hdoj 2246 Interesting Housing Problem 【最大费用最大流 or KM算法】

Interesting Housing Problem

Time Limit: 10000/2000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 2726    Accepted Submission(s): 986

Problem Description

For any school, it is hard to find a feasible accommodation plan with every student assigned to a suitable apartment while keeping everyone happy, let alone an optimal one. Recently the president of University ABC, Peterson, is facing a similar problem. While Peterson does not like the idea of delegating the task directly to the class advisors as so many other schools are doing, he still wants to design a creative plan such that no student is assigned to a room he/she dislikes, and the overall quality of the plan should be maximized. Nevertheless, Peterson does not know how this task could be accomplished, so he asks you to solve this so-called "interesting" problem for him.
Suppose that there are N students and M rooms. Each student is asked to rate some rooms (not necessarily all M rooms) by stating how he/she likes the room. The rating can be represented as an integer, positive value meaning that the student consider the room to be of good quality, zero indicating neutral, or negative implying that the student does not like living in the room. Note that you can never assign a student to a room which he/she has not rated, as the absence of rating indicates that the student cannot live in the room for other reasons.
With limited information available, you've decided to simply find an assignment such that every student is assigned to a room he/she has rated, no two students are assigned to the same room, and the sum of rating is maximized while satisfying Peterson's requirement. The question is … what exactly is the answer?

 

Input

There are multiple test cases in the input file. Each test case begins with three integers, N, M, and E (1 <= N <= 500, 0 <= M <= 500, 0 <= E <= min(N * M, 50000)), followed by E lines, each line containing three numbers, S_i, R_i, V_i, (0 <= S_i < N, 0 <= R_i < M, |V_i| <= 10000), describing the rating V_i given by student S_ifor room R_i. It is guaranteed that each student will rate each room at most once.
Each case is followed by one blank line. Input ends with End-of-File.

 

Output

For each test case, please output one integer, the requested value, on a single line, or -1 if no solution could be found. Use the format as indicated in the sample output.

 

Sample Input

3 5 50 1 50 2 71 1 61 2 32 4 51 1 10 0 01 1 0

 

Sample Output

Case 1: 18Case 2: 0Case 3: -1

 

题意：有N个学生（编号0到N-1）和M个房间（编号0到M-1）。给你E条如(a, b, c)的关系表示学生a对房间b评价为c，当c<0时表示a学生不喜欢b房间，当c为0时表示学生对房间b保持中立态度，当c大于0时表示a学生喜欢房间b。

现在问你存不存在一种方案 保证任意一个学生都不能选上他不喜欢的房间，若存在则输出满足条件的最大评价值之和，若不存在输出-1。

费用流建图：设置超级源点source，超级汇点sink

1，source向所有学生建边，容量为1，费用为0；

2，每个学生向他评价为 喜欢或保持中立的房间建边，容量为1，费用为评价值；

3，所有房间向sink建边，容量为1，费用为0。

费用流1789ms：

#include <cstdio>#include <cstring>#include <queue>#include <algorithm>#define MAXN 1010#define MAXM 200000+10#define INF 0x3f3f3f3fusing namespace std;struct Edge{    int from, to, cap, flow, cost, next;};Edge edge[MAXM];int head[MAXN], edgenum;int pre[MAXN], dist[MAXN];bool vis[MAXN];int N, M, E;int source, sink;//超级源点 超级汇点void init(){    edgenum = 0;    memset(head, -1, sizeof(head));}void addEdge(int u, int v, int w, int c){    Edge E1 = {u, v, w, 0, c, head[u]};    edge[edgenum] = E1;    head[u] = edgenum++;    Edge E2 = {v, u, 0, 0, -c, head[v]};    edge[edgenum] = E2;    head[v] = edgenum++;}void getMap(){    int a, b, c;    source = 0, sink = N+M+1;    for(int i = 0; i < E; i++)    {        scanf("%d%d%d", &a, &b, &c);        if(c >= 0)            addEdge(a+1, b+N+1, 1, c);    }    for(int i = 1; i <= N; i++)        addEdge(source, i, 1, 0);    for(int i = 1; i <= M; i++)        addEdge(i+N, sink, 1, 0);}bool SPFA(int s, int t){    queue<int> Q;    memset(dist, -INF, sizeof(dist));    memset(vis, false, sizeof(vis));    memset(pre, -1, sizeof(pre));    dist[s] = 0;    vis[s] = true;    Q.push(s);    while(!Q.empty())    {        int u = Q.front();        Q.pop();        vis[u] = false;        for(int i = head[u]; i != -1; i = edge[i].next)        {            Edge E = edge[i];            if(dist[E.to] < dist[u] + E.cost && E.cap > E.flow)            {                dist[E.to] = dist[u] + E.cost;                pre[E.to] = i;                if(!vis[E.to])                {                    vis[E.to] = true;                    Q.push(E.to);                }            }        }    }    return pre[t] != -1;}void MCMF(int s, int t, int &cost, int &flow){    cost = flow = 0;    while(SPFA(s, t))    {        int Min = INF;        for(int i = pre[t]; i != -1; i = pre[edge[i^1].to])        {            Edge E = edge[i];            Min = min(Min, E.cap-E.flow);        }        for(int i = pre[t]; i != -1; i = pre[edge[i^1].to])        {            edge[i].flow += Min;            edge[i^1].flow -= Min;            cost += edge[i].cost * Min;        }        flow += Min;    }}int main(){    int k = 1;    while(scanf("%d%d%d", &N, &M, &E) != EOF)    {        init();        getMap();        int cost, flow;        MCMF(source, sink, cost, flow);        printf("Case %d: ", k++);        if(flow == N)            printf("%d\n", cost);        else            printf("-1\n");    }    return 0;}

KM算法：561ms

注意：KM算法 求最大权值匹配时边权赋值为正；求最小权值匹配时边权赋值为负最后取反。

#include <cstdio>#include <cstring>#include <algorithm>#define MAXN 600#define INF 100000000using namespace std;int match[MAXN];int lx[MAXN], ly[MAXN];int slack[MAXN];int Map[MAXN][MAXN];bool visx[MAXN], visy[MAXN];int nx, ny;int DFS(int x){    visx[x] = true;    for(int y = 0; y < ny; y++)    {        if(visy[y]) continue;        int t = lx[x] + ly[y] - Map[x][y];        if(t == 0)        {            visy[y] = true;            if(match[y] == -1 || DFS(match[y]))            {                match[y] = x;                return 1;            }        }        else if(slack[y] > t)            slack[y] = t;    }    return 0;}int k = 1;void KM(){    memset(match, -1, sizeof(match));    memset(ly, 0, sizeof(ly));    for(int x = 0; x < nx; x++)    {        lx[x] = -INF;        for(int y = 0; y < ny; y++)        {            if(Map[x][y] > lx[x])                lx[x] = Map[x][y];        }    }    for(int x = 0; x < nx; x++)    {        for(int i = 0; i < ny; i++)            slack[i] = INF;        while(1)        {            memset(visx, false, sizeof(visx));            memset(visy, false, sizeof(visy));            if(DFS(x)) break;            int d = INF;            for(int i = 0; i < ny; i++)            {                if(!visy[i] && d > slack[i])                    d = slack[i];            }            for(int i = 0; i < nx; i++)            {                if(visx[i])                    lx[i] -= d;            }            for(int i = 0; i < ny; i++)            {                if(visy[i])                    ly[i] += d;                else                    slack[i] -= d;            }        }    }    //判断是否所有学生都匹配到房间    int res = 0;    int num = 0;//匹配数    for(int i = 0; i < ny; i++)//枚举所有人    {        if(match[i] != -1 && Map[match[i]][i] != -INF)        {            num++;            res += Map[match[i]][i];        }    }    printf("Case %d: ", k++);    if(num == nx)        printf("%d\n", res);    else        printf("-1\n");}int main(){    int N, M, E;    int k = 1;    while(scanf("%d%d%d", &N, &M, &E) != EOF)    {        nx = N; ny = M;        for(int i = 0; i < N; i++)            for(int j = 0; j < M; j++)                Map[i][j] = -INF;        int a, b, c;        for(int i = 0; i < E; i++)        {            scanf("%d%d%d", &a, &b, &c);            if(c >= 0)            {                if(c > Map[a][b])                    Map[a][b] = c;            }        }        KM();    }    return 0;}