(spfa专题 1.1)hdu 1535 Invitation Cards(求从1到2~n的点后，再从2~n返回1点的最小距离)

题目：

Invitation Cards

Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 2350    Accepted Submission(s): 1141


[align=left]Problem Description[/align]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. 

 
[align=left]Input[/align]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. 
 
[align=left]Output[/align]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. 
 
[align=left]Sample Input[/align]

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 
[align=left]Sample Output[/align]

46
210 
[align=left]Source[/align]Central Europe 1998 
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题目分析：
              这道题采用spfa来做。有以下2中思路：
       1）首先以1为源点进行一次spfa()算法，求得1到2~n点的最短距离。接下来就是分别以2~n点位源点，进行spfa()，这样既可得到答案。但是这就得进行n次spfa()算法了，在这种数据规模下(节点数n最大可达1000000)，很有可能会TLE。
      2）反向建边。首先以1位源点进行一次spfa算法(),求得1到2~n各个节点的最短距离。然后反向建边，再以1位源点进行一次spfa()算法，这时其实求得的是2~n到1的最短距离。(这道题是有向图，不是无向图)。

以下分别贴上了自己写的三个版本，虽然后面两个这道题没能AC。但是基本体现了用各种数据结构实现的spfa算法的
基本思路及基本使用。

AC代码如下(spfa，用队列+邻接表的实现)：

代码如下：
/*
* hdu_1535_1.cpp
*
* Created on: 2015年3月27日
* Author: Administrator
*/

#include <iostream>
#include <cstdio>
#include <cstring>
#include <queue>
using namespace std;

const int SIZE = 1000001;
const int INF = 0x3f3f3f3f;
int u[2 * SIZE], v[2 * SIZE], w[2 * SIZE], nextt[2 * SIZE];
int startss[2 * SIZE], endss[2 * SIZE], weightss[2 * SIZE];
int head[SIZE], dist[SIZE];
int n, m, cnt;

/**
* 初始化
*/
void init() {
memset(u, 0, sizeof(u));
memset(v, 0, sizeof(v));
memset(w, INF, sizeof(w));
memset(dist, 0, sizeof(dist));
memset(nextt, 0, sizeof(nextt));
memset(head, -1, sizeof(head));
cnt = 0;
}

/**
* 用队列实现的spfa。
*
* 参数说明:
* src: 起点(源点)
*/
void spfa(int src) {
queue<int> q;
bool inq[SIZE] = { 0 };
for (int i = 1; i <= n; i++)
dist[i] = (i == src) ? 0 : INF;
q.push(src);
while (!q.empty()) {
int x = q.front();
q.pop();
inq[x] = 0;
for (int e = head[x]; e != -1; e = nextt[e])
if (dist[v[e]] > dist[x] + w[e]) {
dist[v[e]] = dist[x] + w[e];
if (!inq[v[e]]) {
inq[v[e]] = 1;
q.push(v[e]);
}
}
}
}

/**
* 添加边
*/
void add_edge(int u1, int v1, int w1) {
u[cnt] = u1;
v[cnt] = v1;
w[cnt] = w1;
nextt[cnt] = head[u[cnt]];
head[u[cnt]] = cnt;
cnt++;
}

int main() {
int t;
scanf("%d", &t);
while (t--) {
scanf("%d%d", &n, &m);

int i;
for (i = 0; i < m; ++i) {
scanf("%d%d%d", &startss[i], &endss[i], &weightss[i]);
}

/**
* 使用spfa算法的基本步骤
*/
init();//初始化
for (i = 0; i < m; ++i) {//读入数据
add_edge(startss[i], endss[i], weightss[i]);
}
spfa(1);//进行运算

int ans = 0;
for(i = 2 ; i <= n ; ++i){//计算1到各个节点的最短距离
ans += dist[i];
}

init();
for(i = 0 ; i < m ; ++i){//这里反向建边
add_edge(endss[i],startss[i],weightss[i]);
}
spfa(1);
for(i = 2 ; i <= n ; ++i){//计算回程时各个节点到1的最短距离
ans += dist[i];
}

printf("%d\n",ans);//输出最后的结果..
}

return 0;
}


StackOverFlow的版本：
spfa(用栈+邻接表)
/*
* hdu_1535.cpp
*
* Created on: 2015年3月27日
* Author: Administrator
*/

#include <iostream>
#include <cstdio>
#include <cstring>

using namespace std;

const int INF = 0x3F3F3F3F;
const int V = 1000010;
const int E = 1000010;
int pnt[4*E], cost[4*E], nxt[4*E];
int e, head[V];
int dist[V];
bool vis[V];

int startss[4*E];
int endss[4*E];
int weightss[4*E];

int relax(int u, int v, int c) {
if (dist[v] > dist[u] + c) {
dist[v] = dist[u] + c;
return 1;
}
return 0;
}
inline void addedge(int u, int v, int c) {
pnt[e] = v;
cost[e] = c;
nxt[e] = head[u];
head[u] = e++;
}

/**
* 使用 邻接表+栈 实现的spfa
*
* 参数说明:
* src: 源点
* n: 节点的个数
*/
int SPFA(int src, int n) { // 此处用堆栈实现，有些时候比队列要快
int i;
for (i = 1; i <= n; ++i) { // 顶点1...n
vis[i] = 0;
dist[i] = INF;
}
dist[src] = 0;
int Q[E], top = 1;
Q[0] = src;
vis[src] = true;
while (top) {
int u, v;
u = Q[--top];
vis[u] = false;
for (i = head[u]; i != -1; i = nxt[i]) {
v = pnt[i];
if (1 == relax(u, v, cost[i]) && !vis[v]) {
Q[top++] = v;
vis[v] = true;
}
}
}
return dist
;
}

int main() {
int t;
scanf("%d", &t);
while (t--) {
int n, m;
scanf("%d%d", &n, &m);

memset(head, -1, sizeof(head));
e = 0;
int i;
for (i = 0; i < m; ++i) {
scanf("%d%d%d", &startss[i], &endss[i], &weightss[i]);
}

for(i = 0 ; i < m ; ++i){
addedge(startss[i],endss[i],weightss[i]);
}
int ans = 0;
SPFA(1, n);

for (i = 2; i <= n; ++i) {
ans += dist[i];
}

memset(head, -1, sizeof(head));
e = 0;
for(i = 0 ; i < m ; ++i){
addedge(endss[i],startss[i],weightss[i]);
}
SPFA(1,n);
for(i = 2 ; i <= n ; ++i){
ans += dist[i];
}

printf("%d\n",ans);
}

return 0;
}


CE的版本：
spfa算法(队列+邻接矩阵)
CE的原因估计是：二位矩阵开得太大了。
/*
* hdu1535.cpp
*
* Created on: 2015年3月27日
* Author: Administrator
*/
#include <iostream>
#include <cstring>
#include <queue>
using namespace std;

const int N = 1000001;
const int INF = 0x3f3f3f3f;

int map

, dist
;
bool visit
;
int n, m;

int startss
;
int endss
;
int weightss
;

void init() {//初始化
int i, j;
for (i = 1; i < N; i++) {
for (j = 1; j < N; j++) {
if (i == j) {
map[i][j] = 0;
} else {
map[i][j] = map[j][i] = INF;
}
}
}
}

/**
* SPFA算法.
* 使用spfa算法来求单元最短路径
* 参数说明:
* start:起点
*/
void spfa(int start) {
queue<int> Q;

int i, now;
memset(visit, false, sizeof(visit));
for (i = 1; i <= n; i++){
dist[i] = INF;
}

dist[start] = 0;
Q.push(start);
visit[start] = true;
while (!Q.empty()) {
now = Q.front();
Q.pop();
visit[now] = false;
for (i = 1; i <= n; i++) {
if (dist[i] > dist[now] + map[now][i]) {
dist[i] = dist[now] + map[now][i];
if (visit[i] == 0) {
Q.push(i);
visit[i] = true;
}
}
}
}
}

int main(){
int t;
scanf("%d",&t);
while(t--){
scanf("%d%d",&n,&m);

init();

int i;
for(i = 0 ; i < m ; ++i){
scanf("%d%d%d",&startss[i],&endss[i],&weightss[i]);
}

int ans = 0;

for(i = 0 ; i < m ; ++i){
map[startss[i]][endss[i]] = weightss[i];
}
spfa(1);
for(i = 2 ; i <= n ; ++i){
ans += dist[i];
}

init();
for(i = 0 ; i < m ; ++i){
map[endss[i]][startss[i]] = weightss[i];
}
spfa(1);
for(i = 2 ; i <= n ; ++i){
ans += dist[i];
}

printf("%d\n",ans);
}

return 0;
}