图论trainning-part-1 E. Invitation Cards
E. Invitation Cards
64-bit integer IO format: %lld Java class name: Main
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
Output
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
解题:spfa无疑啊,当然用优先队列优化了的dijkstra也是可以的,Bellman-Ford直接TLE。此题有大坑,使用单源最短路径算法时,需要保证INF足够大,最好比INT的最大值大,否则一直WA,让人摸不着头脑啊。先求1到其他地的最短路之和,再把各边反向,再求一次1到其他站点的最短路之和。
#include <iostream>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <algorithm>
#include <climits>
#include <vector>
#include <queue>
#include <cstdlib>
#include <string>
#include <set>
#define LL long long
#define INF 0xffffffff
using namespace std;
const int maxn = ;
struct arc {
int to,w;
};
int u[maxn],v[maxn],w[maxn],n,m;
LL d[maxn];
vector<arc>g[maxn];
queue<int>qu;
bool used[maxn];
void spfa() {
int i,j;
for(i = ; i <= n; i++)
d[i] = INF;
d[] = ;
memset(used,false,sizeof(used));
while(!qu.empty()) qu.pop();
qu.push();
used[] = true;
while(!qu.empty()) {
int temp = qu.front();
used[temp] = false;
qu.pop();
for(i = ; i < g[temp].size(); i++) {
j = g[temp][i].to;
if(d[j] > d[temp]+g[temp][i].w) {
d[j] = d[temp]+g[temp][i].w;
if(!used[j]) {
qu.push(j);
used[j] = true;
}
}
}
}
}
int main() {
int t;
scanf("%d",&t);
while(t--) {
int i,j;
scanf("%d%d",&n,&m);
for(i = ; i <= n; i++)
g[i].clear();
for(i = ; i <= m; i++) {
scanf("%d%d%d",u+i,v+i,w+i);
g[u[i]].push_back((arc) {v[i],w[i]});
}
LL ans = ;
spfa();
for(i = ; i <= n; i++){
ans += d[i];
g[i].clear();
}
for(i = ; i <= m; i++)
g[v[i]].push_back((arc) {u[i],w[i]});
spfa();
for(i = ; i <= n; i++)
ans += d[i];
printf("%I64d\n",ans);
}
return ;
}
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