hdu3416(最短路+最大流)

最短路+最大流
用Spfa算出 s到各个点的最短路 t到各个点的最短路
if( dis1[i] + dis2[i] + map[i][j] ==dis1[t] )
满足这种情况说明边在最短路上，所以根据这个方法建边

然后最大流解决

这道题很值得体会体会

/** this code is made by LinMeiChen* Problem: hdu 3416 * Type of Problem: 网络流 水题* Thinking:* Feeling:*/#include<iostream>#include<algorithm>#include<stdlib.h>#include<string.h>#include<stdio.h>#include<math.h>#include<string>#include<vector>#include<queue>#include<list>using namespace std;typedef long long lld;typedef unsigned int ud;#define oo 0x3f3f3f3f#define eatline() char chch;while((chch=getchar())!='\n')continue;#define MemsetMax(a) memset(a,0x3f,sizeof a)#define MemsetZero(a) memset(a,0,sizeof a)#define MemsetMin(a) memset(a,-1,sizeof a)#define MemsetFalse(a) MemsetZero(a)#define PQ priority_queue#define Q queue#define maxn 1008#define maxm 200008struct Edge{    int v, f, next;}E[maxm];int level[maxn], h[maxn];int stack[maxn];int head[maxn],mark[maxn];int dis1[maxn], dis2[maxn];int map[maxn][maxn];int c[maxn][maxn];int q[maxm], front, rear;int tol, n, m;void Spfa(int s, int t, int dis[], int f){    fill(dis, dis + maxn, oo);    memset(mark, 0, sizeof mark);    front = rear = 0;    q[rear++] = s;    dis[s] = 0;    mark[s] = 1;    while (front<rear)    {        int u = q[front++];        mark[u] = 0;        for (int i = 1; i <= n; i++)        {            int v = i;            if (dis[v] > dis[u] + ( f==0 ? map[u][v]:map[v][u]))            {                dis[v] = dis[u] + ( f==0 ? map[u][v]:map[v][u]);                if (!mark[v])                {                    q[rear++] = v;                    mark[v] = 1;                }            }        }    }}void add_edge(int u, int v, int f){    E[tol].v = v;    E[tol].f = f;    E[tol].next = head[u];    head[u] = tol++;    E[tol].v = u;    E[tol].f = 0;    E[tol].next = head[v];    head[v] = tol++;}void Build(int s, int t){    memset(head, -1, sizeof head);    tol = 0;    for (int i = 1; i <= n;i++)    for (int j = 1; j <= n; j++)    {        if (dis1[i] + dis2[j] + map[i][j] == dis1[t])            add_edge(i, j, c[i][j]);    }}bool BFS(int s, int t){    memset(level, 0, sizeof level);    level[s] = 1;    front = rear = 0;    q[rear++] = s;    while (front < rear)    {        int u = q[front++];        if (u == t)            return true;        for (int i = head[u]; i != -1; i = E[i].next)        {            int v = E[i].v;            if (!level[v] && E[i].f>0)            {                level[v] = level[u] + 1;                q[rear++] = v;            }        }    }    return false;}int Dinic(int s, int t){    int maxflow = 0;    while (BFS(s, t))    {        memcpy(h, head, sizeof h);        int top = 0, u = s;        while (true)        {            if (u == t)            {                int minflow = oo, flag;                for (int i = 0; i < top; i++)                {                    if (minflow>E[stack[i]].f)                    {                        minflow = E[stack[i]].f;                        flag = i;                    }                }                for (int i = 0; i < top; i++)                {                    E[stack[i]].f -= minflow;                    E[stack[i] ^ 1].f += minflow;                }                top = flag;                maxflow += minflow;                u = E[stack[top] ^ 1].v;//反向边的邻接点            }            for (int i = h[u]; i != -1; i = h[u] = E[i].next)            {                int v = E[i].v;                if (level[v] == level[u] + 1 && E[i].f)                    break;            }            if (h[u] != -1)            {                stack[top++] = h[u];                u = E[h[u]].v;            }            else            {                if (top == 0)                    break;                level[u] = 0;                u = E[stack[--top] ^ 1].v;//反向边的邻接点            }        }    }    return maxflow;}int main(){    int T;    int s, t;    scanf("%d", &T);    while (T--)    {        memset(map, 0x3f, sizeof map);        memset(c, 0, sizeof c);        scanf("%d%d", &n, &m);        for (int i = 1; i <= m; i++)        {            int u, v, w;            scanf("%d%d%d", &u, &v, &w);            if (map[u][v] > w)            {                map[u][v] = w;                c[u][v] = 1;            }            else if (map[u][v] == w)                c[u][v]++;        }        scanf("%d%d", &s, &t);        Spfa(s, t, dis1,0);        Spfa(t, s, dis2,1);        Build(s, t);        printf("%d\n", Dinic(s, t));    }    return 0;}