POJ 1442 Road Construction 最小路径

E - Road Construction

Time Limit:2000MS     Memory Limit:65536KB     64bit IO Format:%I64d & %I64u

Submit Status

Description

It's almost summer time, and that means that it's almost summer construction time! This year, the good people who are in charge of the roads on the tropical island paradise of Remote Island would like to repair and upgrade the various roads that lead between the various tourist attractions on the island.

The roads themselves are also rather interesting. Due to the strange customs of the island, the roads are arranged so that they never meet at intersections, but rather pass over or under each other using bridges and tunnels. In this way, each road runs between two specific tourist attractions, so that the tourists do not become irreparably lost.

Unfortunately, given the nature of the repairs and upgrades needed on each road, when the construction company works on a particular road, it is unusable in either direction. This could cause a problem if it becomes impossible to travel between two tourist attractions, even if the construction company works on only one road at any particular time.

So, the Road Department of Remote Island has decided to call upon your consulting services to help remedy this problem. It has been decided that new roads will have to be built between the various attractions in such a way that in the final configuration, if any one road is undergoing construction, it would still be possible to travel between any two tourist attractions using the remaining roads. Your task is to find the minimum number of new roads necessary.

Input

The first line of input will consist of positive integers n and r, separated by a space, where 3 ≤ n ≤ 1000 is the number of tourist attractions on the island, and 2 ≤ r ≤ 1000 is the number of roads. The tourist attractions are conveniently labelled from 1 to n. Each of the following r lines will consist of two integers, v and w, separated by a space, indicating that a road exists between the attractions labelled v and w. Note that you may travel in either direction down each road, and any pair of tourist attractions will have at most one road directly between them. Also, you are assured that in the current configuration, it is possible to travel between any two tourist attractions.

Output

One line, consisting of an integer, which gives the minimum number of roads that we need to add.

Sample Input

Sample Input 110 121 21 31 42 52 65 63 73 87 84 94 109 10Sample Input 23 31 22 31 3

Sample Output

Output for Sample Input 12Output for Sample Input 20

一个有向无环图，让你用最少的士兵把这个图遍历，每个士兵只能沿一跳路走
ACcode：
#pragma warning(disable:4786)//使命名长度不受限制#pragma comment(linker, "/STACK:102400000,102400000")//手工开栈#include <map>#include <set>#include <queue>#include <cmath>#include <stack>#include <cctype>#include <cstdio>#include <cstring>#include <stdlib.h>#include <iostream>#include <algorithm>#define rd(x) scanf("%d",&x)#define rd2(x,y) scanf("%d%d",&x,&y)#define rds(x) scanf("%s",x)#define rdc(x) scanf("%c",&x)#define ll long long int#define maxn 205#define mod 1000000007#define INF 0x3f3f3f3f //int 最大值#define FOR(i,f_start,f_end) for(int i=f_start;i<=f_end;++i)#define MT(x,i) memset(x,i,sizeof(x))#define PI  acos(-1.0)#define E  exp(1)using namespace std;bool bmap[maxn][maxn];bool bmask[maxn];int pre[maxn];int loop,n,m,x,y;void init(){    MT(bmap,0);    MT(bmask,0);    MT(pre,-1);}int findpath(int u){    FOR(i,1,n)        if(bmap[u][i]&&!bmask[i]){            bmask[i]=1;            if(pre[i]==-1||findpath(pre[i])){                pre[i]=u;                return 1 ;            }        }    return 0;}int Maxmatch(){    int res(0);    FOR(i,1,n){        MT(bmask,false);        res+=findpath(i);    }    return res;}int main(){    rd (loop);    while(loop--){        rd(n);rd(m);init();        FOR(i,1,m){            rd2(x,y);            bmap[x][y]=1;         }        printf("%d\n",n-Maxmatch());    }    return 0;}/*2433 41 32 3331 31 22 3*/