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题意：
最少破坏多少条路，所有的最短路都无法走通。
最多破环多少条路，任然有一条最短路可以走。
分析：对于第二个问题，求出最短路路径数最小的一条即可，第一个问题，用最短路径上所有的边构建网络，然后求出最大流即为所求，因为这题边很多，所以用EK会超时，换成dinic就能AC。
ps：构建网络写的太乱了，不过好歹A了，需要好好整理一下，写一个更清晰的版本。
以下附上代码：
#include <algorithm>
#include <iostream>
#include <sstream>
#include <fstream>
#include <cstring>
#include <cstdio>
#include <vector>
#include <cctype>
#include <cmath>
#include <stack>
#include <queue>
#include <list>
#include <map>
#include <set>using namespace std;typedef long long ll;
typedef unsigned long long ull;const int maxn = 2005;
const int maxm = 600005;
const int inf = 1000000000;int n,m;struct Node{int v,w;int c,f;int next;bool use;
};struct Adlist{int vex[maxn];Node arc[maxm*2];int cnt;void Clear(){for(int i = 0; i <= n; i++){vex[i] = -1;}cnt = 0;}void Insert(int u, int v, int w){ arc[cnt].v = v;arc[cnt].w = w;arc[cnt].next = vex[u];arc[cnt].use = 0;vex[u] = cnt;++cnt;}void Insert(int u, int v){ arc[cnt].v = v;arc[cnt].c = 1;arc[cnt].f = 0;arc[cnt].next = vex[u];arc[cnt].use = 0;vex[u] = cnt;++cnt;}void Augment(int u, int v, int flow){for(int i = vex[u]; i != -1; i = arc[i].next){if(arc[i].v == v && arc[i].f < arc[i].c && arc[i].f + flow < arc[i].c && arc[i].f + flow >= 0){arc[i].f += flow;return;}}}int Have(int u, int v){for(int i = vex[u]; i != -1; i = arc[i].next){if(arc[i].v == v && !arc[i].use){return i;}}return -1;}
};Adlist e;
Adlist Network;
//for spfa
queue<int> q;
int low[maxn];
vector<int> path[maxn];//记录所有路径
int cnt[maxn];//到每个点的最短路径有多少种走法
//for maxFlow
int vis[maxn];
int d[maxn];//层次网路 void input()
{int u,v,w;e.Clear();for(int i = 0; i < m; i++){scanf("%d%d%d",&u,&v,&w);e.Insert(u,v,w);e.Insert(v,u,w);}
}void spfa(int s)
{for(int i = 0; i <= n; i++) low[i] = inf;low[s] = 0,cnt[s] = 0;path[s].clear();q.push(s);int u,v;while(!q.empty()){u = q.front(); q.pop();for(int i = e.vex[u]; i != -1; i = e.arc[i].next){v = e.arc[i].v;if(low[u] + e.arc[i].w < low[v]){low[v] = low[u] + e.arc[i].w;cnt[v] = cnt[u] + 1;path[v].clear();path[v].push_back(u);q.push(v);}else if(low[u] + e.arc[i].w == low[v]){cnt[v] = min(cnt[v],cnt[u]+1);path[v].push_back(u);}} }if(low[n] == inf) cnt[n] = 0;}void build(int t)//构建网络
{int u,v;int pos;for(int i = 0; i < path[t].size(); i++){u = path[t][i]; v = t;pos = e.Have(u,v);if(pos != -1){ Network.Insert(u,v);e.arc[pos].use = 1;}build(u); }
}void print()//
{int u,v;for(int i = 1; i <= n; i++){u = i;cout << u << " : ";for(int j = Network.vex[i]; j != -1; j = Network.arc[j].next){v = Network.arc[j].v;cout << v << " " << Network.arc[j].f << " f ";}cout << endl;}
}int bfs()//bfs构建层次网络
{fill(vis,vis+n+1,0);q.push(1);d[1] = 0;vis[1] = 1;int u,v;while(!q.empty()){u = q.front(); q.pop();for(int i = Network.vex[u]; i != -1; i = Network.arc[i].next){v = Network.arc[i].v;if(!vis[v] && Network.arc[i].f < Network.arc[i].c){// vis[v] = 1;d[v] = d[u] + 1;q.push(v) ;}}}return vis[n];}int dfs(int u, int alpha)//dinic
{if(u == n || alpha == 0) return alpha;int v;int sum = 0;for(int i = Network.vex[u]; i != -1; i = Network.arc[i].next){v = Network.arc[i].v;int flow;if(d[v] == d[u] + 1){flow = dfs(v,min(alpha-sum,Network.arc[i].c-Network.arc[i].f));if(flow > 0){ Network.arc[i].f += flow;Network.Augment(v,u,-flow); sum += flow;if(sum == alpha) break;}}}return sum;
}int maxFlow()//求最大流
{int maximum = 0;int flow;while(bfs()){fill(vis,vis+n+1,0);maximum += dfs(1,inf);}return maximum;
}int main()
{while(scanf("%d%d",&n,&m) != EOF){input();spfa(1);Network.Clear();build(n);printf("%d %d\n",maxFlow(),m-cnt[n]);}return 0;
}
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