[2016-04-05][POJ][1511][Invitation Cards]

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• 时间:2016-04-05 12:57:22 星期二

• 题目编号:[2016-04-05][POJ][1511][Invitation Cards]

• 题目大意:给定一个有向图,从点1出发,分别到各个站点后,又回到点1,问最少需要多少车费,

• 分析:

  • 从1跑一次最短路,然后矩阵转置,再跑一次最短路,两次求和
  • 这里不能用邻接矩阵保存,所以改成邻接表,然后矩阵转置的操作变成重新加一次边

• 遇到的问题:用vector存图超时,改用数组实现

#include <queue>
#include <algorithm>
#include <cstring>
#include <cstdio>
using namespace std;
typedef long long ll;
const int maxn = 1E6 + 20;
const int maxm = 1E6 + 20;
int head[maxn],nxt[maxm],mend[maxm],cost[maxm];
void addedge(int u,int v, int c){
        static int q = 2;
        mend[q] = v;
        nxt[q] = head[u];
        cost[q] = c;
        head[u] = q++;       
}
struct Edge{
        int u,v,c;
        Edge(int _u = 0, int _v = 0,int _c = 0):u(_u),v(_v),c(_c){}
}e[maxn];
struct mNode{
        ll u,c;
        mNode(ll _u = 0,ll _c = 0):u(_u),c(_c){}
        bool operator < (const mNode & a)const{
                return c > a.c;
        }
};
bool vis[maxn];
ll d[maxn];
ll Dijkstra(int s,int n){
        memset(vis,0,sizeof(bool) * (n + 2));
        memset(d,0x3f,sizeof(ll) * (n + 2));
        priority_queue<mNode> q;
        d[s] = 0;
        q.push(mNode(s,0));
        mNode tmp;
        while(!q.empty()){
                tmp = q.top();
                q.pop();
                int u = tmp.u;
                if(vis[u])      continue;
                vis[u] = 1;
                for(int qq = head[u];~qq;qq = nxt[qq]){
                        int v = mend[qq];
                        if(!vis[v] && d[v] > d[u] + cost[qq]){
                                d[v] = d[u] + cost[qq];
                                q.push(mNode(v,d[v]));
                        }
                }
        }
        ll ans = 0;
        for(int i = 2; i <= n ; ++i)
                ans += d[i];
        return ans;
}
inline void ini(int n){
        memset(head,-1,sizeof(int)*(n + 2));
}
int main(){
        int t,n,m,a,b,c;
        scanf("%d",&t);
                while(t--){
                        scanf("%d%d",&n,&m);
                        ini(n);
                        for(int i = 0;i < m ; ++i){
                                scanf("%d%d%d",&a,&b,&c);
                                e[i] = Edge(a,b,c);
                                addedge(a,b,c);
                        }
                        ll ans;
                        ans = Dijkstra(1,n);
                        ini(n);
                        for(int i = 0;i < m ; ++i){
                                addedge(e[i].v,e[i].u,e[i].c);
                        }
                        ans += Dijkstra(1,n);               
                        printf("%I64d\n",ans);
                }
        return 0;
}

[2016-04-05][POJ][1511][Invitation Cards]

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原文地址：http://www.cnblogs.com/qhy285571052/p/c20ae82b86c28c48b4c5b5cd5bc88162.html