SGU 185 Two shortest ★(最短路+网络流)
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题意】给出一个图，求 1 -> n的2条 没有重边的最短路。 真◆神题……卡内存卡得我一脸血= =…… 【
思路】 一开始我的想法是
两遍Dijkstra做一次删一次边不就行了么你们还又Dijkstra预处理又最大流的Too naive……结果事实证明从来都是我naive= =……明显是不行的……最大流可能有好几条……但不重边的更少……也许第一次Dijkstra找到的是最短路但不是最后不重边的最短路，然后就这么把边删了显然不对…… 所以我们还是言归正解吧……这道题就是
ZOJ 2760的升级版吧……也是，可以floyd可以源汇点Dij预处理先筛出最短路上的边加入到网络流中，容量限制为2。那么最大流==2就有解，然后沿着满流边dfs 2遍就找到这两条路了~ 次奥卡内存卡的啊……只能用4000KB内存！！最后还是改了两个short int才勉强过了T_T……
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#define MID(x,y) ((x+y)/2) #define mem(a,b) memset(a,b,sizeof(a)) using namespace std; const int MAXV = 405; const int MAXE = 170055; const int oo = 0x3fffffff; short int map[MAXV][MAXV]; int ds[MAXV], dt[MAXV]; void dijkstra(int *dist, int s, int t, int n){ bool vis[MAXV] = {0}; for (int i = 0; i <= n; i ++){ dist[i] = oo; } dist[s] = 0; for(int p = 1; p <= n; p ++){ int minx = oo; int u; for (int i = 1; i <= n; i ++){ if (!vis[i] && dist[i] < minx){ minx = dist[i]; u = i; } } if (minx == oo) break; vis[u] = 1; for (int v = 1; v <= n; v ++){ if (v == u || vis[v] || map[u][v] == 30000) continue; if (dist[v] > dist[u] + map[u][v]){ dist[v] = dist[u] + map[u][v]; } } } } struct node{ short int u, v; int flow; int next; }; struct Dinic{ node arc[MAXE]; int vn, en, head[MAXV]; //vn点个数(包括源点汇点),en边个数 int cur[MAXV]; //当前弧 int q[MAXV]; //bfs建层次图时的队列 int path[MAXE], top; //存dfs当前最短路径的栈 int dep[MAXV]; //各节点层次 void init(int n){ vn = n; en = 0; mem(head, -1); } void insert_flow(int u, int v, int flow){ arc[en].u = u; arc[en].v = v; arc[en].flow = flow; arc[en].next = head[u]; head[u] = en ++; arc[en].u = v; arc[en].v = u; arc[en].flow = 0; //反向弧 arc[en].next = head[v]; head[v] = en ++; } bool bfs(int s, int t){ mem(dep, -1); int lq = 0, rq = 1; dep[s] = 0; q[lq] = s; while(lq < rq){ int u = q[lq ++]; if (u == t){ return true; } for (int i = head[u]; i != -1; i = arc[i].next){ int v = arc[i].v; if (dep[v] == -1 && arc[i].flow > 0){ dep[v] = dep[u] + 1; q[rq ++] = v; } } } return false; } int solve(int s, int t){ int maxflow = 0; while(bfs(s, t)){ int i, j; for (i = 1; i <= vn; i ++) cur[i] = head[i]; for (i = s, top = 0;;){ if (i == t){ int mink; int minflow = 0x3fffffff; for (int k = 0; k < top; k ++) if (minflow > arc[path[k]].flow){ minflow = arc[path[k]].flow; mink = k; } for (int k = 0; k < top; k ++) arc[path[k]].flow -= minflow, arc[path[k]^1].flow += minflow; maxflow += minflow; top = mink; //arc[mink]这条边流量变为0, 则直接回溯到该边的起点即可(这条边将不再包含在增广路内). i = arc[path[top]].u; } for (j = cur[i]; j != -1; cur[i] = j = arc[j].next){ int v = arc[j].v; if (arc[j].flow && dep[v] == dep[i] + 1) break; } if (j != -1){ path[top ++] = j; i = arc[j].v; } else{ if (top == 0) break; dep[i] = -1; i = arc[path[-- top]].u; } } } return maxflow; } }dinic; int pre[MAXV]; bool vis[MAXV]; stack
path; int ok; void dfs(int u, int n){ if (u == n){ ok = 1; while(!path.empty()){ path.pop(); } while(pre[u] != -1){ path.push(u); dinic.arc[pre[u]].flow = 1; dinic.arc[pre[u]^1].flow = 1; u = dinic.arc[pre[u]].u; } return ; } vis[u] = 1; for (int i = dinic.head[u]; i != -1; i = dinic.arc[i].next){ if (i % 2 == 1 || dinic.arc[i].flow != 0) continue; int v = dinic.arc[i].v; if (vis[v]) continue; pre[v] = i; dfs(v, n); if (ok) return ; pre[v] = -1; } return ; } int main(){ //freopen("test.in", "r", stdin); //freopen("test.out", "w", stdout); int n, m; scanf("%d %d", &n, &m); for (int i = 0; i <= n; i ++){ for (int j = 0; j <= n; j ++){ map[i][j] = 30000; } } for (int i = 0; i < m; i ++){ int u, v, w; scanf("%d %d %d", &u, &v, &w); map[u][v] = map[v][u] = w; if (u == v) map[u][v] = 0; } dijkstra(ds, 1, n, n); dijkstra(dt, n, 1, n); dinic.init(n+2); dinic.insert_flow(n+1, 1, 2); dinic.insert_flow(n, n+2, 2); for (int i = 1; i <= n; i ++){ for (int j = 1; j <= n; j ++){ if (i == j) continue; if (ds[i] + map[i][j] + dt[j] == ds[n]){ dinic.insert_flow(i, j, 1); } } } if (dinic.solve(n+1, n+2) == 2){ mem(pre, -1); mem(vis, 0); ok = 0; dfs(1, n); printf("1"); while(!path.empty()){ printf(" %d", path.top()); path.pop(); } puts(""); mem(pre, -1); mem(vis, 0); ok = 0; dfs(1, n); printf("1"); while(!path.empty()){ printf(" %d", path.top()); path.pop(); } puts(""); } else{ puts("No solution"); } return 0; }