poj 3268 Silver Cow Party & zoj 2008 Invitation Cards

参考 算法导论 p580 单终点最短路径问题

这两题是一样的求法，先正着求一次最短路，再将边反向，再求一次

矩阵表示图的话就是将矩阵转置一次，邻接表表示的话就是读入边的时候，建两个表

poj3268

dijkstra求最短路

#include <iostream>
#include <string.h>
using namespace std;
const int MAXN =1001;
int n,m,x;
const int INF = 0x7FFF;
int G[MAXN][MAXN];
int dist1[MAXN];
int sum[MAXN];
bool used[MAXN];

void in(){
int i,j,from,to,t;
cin >> n >> m >> x;
for(i = 1; i <= n;i++){
for(j = 1; j <= n; j++){
if(i != j){
G[i][j] = INF;
} else {
G[i][j] = 0;
}
}
}
for(i = 1; i <= m;i++){
cin >> from >> to >> t;
G[from][to] = t;
}
memset(sum,0,sizeof(sum));
}

void dijkstra(){
int i,j;
for(i = 1; i <= n;i++){
dist1[i] = G[x][i];
used[i] = false;
}
used[x] = true;
dist1[x] = 0;
for(i = 1; i < n; i++){
int min = INF,minIndex = x;
for(j = 1; j <= n; j++)
if(!used[j] && dist1[j] < min){
min = dist1[j];
minIndex = j;
}
used[minIndex] = true;
sum[minIndex] += dist1[minIndex];
for(j = 1; j <= n; j++){
if(!used[j] && dist1[j] > dist1[minIndex]+G[minIndex][j]){
dist1[j] = dist1[minIndex]+G[minIndex][j];
}
}
}
}

void reverse(){
int i,j,temp;
for(i = 1; i <= n; i++){
for(j = 1; j < i; j++){
temp = G[i][j];
G[i][j] = G[j][i];
G[j][i] = temp;
}
}
}

int main(){
int max = 0,i;
in();
dijkstra();
reverse();
dijkstra();
for(i = 1; i <= n; i++)
if(max < sum[i])
max = sum[i];
cout << max << endl;
return 0;
}

zoj2008

spfa 求最短路

#include<stdio.h>
#include<queue>
using namespace std;
const int MAXN = 1000000;
const int INF = 0x3f3f3f3f;
struct edge{
int u, v, w, next, next2;
}e[MAXN];
int first[MAXN], first2[MAXN], dist[MAXN];
queue<int> que;
int p, q, cnt;
bool inq[MAXN];
void spfa(){
dist[1] = 0;
inq[1] = true;
while(!que.empty())
que.pop();
que.push(1);
while(!que.empty()){
int x = que.front();
que.pop();
inq[x] = false;
for(int i = first[x]; i != -1; i = e[i].next){
int u = e[i].u, v =e[i].v, w = e[i].w;
if(dist[v] > dist[u] + w){
dist[v] = dist[u] + w;
if(!inq[v]){
inq[v] = true;
que.push(v);
}
}
}
}
}
void spfa2(){
for(int i = 1; i <= p; ++i){
inq[i] = false;
dist[i] = INF;
}
dist[1] = 0;
inq[1] = true;
while(!que.empty())
que.pop();
que.push(1);
while(!que.empty()){
int x = que.front();
que.pop();
inq[x] = false;
for(int i = first2[x]; i != -1; i = e[i].next2){
int u = e[i].v, v =e[i].u, w = e[i].w;
if(dist[v] > dist[u] + w){
dist[v] = dist[u] + w;
if(!inq[v]){
inq[v] = true;
que.push(v);
}
}
}
}
}
int main(){
int t;
scanf("%d",&t);
while(t--){
scanf("%d%d",&p, &q);
for(int i = 1; i <= p; ++i){
first[i] = -1;
first2[i] = -1;
dist[i] = INF;
inq[i] = false;
}
for(int i = 0; i < q; ++i){
scanf("%d%d%d",&e[i].u, &e[i].v, &e[i].w);
e[i].next = first[e[i].u];
first[e[i].u] = i;
e[i].next2 = first2[e[i].v];//建立两个邻接表
first2[e[i].v] = i;
}
spfa();
int ans = 0;
for(int i = 1; i <= p; ++i)
ans += dist[i];
spfa2();
for(int i = 1; i <= p; ++i)
ans += dist[i];
printf("%d\n",ans);
}
return 0;
}