Invitation Cards
Time Limit: 8000MS Memory Limit: 262144K
Total Submissions: 25066 Accepted: 8277
Time Limit: 8000MS Memory Limit: 262144K
Total Submissions: 25066 Accepted: 8277
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 50
Sample Output
46
210
给你一个N个点的图，求1点到其他每个点最短路权值之和和其他所有点到1点最短距离路权值之和。
题目数据给到了1000000之大，很明显让我们用高效率的SPFA算法去做，Dijkstra弄不好就会超时，O(N^3)的Floyd更不用说。
#include<iostream>
#include<cstdio>
#include<cstring>
#include<queue>
using namespace std;
const int N = 1000010;
const long long MAX = 0xFFFFFFFF; //数据太大。10亿*100W
queue<int> q;
int first[N], next[N], w[N], Num[N], num;
long long dis[N];
bool vis[N];
int a, b;
struct Edge //结构体存图，方便反向建边
{
int u, v;
int ww;
}ee[N];
void add(int u, int v, int cost) //u,v之间加入权值为cost的边
{
w[num] = cost;
next[num] = first[u];
Num[num] = v;
first[u] = num++;
}
void init()
{
num = 0;
memset(first, -1, sizeof(first));
for(int i = 1; i <= a; ++i)
dis[i] = MAX;
}
long long spfa(int start)
{
long long total = 0;
int temp;
memset(vis,false,sizeof(vis));
dis[start] = 0;
vis[start] = true;
q.push(start);
while(!q.empty())
{
temp = q.front();
q.pop();
vis[temp] = false;
for(int i = first[temp]; i != -1; i = next[i]) //邻接表实现
{
if(dis[Num[i]] > dis[temp] + w[i])
{
dis[Num[i]] = dis[temp] + w[i];
if(!vis[Num[i]])
{
q.push(Num[i]);
vis[Num[i]] = true;
}
}
}
}
for(int i = 1; i <= a; ++i)
total += dis[i];
return total;
}
int main()
{
int T;
int top;
long long ans;
scanf("%d", &T);
while(T--)
{
top = ans = 0;
scanf("%d%d", &a, &b);//输入点数和要处理的边
init(); //初始化dis和vis数组
for(int i = 0; i < b; i++)
{
scanf("%d %d %d", &ee[top].u, &ee[top].v, &ee[top].ww);
add(ee[top].u, ee[top].v, ee[top].ww);
top++;
}
ans += spfa(1);
init();
for(int i = 0; i < top; ++i) //反向图
add(ee[i].v, ee[i].u, ee[i].ww);
ans += spfa(1);
printf("%lld
", ans);
}
return 0;
}