“浪潮杯”第九届山东省ACM大学生程序设计竞赛 B - Bullet[二分 + 最大流匹配]

版权声明： https://blog.csdn.net/weixin_39792252/article/details/82148205

题目：n*n的方格中可能有monster，Shino只能在一行一列只能杀一个，求尽可能多的杀monster获得经验的最小值最大。

题解：每次二分最小值，二分匹配，我习惯用最大流跑。。。具体一点就是说一开始先确定杀得怪物最多有多少个，如果是0个，那么输出0,否则二分最小值，最大流匹配，判断满足杀最多的monster获得经验的最小值最大，最后就得到答案啦。

代码：

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#define ll long long
#define INF 1e9+7;
using namespace std;
const int maxn = 1100;


/*
* SAP 算法(矩阵形式)
* 结点编号从 0 开始
*/

int maze[maxn][maxn], n, path[maxn][maxn]; //
int gap[maxn], dis[maxn], pre[maxn], cur[maxn];
//gap[];
//dis[] 某点到汇点最少弧的数量;
//pre[];
//cur[];

int spa(int start, int end, int nodenum)//start 起点, end 终点 ,nodenum 结点个数;
{
    memset(cur, 0, sizeof(cur));
    memset(dis, 0, sizeof(dis));
    memset(gap, 0, sizeof(gap));
    memset(pre, 0, sizeof(pre));
    memset(cur, 0, sizeof(cur));

    int u = pre[start] = start, maxflow=0, aug = -1;   //maxflow最大流;
    gap[0] = nodenum; //距离标号;

    while(dis[start] < nodenum) {
        loop:
            for(int v = cur[u]; v < nodenum; v++)
                if(maze[u][v] && dis[u] == dis[v] + 1) {   //找可行弧;
                    if(aug == -1 || aug > maze[u][v]) aug = maze[u][v];
                    pre[v] = u;
                    u = cur[u] = v;
                    if(v == end) {
                        maxflow += aug;
                        for(u = pre[u]; v != start; v = u, u = pre[u]){
                            maze[u][v] -= aug;
                            maze[v][u] += aug;
                        }
                        aug = -1;
                    }
                    goto loop;
                }

            int mindis = nodenum - 1;

            for(int v = 0; v < nodenum; v++)
                if(maze[u][v]&&mindis>dis[v]) {
                    cur[u] = v;
                    mindis = dis[v];
                }
            if((--gap[dis[u]]) == 0) break;
            gap[dis[u] = mindis+1]++;
            u=pre[u];
    }
    return maxflow;
}


void solve(int x) {
    memset(maze, 0, sizeof(maze));
    for(int i = 1; i <= n; i++) maze[0][i] = 1;
    for(int i = 1; i <= n; i++) maze[n+i][2*n+1] = 1;

    for(int i = 1; i <= n; i++)
        for(int j = 1; j <= n; j++)
            if(path[i][j]>=x&&path[i][j] != 0) maze[i][n+j] = 1;
}


int main()
{
    //freopen("in.txt", "r", stdin);
    while(scanf("%d", &n) != EOF)
    {
        memset(maze, 0, sizeof(maze));
        memset(path, 0, sizeof(path));

        for(int i = 1; i <= n; i++) maze[0][i] = 1;
        for(int i = 1; i <= n; i++) maze[n+i][2*n+1] = 1;

        for(int i = 1; i <= n; i++)
            for(int j = 1; j <= n; j++)
            {
                scanf("%d", &path[i][j]);
                if(path[i][j] != 0) maze[i][n+j] = 1;
            }

        int temp = spa(0, 2*n+1, 2*n+2);

        //printf("%d\n", temp);
        if(temp == 0) {
            printf("0\n"); continue;
        }

        int a, b, mid, ans;
        a = -INF;
        b = INF;
        while(a<=b) {
            mid = (b+a)/2;
            solve(mid);
            int cas = spa(0, 2*n+1, 2*n+2);

            //printf("%d\n", cas);
            if(cas == temp)   {
                ans = mid;
                a = mid+1;
            }
            else  b = mid-1;
        }
        printf("%d\n", ans);
    }
    return 0;
}