SD第九届省赛B题 Bullet

最新推荐文章于 2025-05-14 14:35:28 发布

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最新推荐文章于 2025-05-14 14:35:28 发布

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原文链接：http://www.cnblogs.com/WTSRUVF/p/10715685.html

本文介绍了一种解决GGO游戏中玩家Shino如何在限定条件下获取最大经验值的问题。通过使用二分法和匹配算法，文章详细阐述了如何在n×n的地图中，玩家在每列和每行最多只能击杀一个怪物的前提下，实现经验值的最大化。

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Bullet

Time Limit: 1000 ms Memory Limit: 65536 KiB

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Problem Description

In GGO, a world dominated by gun and steel, players are fighting for the honor of being the strongest gunmen. Player Shino is a sniper, and her aimed shot kills one monster at a time. Now she is in an $\times nn×n map, and there are monsters in some grids. Each monster has an experience. As a master, however, Shino has a strange self-restrain. She would kill at most one monster in a column, and also at most one in a row. Now she wants to know how to get max experience, under the premise of killing as many monsters as possible.$

Input

The first line contains an integer

The experience is the minimal experience of all the monster which are killed.

It guaranteed that the maximum of the experience of the monster is not larger than 10^9

Output

One integer, the value of max experience.

Sample Input

2
2 0
1 8

Sample Output

Hint

Source

“浪潮杯”山东省第九届ACM大学生程序设计竞赛（感谢山东财经大学）

求获得最大匹配用到的经验值的最小值最大化

就是一个水二分 + 匹配

记得只对w不为0的i和j建边

先求出最大匹配max_flow

然后二分最小值就好了

我的最小值最大化板子翻车了。。。。。。

我用的Dinic写的匈牙利就好了

#include <iostream>
#include <cstdio>
#include <sstream>
#include <cstring>
#include <map>
#include <cctype>
#include <set>
#include <vector>
#include <stack>
#include <queue>
#include <algorithm>
#include <cmath>
#include <bitset>
#define rap(i, a, n) for(int i=a; i<=n; i++)
#define rep(i, a, n) for(int i=a; i<n; i++)
#define lap(i, a, n) for(int i=n; i>=a; i--)
#define lep(i, a, n) for(int i=n; i>a; i--)
#define rd(a) scanf("%d", &a)
#define rlld(a) scanf("%lld", &a)
#define rc(a) scanf("%c", &a)
#define rs(a) scanf("%s", a)
#define rb(a) scanf("%lf", &a)
#define rf(a) scanf("%f", &a)
#define pd(a) printf("%d\n", a)
#define plld(a) printf("%lld\n", a)
#define pc(a) printf("%c\n", a)
#define ps(a) printf("%s\n", a)
#define MOD 2018
#define LL long long
#define ULL unsigned long long
#define Pair pair<int, int>
#define mem(a, b) memset(a, b, sizeof(a))
#define _  ios_base::sync_with_stdio(0),cin.tie(0)
//freopen("1.txt", "r", stdin);
using namespace std;
const int maxn = 1e6 + 10, INF = 0x7fffffff;
int n, s, t;
int head[maxn], cur[maxn], vis[maxn], d[maxn], cnt, nex[maxn << 1];

struct node
{
    int u, v, c;
}Node[maxn << 1];

void add_(int u, int v, int c)
{
    Node[cnt].u = u;
    Node[cnt].v = v;
    Node[cnt].c = c;
    nex[cnt] = head[u];
    head[u] = cnt++;
}

void add(int u, int v, int c)
{
    add_(u, v, c);
    add_(v, u, 0);
}

bool bfs()
{
    queue<int> Q;
    mem(d, 0);
    Q.push(s);
    d[s] = 1;
    while(!Q.empty())
    {
        int u = Q.front(); Q.pop();
        for(int i = head[u]; i != -1; i = nex[i])
        {
            int v = Node[i].v;
            if(!d[v] && Node[i].c > 0)
            {
                d[v] = d[u] + 1;
                Q.push(v);
                if(v == t) return 1;
            }
        }
    }
    return d[t] != 0;
}

int dfs(int u, int cap)
{
    int ret = 0;
    if(u == t || cap == 0)
        return cap;
    for(int &i = cur[u]; i != -1; i = nex[i])
    {
        int v = Node[i].v;
        if(d[v] == d[u] + 1 && Node[i].c > 0)
        {
            int V = dfs(v, min(cap, Node[i].c));
            Node[i].c -= V;
            Node[i ^ 1].c += V;
            ret += V;
            cap -= V;
            if(cap == 0) break;
        }
    }
    if(cap > 0) d[u] = -1;
    return ret;
}

int Dinic()
{
    int ans = 0;
    while(bfs())
    {
        memcpy(cur, head, sizeof head);
        ans += dfs(s, INF);
    }
    return ans;
}

int w[550][550];

void build(int m)
{
    mem(head, -1);
    cnt = 0;
    for(int i = 1; i <= n; i++)
    {
        add(s, i, 1);
        add(n + i, t, 1);
        for(int j = 1; j <= n; j++)
            if(w[i][j] >= m)
                add(i, n + j, 1);
    }


}


int main()
{
    while(cin >> n)
    {
        int sum = 0;
        mem(head, -1);
        cnt = 0;
        s = 0, t = maxn - 1;
        int l = 1e9, r = 0;
        for(int i = 1; i <= n; i++)
        {
            add(s, i, 1);
            add(n + i, t, 1);
            for(int j = 1; j <= n; j++)
            {
                rd(w[i][j]);
                if(w[i][j])
                    add(i, n + j, 1);
                l = min(l, w[i][j]);
                r = max(r, w[i][j]);
            }
        }
        int max_flow = Dinic();

        while(l < r)
        {
            int m = (l + r + 1) / 2;
            build(m);
            if(Dinic() < max_flow) r = m - 1;
            else l = m;
        }
        cout << r << endl;
    }

    return 0;
}