LightOJ 1176 - Getting a T-shirt【最大流】

版权声明：转载请注明出处。 https://blog.csdn.net/u014427196/article/details/48167669

题目链接： http://www.lightoj.com/volume_showproblem.php?problem=1176

建图：源点向人，人向6*n个衣服，衣服向汇点，容量均为1。

代码：

#include <iostream>  
#include <algorithm>  
#include <set>  
#include <map>  
#include <string.h>  
#include <queue>  
#include <sstream>  
#include <stdio.h>  
#include <math.h>  
#include <stdlib.h>  
#include <string>

using namespace std;

const int MAXN = 1010;//点数的最大值
const int MAXM = 400100;//边数的最大值
const int INF = 0x3f3f3f3f;

struct Edge
{
    int to, next, cap, flow;
}edge[MAXM];//注意是MAXM

int tol;
int head[MAXN];
int gap[MAXN], dep[MAXN], pre[MAXN], cur[MAXN];

void init()
{
    tol = 0;
    memset(head, -1, sizeof(head));
}
//加边，单向图三个参数，双向图四个参数
void addedge(int u, int v, int w, int rw = 0)
{
    edge[tol].to = v; edge[tol].cap = w; edge[tol].next = head[u];
    edge[tol].flow = 0; head[u] = tol++;
    edge[tol].to = u; edge[tol].cap = rw; edge[tol].next = head[v];
    edge[tol].flow = 0; head[v] = tol++;
}
//输入参数：起点、终点、点的总数
//点的编号没有影响，只要输入点的总数
int sap(int start, int end, int N)
{
    memset(gap, 0, sizeof(gap));
    memset(dep, 0, sizeof(dep));
    memcpy(cur, head, sizeof(head));
    int u = start;
    pre[u] = -1;
    gap[0] = N;
    int ans = 0;
    while (dep[start] < N)
    {
        if (u == end)
        {
            int Min = INF;
            for (int i = pre[u]; i != -1; i = pre[edge[i ^ 1].to])
                if (Min > edge[i].cap - edge[i].flow)
                    Min = edge[i].cap - edge[i].flow;
            for (int i = pre[u]; i != -1; i = pre[edge[i ^ 1].to])
            {
                edge[i].flow += Min;
                edge[i ^ 1].flow -= Min;
            }
            u = start;
            ans += Min;
            continue;
        }
        bool flag = false;
        int v;
        for (int i = cur[u]; i != -1; i = edge[i].next)
        {
            v = edge[i].to;
            if (edge[i].cap - edge[i].flow && dep[v] + 1 == dep[u])
            {
                flag = true;
                cur[u] = pre[v] = i;
                break;
            }
        }
        if (flag)
        {
            u = v;
            continue;
        }
        int Min = N;
        for (int i = head[u]; i != -1; i = edge[i].next)
            if (edge[i].cap - edge[i].flow && dep[edge[i].to] < Min)
            {
                Min = dep[edge[i].to];
                cur[u] = i;
            }
        gap[dep[u]]--;
        if (!gap[dep[u]])return ans;
        dep[u] = Min + 1;
        gap[dep[u]]++;
        if (u != start) u = edge[pre[u] ^ 1].to;
    }
    return ans;
}

int m, n, ans;
string a, b;

map<string, int> Mp;

int main()
{
    Mp.clear();
    Mp.insert(make_pair("XXL", 1));
    Mp.insert(make_pair("XL", 2));
    Mp.insert(make_pair("L", 3));
    Mp.insert(make_pair("M", 4));
    Mp.insert(make_pair("S", 5));
    Mp.insert(make_pair("XS", 6));

    int T, cases = 1;
    cin >> T;
    while (T--)
    {
        init();
        cin >> n >> m;


        for (int i = 1;i <= m;i++) addedge(0, i, 1);

        for (int i = m + 1;i <= m + 6 * n;i++) addedge(i, m + 6 * n + 1, 1);

        for (int i = 1;i <= m;i++)
        {
            cin >> a >> b;
            int t1 = Mp[a];
            int t2 = Mp[b];
            for (int j = 0;j < n;j++)
            {
                addedge(i, m + j * 6 + t1, 1);
                addedge(i, m + j * 6 + t2, 1);
            }
        }
        ans = sap(0, 6 * n + m + 1, 6 * n + m + 2);
        printf("Case %d: ", cases++);
        if (ans == m) puts("YES");
        else puts("NO");
    }
    return 0;
}