HDU3416 Marriage Match IV(最大流+最短路)

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城市A和B之间有一些有向边，求A，B之间的走最短路有多少种方法，即点可以重复走，而边不能。
首先求最短路，然后把在最短路上的边拖到新图里，容量为1，求最大流。判断一条边是否在最短路上：d1[edge[i].v]+d2[edge[i].v]+edge[i].w=d1[t] , d1[v]是起点到v的最短路，d2[v]是终点到v的最短路。也就是说在正向图里做一遍SPFA或dijkstra求出d1，再在反向图里以t为起点做一次求出d2。

#include<cstdio>#include<cstring>#include<queue>#include<vector>#define MAXN 1010#define MAXM  200010using namespace std;inline int Min(int a,int b){return a<b?a:b;}struct E{    int v,w,op;    E(){}    E(int a,int b,int c)    {v = a; w = b; op = c;}};vector<E> g[MAXM];struct Edge{    int u,v,w,next;}edge[MAXM],edge2[MAXM];//把MAXM打成MAXN，调了一晚上，真是R了整个动物园了！int cnt,head[MAXN],cnt2,head2[MAXN];void add_edge(int u,int v,int w){    edge[cnt].u = u;    edge[cnt].v = v;    edge[cnt].w = w;    edge[cnt].next = head[u];    head[u] = cnt++;}void add_edge2(int u,int v,int w){    edge2[cnt2].u = u;    edge2[cnt2].v = v;    edge2[cnt2].w = w;    edge2[cnt2].next = head2[u];    head2[u] = cnt2++;}int n,m,s,t,d1[MAXN],d2[MAXN],a,b,c;void init(){    memset(head,-1,sizeof head);    memset(head2,-1,sizeof head2);    cnt = cnt2 = 0;}bool inque[MAXN];void SPFA1(){    memset(d1,0x3f,sizeof d1);    memset(inque,0,sizeof inque);    queue<int> Q;    inque[s] = 1,d1[s] = 0;    Q.push(s);    while(!Q.empty())    {        int u = Q.front();        Q.pop(),inque[u] = 0;        for(int i = head[u]; i != -1; i = edge[i].next)        {            int v = edge[i].v;            if(d1[u]+edge[i].w < d1[v])            {                    d1[v] = d1[u]+edge[i].w;                if(!inque[v]) inque[v] = 1,Q.push(v);            }        }    }}void SPFA2(){    memset(d2,0x3f,sizeof d2);    memset(inque,0,sizeof inque);    queue<int> Q;    inque[t] = 1,d2[t] = 0;    Q.push(t);    while(!Q.empty())    {        int u = Q.front();        Q.pop(),inque[u] = 0;        for(int i = head2[u]; i != -1; i = edge2[i].next)        {            int v = edge2[i].v;            if(d2[u]+edge2[i].w < d2[v])            {                d2[v] = d2[u]+edge2[i].w;                if(!inque[v]) inque[v] = 1,Q.push(v);            }        }    }}int flow,d[MAXN],vd[MAXN];int aug(int i,int augco){    int j,augc = augco,mind = n-1,delta,sz = g[i].size();    if(i == t) return augco;    for(j = 0; j < sz; j++)    {        int v = g[i][j].v;        if(g[i][j].w)        {            if(d[i] == d[v]+1)            {                delta = Min(augc,g[i][j].w);                delta = aug(v,delta);                g[i][j].w -= delta;                g[v][g[i][j].op].w += delta;                augc -= delta;                if(d[s] >= n) return augco - augc;                if(augc == 0) break;            }            if(d[v] < mind) mind = d[v];        }    }    if(augco == augc)    {        vd[d[i]]--;        if(vd[d[i]] == 0) d[s] = n;        d[i] = mind+1;        vd[d[i]]++;    }    return augco - augc;}void sap(){    flow = 0;    memset(d,0,sizeof d);    memset(vd,0,sizeof vd);    vd[0] = n;    while(d[s] < n)        flow += aug(s,0x3f3f3f3f);}int main(){    int T;    scanf("%d",&T);    while(T--)    {        scanf("%d%d",&n,&m);        init();        for(int i = 1; i <= m; i++)        {            scanf("%d%d%d",&a,&b,&c);            add_edge(a,b,c);            add_edge2(b,a,c);        }        scanf("%d%d",&s,&t);        SPFA1();        SPFA2();        for(int i = 0 ; i < cnt; i++)        {                int u = edge[i].u,v = edge[i].v;            if(d1[u]+d2[v]+edge[i].w == d1[t])            {                g[u].push_back(E(v,1,g[v].size()));                g[v].push_back(E(u,0,g[u].size()-1));            }        }        sap();        printf("%d\n",flow);        for(int i = 1; i <= n; i++)            g[i].clear();    }}

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