are as many volunteers as stops. At the end of the day, all students travel back to CCS. You are to write a computer program that helps ACM to minimize the amount of money to pay every day for the transport of their employees.

Input

The input consists of N cases. The first line of the input contains only positive integer N. Then follow the cases. Each case begins with a line containing exactly two integers P and Q, 1 <= P,Q <= 1000000. P is the number of stops including CCS and Q the number of bus lines. Then there are Q lines, each describing one bus line. Each of the lines contains exactly three numbers - the originating stop, the destination stop and the price. The CCS is designated by number 1. Prices are positive integers the sum of which is smaller than 1000000000. You can also assume it is always possible to get from any stop to any other stop.

Output

For each case, print one line containing the minimum amount of money to be paid each day by ACM for the travel costs of its volunteers.

Sample Input

2 2 2 1 2 13 2 1 33 4 6 1 2 10 2 1 60 1 3 20 3 4 10 2 4 5 4 1 50Sample Output

46 210Source

Central Europe 1998

题意：给出一个有向图的边，求原点到所有点的来回最短路径之和。

看这数据范围都知道，直接用bellman-ford, dijkstra或者floyd是不行的啦，所以要用到SPFA,,,但是我之前并没有学过SPFA，所以只好找网上大神的代码请教一下Orz 此外这道题不能用邻接矩阵（因为会爆...），所以得用邻接表，我这里因为想多学点东西所以又用了静态邻接表（链式前向星，都可以啦..）

思路：静态邻接表 + SPFA

注意这道题几个坑点！！！

1. INF一定要够大！！我的是0xffffffff
2. ans一定要是long long!!!我为了保险起见连dis都用了long long
3. 初始化啊..
4. 清空...
5. （反正以上错误我基本都犯了，大神笑笑看过就好..）

代码如下：

/*Problem: 1511		User: aclolicon
Memory: 40740K		Time: 1766MS
Language: G++		Result: Accepted
Source Code*/
#include
#include
#define MAXN 1000010
#define INF 0xffffffff
using namespace std;
struct edge{
	int to, next, len;
};
edge e[2][MAXN];
int n, p, q;
long long dis[MAXN], ans;
int head[2][MAXN];
bool vis[MAXN];
void spfa(int r){
	for (int i = 1; i <= p; i++){
		dis[i] = INF;
		vis[i] = 0;
	}
	dis[1] = 0;
	vis[1] = 1;
	queue  qu;
	qu.push(1);
	while(!qu.empty()){
		int now = qu.front();
		qu.pop();
		vis[now] = 0;
		for (int i = head[r][now];i != -1;i = e[r][i].next){
			int to = e[r][i].to;
			if (dis[now] + e[r][i].len < dis[to]){
				dis[to] = dis[now] + e[r][i].len;
				if (!vis[to]){
					vis[to] = 1;
					qu.push(to);
				}
			}
		}
	}
} 

void solve(){
	ans = 0;
	spfa(0);
	for (int i = 1; i <= p; i++) ans += dis[i];
	spfa(1);
	for (int i = 1; i <= p; i++) ans += dis[i];
	printf("%lldn", ans);
}

void init(){
	scanf("%d%d", &p, &q);
	for (int i = 1; i <= p; i++){
		head[0][i] = -1;
		head[1][i] = -1;
	}
	int u, v, l;
	for (int i = 0; i < q; i++){
		scanf("%d%d%d", &u, &v, &l);
		e[0][i].to = v;
		e[0][i].next = head[0][u];
		e[0][i].len = l;
		e[1][i].to = u;
		e[1][i].next = head[1][v];
		e[1][i].len = l;	
		head[0][u] = i;
		head[1][v] = i;	
	}	
}

int main(){
	int n;
	scanf("%d", &n);
	while(n--){
		init();
		solve();
	}
	return 0;
}

Source Code

WA了N次啊..希望大家不要学我这种SB...