HDU5294(最短路，最大流)

题意是给一个n点m边的无向图，求最少去掉多少边使得最短路不存在，和最短路存在的情况下最多去掉多少边。

把所有最短路上的边搞进来，流量为1跑最大流，去掉多少边只要搞个DP就能找到最短路图上得最短的路，用m去减就行。

因为n<=2000，n^2复杂度随便搞都行。

坑点是有重边。

#include 
#include 
#include 
#include 
#include 
#include 
#include 
using namespace std;
#define maxn 2111
#define maxm 121111
#define INF 1000000000

struct node {
    int from, to, next, f;
} edge[maxm], edge1[maxm];
int cnt;
int head[maxn];
int n, m, u, v, w;
int mp[maxn][maxn], num[maxn][maxn]; //记录重边
long long d1[maxn], d2[maxn];
bool vis[maxn];

void init () {
    memset (head, -1, sizeof head);
    memset (num, 0, sizeof num);
    cnt = 0;
    for (int i = 1; i <= n; i++) {
        for (int j = 1; j <= n; j++) {
            if (i == j)
                mp[i][j] = 0;
            else 
                mp[i][j] = INF;
        }
    }
}

void dij (int from, int to) {
    memset (vis, 0, sizeof vis);
    for (int i = 1; i <= n; i++) 
        d1[i] = (i == from ? 0 : INF);
    for (int i = 1; i <= n; i++) {
        long long cur = INF;
        int x;
        for (int j = 1; j <= n; j++) if (!vis[j] && d1[j] <= cur) {
            cur = d1[j];
            x = j;
        }
        vis[x] = 1;
        for (int j = 1; j <= n; j++) {
            d1[j] = min (d1[j], d1[x]+mp[x][j]);
        }
    }
    return ;
}

void anti_dij (int from, int to) {
    memset (vis, 0, sizeof vis);
    for (int i = 1; i <= n; i++) 
        d2[i] = (i == from ? 0 : INF);
    for (int i = 1; i <= n; i++) {
        long long cur = INF;
        int x;
        for (int j = 1; j <= n; j++) if (!vis[j] && d2[j] <= cur) {
            cur = d2[j];
            x = j;
        }
        vis[x] = 1;
        for (int j = 1; j <= n; j++) {
            d2[j] = min (d2[j], d2[x]+mp[x][j]);
        }
    }
    return ;
}


void add_edge (int from, int to, int cap, int i) {
    edge[i].from = from;
    edge[i].to = to;
    edge[i].next = head[from];
    edge[i].f = cap; //边上剩余流量
    head[from] = i;
    return ;
}

int d[maxn]; //从起点到i的距离
bool vis_edge[maxm];

bool bfs (int s, int t) {
    memset (d, -1, sizeof (d));
    queue  gg;
    while (!gg.empty ())
        gg.pop ();
    gg.push (s);
    d[s] = 0;
    while (!gg.empty ()) {
        int x = gg.front (); gg.pop ();
        if (x == t)
            return 1;
        for (int i = head[x]; i != -1; i = edge[i].next) {
            node &e = edge[i];
            if (d[e.to] == -1 && e.f) {
                d[e.to] = d[x] + 1;
                gg.push (e.to);
            }
        }
    }
    return 0;
}

int dfs (int u, int a, int t) { //来源 最短边 汇点
    if (u == t || a == 0)
        return a;
    int flow = 0, f; //汇点增加的流量 最短边
    for (int i = head[u]; i != -1; i = edge[i].next) {
        node &e = edge[i];
        if (e.f && d[u] == d[e.to]-1) {
            f = dfs (e.to, min (a-flow, e.f), t);
            e.f -= f;
            edge[i^1].f += f; 
            flow += f;
            if (flow == a)
                return flow;
        }
    }
    return flow;
}

long long maxflow (int s, int t) {
    long long ans = 0;
    while (bfs (s, t)) {
        ans += dfs (s, INF, t);
    }
    return ans;
}

int cnt1;
int head1[maxn];
int son[maxn];

void dfs1 (int u) {
    son[u] = INF;
    int child = 0;
    for (int i = head1[u]; i != -1; i = edge1[i].next) {
        int v = edge1[i].to; child++;
        dfs1 (v);
        son[u] = min (son[u], son[v]+1);
    }
    if (child == 0)
        son[u] = 0;
}

void add_edge1 (int from, int to, int i) {
    edge1[i].from = from, edge1[i].to = to, edge1[i].next = head1[from], head1[from] = i;
}


int main () {
    while (scanf ("%d%d", &n, &m) == 2) {
        init ();
        for (int i = 1; i <= m; i++) {
            scanf ("%d%d%d", &u, &v, &w);
            if (mp[u][v] > w) {
                mp[u][v] = mp[v][u] = w;
                num[u][v] = num[v][u] = 1;
            }
            else if (mp[u][v] == w) {
                num[u][v]++;
                num[v][u]++;
            }
        }
        dij (1, n); 
        anti_dij (n, 1);
        long long Min = d1[n];
        cnt1 = 0;
        memset (head1, -1, sizeof head1);
        for (int i = 1; i <= n; i++) {
            for (int j = 1; j <= n; j++) {
                if (i == j)
                    continue;
                if (mp[i][j] != INF) {
                    if (d1[i]+d2[j]+mp[i][j] == Min) { 
                        for (int t = 1; t <= num[i][j]; t++) {
                            add_edge (i, j, 1, cnt++);
                            add_edge (j, i, 0, cnt++);
                            add_edge1 (i, j, cnt1++);
                        }
                    }
                }
            }
        }
        dfs1 (1); 
        printf ("%lld %d\n", maxflow (1, n), m-son[1]);
    }
    return 0;
}
/*
6 6 
1 2 1
2 3 2
3 6 3
1 4 3
4 5 2
5 6 1
2 2 
1 2 1
1 2 1
8 9
1 2 1
2 3 1
3 4 1
4 8 1
2 7 1
1 5 1
5 6 1
6 7 1
7 8 1
*/