SGU 185 Two shortest 最短路+最小费用最大流 或者 最短路+最大流

真心服了此题了

此题最贱之处在于内存只给了4M，也就是你只能开100W左右的int

我开了各种short 最后发现 short会莫名其妙变成4字节的去

然后就杯具了，MLE，开小了就会RE  

这题明显不能用裸费用流去做了，裸费用流加的边太多了。任意两个点之间都来俩边的话一下子就给超内存了。

所以只好用最短路+费用流的方法去做。

不想写最短路+最大流的原因是 因为我的最大流模板太臃肿了 ，虽然至今没被卡过时，但是还是编程复杂度太高

于是决定偷了庄神 lost神牛的ISAP模板一个 http://www.zlinkin.com/?p=34  以后改造一下自己用

目测很短吧，  那是因为使用DFS的原因，而我那个模板是BFS无递归版本的。

#include <iostream>#include <algorithm>#include <cstring>#include <string>#include <cstdio>#include <cmath>#include <queue>#include <map>#include <set>#define eps 1e-5#define MAXN 404#define MAXM 160005#define INF 10007using namespace std;typedef int type;struct EDGE{    int v, next;    type w;} edge[MAXM];int head[MAXN], e;inline void init(){    memset(head, -1, sizeof(head));    e = 0;}inline void add(int u, int v, type w){    edge[e].v = v;    edge[e].w = w;    edge[e].next = head[u];    head[u] = e++;    edge[e].v = u;    edge[e].w = 0;    edge[e].next = head[v];    head[v] = e++;}int n;int h[MAXN];int gap[MAXN];int src, des;inline type dfs(int pos, type cost){    if (pos == des) return cost;    int j, minh = n - 1;    type lv = cost, d;    for (j = head[pos]; j != -1; j = edge[j].next)    {        int v = edge[j].v;        type w = edge[j].w;        if(w > 0)        {            if (h[v] + 1 == h[pos])            {                if (lv < edge[j].w) d = lv;                else d = edge[j].w;                d = dfs(v, d);                edge[j].w -= d;                edge[j ^ 1].w += d;                lv -= d;                if (h[src] >= n) return cost - lv;                if (lv == 0) break;            }            if (h[v] < minh) minh = h[v];        }    }    if (lv == cost)    {        --gap[h[pos]];        if (gap[h[pos]] == 0) h[src] = n;        h[pos] = minh + 1;        ++gap[h[pos]];    }    return cost - lv;}type sap(){    type ret = 0;    memset(gap, 0, sizeof(gap));    memset(h, 0, sizeof(h));    gap[0] = n;    while (h[src] < n) ret += dfs(src, INF);    return ret;}

整个模板里先init初始化，然后src,des, n 自己指定， 一般标号从1开始

 然后这题我先用了一次spfa

将最短路求出来

然后跑一次最小费用最大流，看是否等于最短路的二倍，然后DFS之类的。

建最小费用的图时，注意使用SPFA求出来的最短路边集去建立，就是dis[i] + g[i][j] = dis[j]这样的边

可以想到的是这样的边明显会比整个图的边要少一半以上 而不至于去MLE

#include <iostream>#include <algorithm>#include <cstring>#include <string>#include <cstdio>#include <cmath>#include <queue>#include <map>#include <set>#define eps 1e-5#define MAXN 404#define MAXM 160005#define INF 10007using namespace std;struct EDGE{    short v, cap, cost;    int next;} edge[MAXM];int n, m, ans, flow, src, des;int e, head[MAXN];int que[MAXN], pre[MAXN], dis[MAXN];bool vis[MAXN];short g[MAXN][MAXN];void init(){    e = ans = flow = 0;    memset(head, -1, sizeof(head));}void add(short u, short v, short cap, short cost){    edge[e].v = v;    edge[e].cap = cap;    edge[e].cost = cost;    edge[e].next = head[u];    head[u] = e++;}void addEdge(short u, short v, short cap, short cost){    add(u, v, cap, cost);    add(v, u, 0, -cost);}bool spfa(){    int i, h = 0, t = 1;    for(i = 0; i <= n; i ++)    {        dis[i] = 1000000007;        vis[i] = false;    }    dis[src] = 0;    que[0] = src;    vis[src] = true;    while(t != h)    {        int u = que[h++];        h %= n;        vis[u] = false;        for(i = head[u]; i != -1; i = edge[i].next)        {            int v = edge[i].v;            if(edge[i].cap && dis[v] > dis[u] + edge[i].cost)            {                dis[v] = dis[u] + edge[i].cost;                pre[v] = i;                if(!vis[v])                {                    vis[v] = true;                    que[t++] = v;                    t %= n;                }            }        }    }    if(dis[des] == 1000000007) return false;    return true;}void end(){    int u, p;    short mi = 100;    for(u = des; u != src; u = edge[p ^ 1].v)    {        p = pre[u];        mi = min(mi, edge[p].cap);    }    for(u = des; u != src; u = edge[p ^ 1].v)    {        p = pre[u];        edge[p].cap -= mi;        edge[p ^ 1].cap += mi;        ans += mi * edge[p].cost;     //  cost记录的为单位流量费用，必须得乘以流量。    }    flow += mi;}int nt;void build1(){    short x, y, w;    scanf("%d%d", &nt, &m);    for(int i = 1; i <= m; i++)    {        scanf("%hd%hd%hd", &x, &y, &w);        add(x, y, 1, w);        add(y, x, 1, w);        g[x][y] = g[y][x] = w;    }    src = 1;    des = nt;    n = des;}void build2(){    for(short i = 1; i <= nt; i++)        for(short j = 1; j <= nt; j++)            if(g[i][j] && dis[i] + g[i][j] == dis[j])            {                addEdge(i, j, 1, g[i][j]);               // printf("%d %d\n", i, j);            }    src = nt + 1;    des = nt + 2;    n = des;    addEdge(src, 1, 2, 0);    addEdge(nt, des, 2, 0);}void get(){    init();    build1();    spfa();}void MCMF2(){    init();    build2();    while(spfa()) end();}int flag;void dfs(int u){    printf(" %d", u);    if(u == nt)    {        printf("\n");        flag = 1;        return;    }    for(int i = head[u]; i != -1 && !flag; i = edge[i].next)        if(edge[i].cap == 0 && i % 2 == 0)        {            if(edge[i].v > nt) return;            edge[i].cap = -1;            dfs(edge[i].v);        }}int main(){    get();    int tmp = dis[des];    //printf("%d\n", tmp);    MCMF2();    //printf("%d %d\n", tmp, ans);    if(tmp * 2 == ans)    {        for(int i = head[1]; i != -1; i = edge[i].next)            if(edge[i].cap == 0 && i % 2 == 0)            {                flag = 0;                printf("1");                dfs(edge[i].v);            }    }    else printf("No solution\n");    return 0;}