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lightoj 1429 - Assassin`s Creed (II)

2016-01-31 17:26 330 查看
题意就是一个组织要杀人,道路是单向的,一个杀人者可以沿着某条路把所经过的点上的人全部杀完,点是可以重复经过的。

问最少需要几个杀人者可以把所有的n个点上的人杀完。。。

这题的难点在于点可以重复经过,而且图中含有环。。。。。

环其实容易想到缩点,,,只是点重复经过怎么处理,因为要用二分图的性质的话,里面的边石不能重复的,所以这里的处理就是添加边进去,如果原图中u到v是可达的,那么u到v就新建一条边。。。。

这样以来就可以用二分图的性质了。。。。

/*****************************************
Author      :Crazy_AC(JamesQi)
Time        :2015
File Name   :
*****************************************/
// #pragma comment(linker, "/STACK:1024000000,1024000000")
#include <iostream>
#include <algorithm>
#include <iomanip>
#include <sstream>
#include <string>
#include <stack>
#include <queue>
#include <deque>
#include <vector>
#include <map>
#include <set>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <cstdlib>
#include <climits>
using namespace std;
#define MEM(x,y) memset(x, y,sizeof x)
#define pk push_back
typedef long long LL;
typedef unsigned long long ULL;
typedef pair<int,int> ii;
typedef pair<ii,int> iii;
const double eps = 1e-10;
const int inf = 1 << 30;
const int INF = 0x3f3f3f3f;
const int MOD = 1e9 + 7;
/**********************Point*****************************/
struct Point{
double x,y;
Point(double x=0,double y=0):x(x),y(y){}
};
typedef Point Vector;
Vector operator + (Vector A,Vector B){
return Vector(A.x + B.x,A.y + B.y);
}
Vector operator - (Vector A,Vector B){//向量减法
return Vector(A.x - B.x,A.y - B.y);
}
Vector operator * (Vector A,double p){//向量数乘
return Vector(A.x * p,A.y * p);
}
Vector operator / (Vector A,double p){//向量除实数
return Vector(A.x / p,A.y / p);
}
int dcmp(double x){//精度正负、0的判断
if (fabs(x) < eps) return 0;
return x < 0?-1:1;
}
bool operator < (const Point& A,const Point& B){//小于符号的重载
return A.x < B.x || (A.x == B.x && A.y < B.y);
}
bool operator == (const Point& A,const Point& B){//点重的判断
return dcmp(A.x - B.x) == 0&& dcmp(A.y - B.y) == 0;
}
double Dot(Vector A,Vector B){//向量的点乘
return A.x * B.x + A.y * B.y;
}
double Length(Vector A){//向量的模
return sqrt(Dot(A,A));
}
double Angle(Vector A,Vector B){//向量的夹角
return acos(Dot(A,B) / Length(A) / Length(B));
}
double Cross(Vector A,Vector B){//向量的叉积
return A.x * B.y - A.y * B.x;
}
double Area2(Point A,Point B,Point C){//三角形面积
return Cross(B - A,C - A);
}
Vector Rotate(Vector A,double rad){//向量的旋转
return Vector(A.x * cos(rad) - A.y * sin(rad),A.x * sin(rad) + A.y * cos(rad));
}
Vector Normal(Vector A){//法向量
int L = Length(A);
return Vector(-A.y / L,A.x / L);
}
double DistanceToLine(Point p,Point A,Point B){//p到直线AB的距离
Vector v1 = B - A,v2 = p - A;
return fabs(Cross(v1,v2)) / Length(v1);
}
double DistanceToSegment(Point p,Point A,Point B){//p到线段AB的距离
if (A == B) return Length(p - A);
Vector v1 = B - A, v2 = p - A,v3 = p - B;
if (dcmp(Dot(v1,v2) < 0)) return Length(v2);
else if (dcmp(Dot(v1,v3)) > 0) return Length(v3);
else return DistanceToLine(p,A,B);
}
bool SegmentProperIntersection(Point A1,Point A2,Point B1,Point B2){//线段相交
double c1 = Cross(A2 - A1,B1 - A1),c2 = Cross(A2 - A1,B2 - A1);
double c3 = Cross(B2 - B1,A1 - B1),c4 = Cross(B2 - B1,A2 - B1);
return dcmp(c1)*dcmp(c2) < 0 && dcmp(c3)*dcmp(c4) < 0;
}
const int N = 1010;
vector<int> G1
, G2
, G3
;//原图,扩张图,缩点后的图。
bool vis
;
int pre
, low
, Belong
, scc_cnt, Times;
int n, m;
stack<int> st;
void Tarjan(int u) {
pre[u] = low[u] = ++Times;
st.push(u);
for (int i = 0;i < (int)G2[u].size();++i) {
int v = G2[u][i];
if (!pre[v]) {
Tarjan(v);
low[u] = min(low[u], low[v]);
}else if (!Belong[v]) low[u] = min(low[u], pre[v]);
}
if (pre[u] == low[u]) {
scc_cnt++;
while(true) {
int x = st.top();
st.pop();
Belong[x] = scc_cnt;
if (x == u) break;
}
}
}
void FindSCC() {
memset(pre, 0,sizeof pre);
memset(Belong, 0,sizeof Belong);
Times = scc_cnt = 0;
for (int i = 1;i <= n;++i) {
if (!pre[i]) Tarjan(i);
}
}
void BFS(int st) {
queue<int> que;
que.push(st);
memset(vis, false,sizeof vis);
vis[st] = true;
while(!que.empty()) {
int u = que.front();
que.pop();
for (int i = 0;i < G1[u].size();++i) {
int v = G1[u][i];
if (vis[v]) continue;
vis[v] = true;
G2[st].push_back(v);//对原图进行这种扩展是不会形成可行环的。
que.push(v);
}
}
}
void Initation() {
for (int i = 1;i <= n;++i)
BFS(i);
}
void Input() {
scanf("%d%d",&n,&m);
for (int i = 1;i <= n;++i)
G1[i].clear(),G2[i].clear(),G3[i].clear();
int u, v;
for (int i = 1;i <= m;++i) {
scanf("%d%d",&u,&v);
G1[u].push_back(v);
}
}
void Trans() {
for (int u = 1;u <= n;++u) {
for (int i = 0;i < G2[u].size();++i) {
int v = G2[u][i];
if (Belong[u] != Belong[v])
G3[Belong[u]].push_back(Belong[v]);
}
}
}
int linker
;
bool Search(int u) {
for (int i = 0;i < G3[u].size();++i) {
int v = G3[u][i];
if (vis[v]) continue;
vis[v] = true;
if (linker[v] == -1 || Search(linker[v])) {
linker[v] = u;
return true;
}
}
return false;
}
int Hungary() {
int ret = 0;
memset(linker, -1,sizeof linker);
for (int i = 1;i <= scc_cnt;++i) {
memset(vis, false,sizeof vis);
if (Search(i)) ret++;
}
return ret;
}
int main()
{
freopen("in.txt","r",stdin);
// freopen("out.txt","w",stdout);
int t, icase = 0;
scanf("%d",&t);
while(t--) {
Input();
Initation();
FindSCC();
// printf("SCC = %d\n", scc_cnt);
// for (int i = 1;i <= n;++i)
// printf("%d ", Belong[i]);
// puts("");
Trans();

// for (int i = 1;i <= n;++i) {
// printf("<%d>::", i);
// for (int j = 0;j < G2[i].size();++j)
// printf("%d ", G2[i][j]);
// puts("");
// }

// for (int i = 1;i <= scc_cnt;++i) {
// 	printf("i = %d::::", i);
// 	for (int j = 0;j < G3[i].size();++j)
// 		printf("%d ",G3[i][j]);
// 	puts("");
// }
printf("Case %d: %d\n", ++icase, scc_cnt - Hungary());
}
return 0;
}
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