Harry Potter and the Forbidden Forest(割边最小的最小割)

Harry Potter and the Forbidden Forest

Time Limit:3000MS    Memory Limit:65536KB    64bit IO Format:%I64d & %I64u

Description

Harry Potter notices some Death Eaters try to slip into Castle. The Death Eaters hide in the most depths of Forbidden Forest. Harry need stop them as soon as.

The Forbidden Forest is mysterious. It consists of N nodes numbered from 0 to N-1. All of Death Eaters stay in the node numbered 0. The position of Castle is node n-1. The nodes connected by some roads. Harry need block some roads by magic and he want to minimize the cost. But it’s not enough, Harry want to know how many roads are blocked at least.

 

Input

Input consists of several test cases.

The first line is number of test case.

Each test case, the first line contains two integers n, m, which means the number of nodes and edges of the graph. Each node is numbered 0 to n-1.

Following m lines contains information about edges. Each line has four integers u, v, c, d. The first two integers mean two endpoints of the edges. The third one is cost of block the edge. The fourth one means directed (d = 0) or undirected (d = 1).

Technical Specification

1. 2 <= n <= 1000
2. 0 <= m <= 100000
3. 0 <= u, v <= n-1
4. 0 < c <= 1000000
5. 0 <= d <= 1

 

Output

For each test case:
Output the case number and the answer of how many roads are blocked at least.

 

Sample Input

     
     
     
     

      
      
      
      3

4 5
0 1 3 0
0 2 1 0
1 2 1 1
1 3 1 1
2 3 3 1

6 7
0 1 1 0
0 2 1 0
0 3 1 0
1 4 1 0
2 4 1 0
3 5 1 0
4 5 2 0

3 6
0 1 1 0
0 1 2 0
1 1 1 1
1 2 1 0
1 2 1 0
2 1 1 1 
     
     
     
     

 

Sample Output

   
   
   
   

    
    
    
    Case 1: 3
Case 2: 2
Case 3: 2 

题意: 
      给出一个混合图,求要使s不能到达t,至少应该删除哪些边,使得删除的边的边权和最小,而且要求删除的边数达到最小;

题解:
          很典型的最小割题目,不过题目在最小割的基础上添加了一个条件,就是要求割边尽可能的小. 首先我们可以求出混合图的最大流,最小割等于最大流,
      然后对于割边的处理,可以这样做.将每条边的边权加1,再跑一次最大流. 这样可以求出割边最小的那个割集,因为割边越多,最小割会越大,假设存在割边
      更小的割集,那么就你跑最大流就会跑出更小的流量出来. 最后直接两次最大流的结果相减即为答案;

AC代码:

    
    
    
    
/* ***********************************************
Author        :xdlove
Created Time  :2015年07月31日 星期五 12时37分29秒
File Name     :a.cpp
 ************************************************ */

#include <stdio.h>
#include <string.h>
#include <iostream>
#include <algorithm>
#include <vector>
#include <queue>
#include <set>
#include <map>
#include <string>
#include <math.h>
#include <stdlib.h>
#include <time.h>
using namespace std;

/**宏定义类
 * **/
#define FOR(i,s,t) for(int i = (s); i < (t); i++)
#define FOR_REV(i,s,t) for(int i = (s - 1); i >= (t); i--)
#define mid ((l + r) >> 1)
#define lson l,mid,u<<1
#define rson mid+1,r,u<<1|1
#define ls u<<1
#define rs u<<1|1

typedef long long ll;
typedef unsigned long long ull;
const int INF = 0x3f3f3f3f;
const double pi = acos(-1.0);

/**输入输出挂类模板
 * **/

class Fast
{
    public:
        inline void rd(int &ret)
        {
            char c;
            int sgn;
            if(c = getchar(),c == EOF) return;
            while(c != '-' && (c < '0' || c > '9')) c = getchar();
            sgn = (c == '-') ? -1 : 1;
            ret = (c == '-') ? 0 : (c - '0');
            while(c = getchar(),c >= '0' && c <= '9')
                ret = ret * 10 + c - '0';
            ret *= sgn;
        }

    public:
        inline void pt(int x)
        {
            if(x < 0)
            {
                putchar('-');
                x = -x;
            }
            if(x > 9) pt(x / 10);
            putchar(x % 10 + '0');
        }
};
Fast in,out;

const int MAXN = 1e3 + 5;
const int MAXM = 1e5 * 2 + 5;


struct Node
{
    int from,to,next;
    int cap;
}edge[MAXM * 2];
int tol;
int Head[MAXN];
int que[MAXN];
int dep[MAXN]; //dep为点的层次
int stack[MAXN];//stack为栈，存储当前增广路
int cur[MAXN];//存储当前点的后继

void Init()
{
    tol = 0;
    memset(Head,-1,sizeof(Head));
}

void add_edge(int u, int v, int w)
{
    edge[tol].from = u;
    edge[tol].to = v;
    edge[tol].cap = w;
    edge[tol].next = Head[u];
    Head[u] = tol++;
    edge[tol].from = v;
    edge[tol].to = u;
    edge[tol].cap = 0;
    edge[tol].next = Head[v];
    Head[v] = tol++;
}

int BFS(int start, int end)
{
    int front, rear;
    front = rear = 0;
    memset(dep, -1, sizeof(dep));
    que[rear++] = start;
    dep[start] = 0;
    while (front != rear)
    {
        int u = que[front++];
        if (front == MAXN)front = 0;
        for (int i = Head[u]; i != -1; i = edge[i].next)
        {
            int v = edge[i].to;
            if (edge[i].cap > 0 && dep[v] == -1)
            {
                dep[v] = dep[u] + 1;
                que[rear++] = v;
                if (rear >= MAXN)rear = 0;
                if (v == end)return 1;
            }
        }
    }
    return 0;
}

int dinic(int start, int end)
{
    int res = 0;
    int top;
    while (BFS(start, end))
    {
        memcpy(cur, Head, sizeof(Head));
        int u = start;
        top = 0;
        while (true)
        {
            if (u == end)
            {
                int min = INF;
                int loc;
                for (int i = 0; i < top; i++)
                    if (min > edge[stack[i]].cap)
                    {
                        min = edge[stack[i]].cap;
                        loc = i;
                    }
                for (int i = 0; i < top; i++)
                {
                    edge[stack[i]].cap -= min;
                    edge[stack[i] ^ 1].cap += min;
                }
                res += min;
                top = loc;
                u = edge[stack[top]].from;
            }
            for (int i = cur[u]; i != -1; cur[u] = i = edge[i].next)
                if (edge[i].cap != 0 && dep[u] + 1 == dep[edge[i].to])
                    break;
            if (cur[u] != -1)
            {
                stack[top++] = cur[u];
                u = edge[cur[u]].to;
            }
            else
            {
                if (top == 0)break;
                dep[u] = -1;
                u = edge[stack[--top]].from;
            }
        }
    }
    return res;
}

struct RNode
{
    int u,v,c,d;
    RNode(int u,int v,int c,int d) : u(u),v(v),c(c),d(d) {}
    RNode(){}
}p[MAXM];

void Creatgraph(int m)
{
    Init();
    FOR(i, 0, m)
    {
       add_edge(p[i].u,p[i].v,p[i].c + 1);
       if(p[i].d) 
           add_edge(p[i].v,p[i].u,p[i].c + 1);
    }
}



int main()
{
    //freopen("in.txt","r",stdin);
    //freopen("out.txt","w",stdout);
    int T,cnt = 0;
    cin>>T;
    while(T--)
    {
        int n,m;
        Init();
        scanf("%d %d",&n,&m);
        FOR(i, 0, m)
        {
            int u,v,c,d;
            in.rd(u),in.rd(v),in.rd(c),in.rd(d);
            add_edge(u,v,c);
            if(d)
                add_edge(v,u,c);
            p[i] = RNode(u,v,c,d);
        }
        int ans = dinic(0,n - 1);
        //cout<<ans<<endl;
        Creatgraph(m);
        printf("Case %d: %d\n",++cnt,dinic(0,n - 1) - ans);
    }
    return 0;
}