SGU 185 Two shortest ★(最短路+网络流)

【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……  

#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#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;
}