【图论】【最短路】 pku1511 Invitation Cards

【题目】：Invitation Cards

【类型】：图论

【难度】：中等

【来源】：pku1511

【关键字】：SPFA

【题目大意】：给N个点的路径信息，给定一个点作为起点，求该点到每个点再回来总路程最小值。

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【自己的分析】：

【正确的算法分析】：之前用DIJ做过类似的题，但这题数据量很大，用邻接表+SPFA。出去每个点直接用SPFA，求返回的路径的解决方法是将所有边反向保存，再对起点做一次SPFA，即是返回路径长度。

【数据结构】：邻接表

【其他优化】：

【时间复杂度】：

【学到了什么知识经验】：第一次写用正反两个邻接表边数组edge[MAXE]保存，后来修修补补，看别人的把两个边数组变成一个数组edge[2*MAXE]，就AC了。不知何故。最近经常这样，错的代码似乎没改什么，交第二次就能AC。。。

【同类型题目】：pku2387的强化版

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【其他】：

 http://acm.pku.edu.cn/JudgeOnline/problem?id=1511

 

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【代码】：

Source CodeProblem: 1511 User: mofixroot Memory: 49128K Time: 2000MS Language: C++ Result: Accepted Source Code #include<iostream>using namespace std;#define inf 1000000005#define MAXV 1000005#define MAXE 1000005int head[MAXV], rhead[MAXV];struct edge_t{ int s, t, w; int next;};edge_t edge[MAXE*2];int nEdge;int nv, ne;int dis[MAXV];char vstd[MAXV];__int64 sum;int que[MAXV];char inque[MAXV];void Addedge(int u, int v, int w) //建立邻接表,有向边 { edge[nEdge].s = u; edge[nEdge].t = v; edge[nEdge].w = w; edge[nEdge].next = head[u]; head[u] = nEdge++; edge[nEdge].s = v; edge[nEdge].t = u; edge[nEdge].w = w; edge[nEdge].next = rhead[v]; rhead[v] = nEdge++;}void spfa(int s){ int i, j, k, v, u, ie, tag; for(i = 1; i <= nv; i++) dis[i]=inf, vstd[i]=0; memset(que, 0, sizeof(que)); memset(inque, 0, sizeof(inque)); int qhead=0, qtail=1; que[qhead] = s; inque[s] = 1; dis[s] = 0; while(qhead < qtail) { v = que[qhead]; inque[v] = 0; for(ie = head[v]; ie!=-1; ie=edge[ie].next) { u = edge[ie].t; if(dis[u] > dis[v] + edge[ie].w) { dis[u] = dis[v] + edge[ie].w; if(!inque[u]) { inque[u]=1; que[qtail++] = u; } } } qhead++; } for(i = 1; i <= nv; i++) { sum += dis[i]; }}void rspfa(int s){ int i, j, k, v, u, ie, tag; for(i = 1; i <= nv; i++) dis[i]=inf, vstd[i]=0; memset(que, 0, sizeof(que)); memset(inque, 0, sizeof(inque)); int qhead=0, qtail=1; que[qhead] = s; inque[s] = 1; dis[s] = 0; while(qhead < qtail) { v = que[qhead]; inque[v] = 0; for(ie = rhead[v]; ie!=-1; ie=edge[ie].next) { u = edge[ie].t; if(dis[u] > dis[v] + edge[ie].w) { dis[u] = dis[v] + edge[ie].w; if(!inque[u]) { inque[u]=1; que[qtail++] = u; } } } qhead++; } for(i = 1; i <= nv; i++) { sum += dis[i]; }}int main(){ int cas, i, s, t, d; scanf("%d", &cas); while(cas--) { memset(head, -1, sizeof(head)); memset(edge, 0, sizeof(edge)); memset(rhead, -1, sizeof(rhead)); nEdge = 0; scanf("%d %d", &nv, &ne); for(i = 1; i <= ne; i++) { scanf("%d %d %d", &s, &t, &d); Addedge(s, t, d); } sum = 0; spfa(1); rspfa(1); printf("%I64d/n", sum); } return 0;}