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;
}