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POJ 3308 Paratroopers (二分图最小点权覆盖 -> 最小割 -> 最大流)

时间:2014-08-29 20:06:28      阅读:404      评论:0      收藏:0      [点我收藏+]

标签:网络流   最小割   图论   poj   二分图最小点权覆盖   

POJ 3308 Paratroopers


题意:有一个N*M的方阵,有L个伞兵降落在方阵上。现在要将所有的伞兵都消灭掉,可以在每行每列装一个高射炮,如果在某行(某列)装上高射炮之后,能够消灭所有落在该行(该列)的伞兵。每行每列安高射炮有费用,问如何安装能够使得费用之积最小。

思路:首先题目要求乘积最小,将乘积对e取对数,会发现就变成了求和。然后抽象出一个二分图,每一行是x部的一个点,每个点有权值,权值为费用取ln。每一列是y部的一点,费用计算相同。如果有伞兵降落在方格上,那么将x部与y部连边。问题就成了求该二分图的最小点权覆盖。通过求最小割即可得到。而最小割就是最大流。所以构造一个流图。建立一个源点,源点向每一行连边,流量为费用,建立一个汇点,每列向汇点连边,流量为费用。二分图中原先的边的流量是INF。最后答案就是exp(最大流)。

细节:INF不能取太大,否则精度会有问题。

代码:
/*
ID: wuqi9395@126.com
PROG:
LANG: C++
*/
#include<map>
#include<set>
#include<queue>
#include<stack>
#include<cmath>
#include<cstdio>
#include<vector>
#include<string>
#include<fstream>
#include<cstring>
#include<ctype.h>
#include<iostream>
#include<algorithm>
using namespace std;
#define LINF (1LL<<60)
#define INF 1e8
#define PI acos(-1.0)
#define mem(a, b) memset(a, b, sizeof(a))
#define rep(i, a, n) for (int i = a; i < n; i++)
#define per(i, a, n) for (int i = n - 1; i >= a; i--)
#define eps 1e-6
#define debug puts("===============")
#define pb push_back
//#define mp make_pair
#define all(x) (x).begin(),(x).end()
#define fi first
#define se second
#define SZ(x) ((int)(x).size())
#define POSIN(x,y) (0 <= (x) && (x) < n && 0 <= (y) && (y) < m)
typedef long long ll;
typedef unsigned long long ULL;
const int maxn = 110;
const int maxm = 20000;
int st, ed, n, m, l;
struct node {
    int v;    // vertex
    double cap;    // capacity
    double flow;   // current flow in this arc
    int nxt;
} e[maxm * 2];
int g[maxn], cnt;
void add(int u, int v, double c) {
    e[++cnt].v = v;
    e[cnt].cap = c;
    e[cnt].flow = 0;
    e[cnt].nxt = g[u];
    g[u] = cnt;

    e[++cnt].v = u;
    e[cnt].cap = 0;
    e[cnt].flow = 0;
    e[cnt].nxt = g[v];
    g[v] = cnt;
}
void init() {
    mem(g, 0);
    cnt = 1;
    st = 0;
    ed = m + n + 1;
    double w;
    int u, v;
    for (int i = 1; i <= m; i++) {
        scanf("%lf", &w);
        add(st, i, log(w));
    }
    for (int i = m + 1; i < ed; i++) {
        scanf("%lf", &w);
        add(i, ed, log(w));
    }
    for (int i = 1; i <= l; i++) {
        scanf("%d%d", &u, &v);
        add(u, m + v, INF);
    }
    n = ed + 3;
}

int dist[maxn], numbs[maxn], q[maxn];
void rev_bfs() {
    int font = 0, rear = 1;
    for (int i = 0; i <= n; i++) { //n为总点数
        dist[i] = maxn;
        numbs[i] = 0;
    }
    q[font] = ed;
    dist[ed] = 0;
    numbs[0] = 1;
    while(font != rear) {
        int u = q[font++];
        for (int i = g[u]; i; i = e[i].nxt) {
            if (e[i ^ 1].cap == 0 || dist[e[i].v] < maxn) continue;
            dist[e[i].v] = dist[u] + 1;
            ++numbs[dist[e[i].v]];
            q[rear++] = e[i].v;
        }
    }
}
double maxflow() {
    rev_bfs();
    int u;
    double totalflow = 0;
    int curg[maxn], revpath[maxn];
    for(int i = 0; i <= n; ++i) curg[i] = g[i];
    u = st;
    while(dist[st] < n) {
        if(u == ed) {   // find an augmenting path
            double augflow = INF;
            for(int i = st; i != ed; i = e[curg[i]].v)
                augflow = min(augflow, e[curg[i]].cap);
            for(int i = st; i != ed; i = e[curg[i]].v) {
                e[curg[i]].cap -= augflow;
                e[curg[i] ^ 1].cap += augflow;
                e[curg[i]].flow += augflow;
                e[curg[i] ^ 1].flow -= augflow;
            }
            totalflow += augflow;
            u = st;
        }
        int i;
        for(i = curg[u]; i; i = e[i].nxt)
            if(e[i].cap > 0 && dist[u] == dist[e[i].v] + 1) break;
        if(i) {   // find an admissible arc, then Advance
            curg[u] = i;
            revpath[e[i].v] = i ^ 1;
            u = e[i].v;
        } else {    // no admissible arc, then relabel this vertex
            if(0 == (--numbs[dist[u]])) break;    // GAP cut, Important!
            curg[u] = g[u];
            int mindist = n;
            for(int j = g[u]; j; j = e[j].nxt)
                if(e[j].cap > 0) mindist = min(mindist, dist[e[j].v]);
            dist[u] = mindist + 1;
            ++numbs[dist[u]];
            if(u != st)
                u = e[revpath[u]].v;    // Backtrack
        }
    }
    return totalflow;
}

int main () {
    int T;
    scanf("%d", &T);
    while(T--) {
        scanf("%d%d%d", &m, &n, &l);
        init();
        printf("%.4f\n", exp(maxflow()));
    }
    return 0;
}


POJ 3308 Paratroopers (二分图最小点权覆盖 -> 最小割 -> 最大流)

标签:网络流   最小割   图论   poj   二分图最小点权覆盖   

原文地址:http://blog.csdn.net/sio__five/article/details/38929703

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