hdu 1513 Invitation Cards【spfa翻转边】

来源：互联网发布：fgo花嫁尼禄宝具本数据编辑：程序博客网时间：2024/05/20 15:41

题目链接：http://acm.acmcoder.com/showproblem.php?pid=1535

题意：有向图，求源点到各个点最短路径和+各个点到源点最短路径和。

spfa求单源最短路径，求各个点到源点最短路径翻转边。

#include <iostream>#include <stdio.h>#include <string.h>#include <algorithm>#include <math.h>#include <string>#include <set>#include <queue>#include <stack>#include <vector>#include <map>using namespace std;const int MAXN = 2000010;const int INF = 1e9+10;struct{    int a, b, c;}Map[MAXN];int t;int n, m, a, b, c;struct Edge{    int v;    int cost;    Edge(int _v = 0, int _cost = 0)    {        v = _v;        cost = _cost;    }};vector<Edge> E[MAXN];void addedge(int u, int v, int cost){    E[u].push_back(Edge(v, cost));}bool vis[MAXN];int cnt[MAXN];int dist[MAXN];void SPFA(int start, int n){    memset(vis, false, sizeof(vis));    for (int i = 1; i <= n; i++) dist[i] = INF;    vis[start] = true;    dist[start] = 0;    queue<int> que;    while (!que.empty()) que.pop();    que.push(start);    memset(cnt, 0, sizeof(cnt));    cnt[start] = 1;    while (!que.empty())    {        int u = que.front(); que.pop();        vis[u] = false;        for (int i = 0; i<E[u].size(); i++)        {            int v = E[u][i].v;            if (dist[v]>dist[u] + E[u][i].cost)            {                dist[v] = dist[u] + E[u][i].cost;                if (!vis[v])                {                    vis[v] = true;                    que.push(v);                    if (++cnt[v] > n) return;                }            }        }    }}int main(){    scanf("%d",&t);    while (t--)    {        scanf("%d%d", &n, &m);        for (int i = 1; i <= n; i++)            E[i].clear();        for (int i = 1; i <= m; i++)        {            scanf("%d%d%d", &Map[i].a, &Map[i].b, &Map[i].c);            addedge(Map[i].a, Map[i].b, Map[i].c);        }        int ans = 0;        SPFA(1, n);        for (int i = 1; i <= n; i++)            ans += dist[i];        for (int i = 1; i <= n; i++)            E[i].clear();        for (int i = 1; i <= m; i++)            addedge(Map[i].b, Map[i].a, Map[i].c);        SPFA(1, n);        for (int i = 1; i <= n; i++)            ans += dist[i];        printf("%d\n", ans);    }    return 0;}

0 0