HDU 4322 Candy

ProblemDescription

There are Ncandies and M kids, the teacher will give this N candies to the M kids. Thei-th kids for the j-th candy has a preference for like[i][j], if he like thesugar, like[i][j] = 1, otherwise like[i][j] = 0. If the i-th kids get the candywhich he like he will get K glad value. If he or she do not like it. He willget only one glad value. We know that the i-th kids will feel happy if he thesum of glad values is equal or greater than B[i]. Can you tell me whetherreasonable allocate this N candies, make every kid feel happy.

 

 

Input

The Input consistsof several cases .The first line contains a single integer t .the number oftest cases.
For each case starts with a line containing three integers N, M, K(1<=N<=13, 1<=M<=13, 2<=K<=10)
The next line contains M numbers which is B[i](0<=B[i]<=1000). Separatedby a single space.
Then there are M*N like[i][j] , if the i-th kids like the j-th sugarlike[i][j]=1 ,or like[i][j]=0.

 

 

Output

If there have areasonable allocate make every kid feel happy, output "YES", or"NO".

 

 

Sample Input

2

3 2 2

2 2

0 0 0

0 0 1

 

3 2 2

2 2

0 0 0

0 0 0

 

 

Sample Output

Case #1: YES

Case #2: NO

 

集训期间，实在太累，不写了。。。。。。待到开学再来慢慢写吧

#include <iostream>#include<cstdio>#include<cstring>#include<cstdlib>#include<algorithm>>#define INF 1000000000using namespace std;const int M = 4 * 10010;//边struct Node//边，点f到点t，流量为c，费用为w{int st, ed;int flow, cost;int next;}edge[M];int head[M], dis[M], q[M], pre[M], cnt;//cnt为已添加的边数，head为邻接表,dis为花费，pre为父亲节点bool vis[M];int b[15] ,like[15][15];int n, m, k;void init(){memset(head, -1, sizeof(head));cnt = 0;}void add_edge(int f, int t, int d1, int d2, int w){//f到t的一条边，流量为d1,反向流量d2,花费w,反向边花费-w（可以反悔)edge[cnt].st = f;edge[cnt].ed = t;edge[cnt].flow = d1;edge[cnt].cost = w;edge[cnt].next = head[f];head[f] = cnt++;edge[cnt].st = t;edge[cnt].ed = f;edge[cnt].flow = d2;edge[cnt].cost = -w;edge[cnt].next = head[t];head[t] = cnt++;}bool spfa(int s, int t, int n){int i, tmp, l, r;memset(pre, -1, sizeof(pre));for(i = 0; i < n; ++i)    {        dis[i] = INF;    }dis[s] = 0;q[0] = s;l = 0, r = 1;vis[s] = true;while(l != r)    {tmp = q[l];l = (l + 1) % (n + 1);vis[tmp] = false;for(i = head[tmp]; i!=-1; i = edge[i].next){if(edge[i].flow && dis[edge[i].ed] > dis[tmp] + edge[i].cost){dis[edge[i].ed] = dis[tmp] + edge[i].cost;pre[edge[i].ed] = i;if(!vis[edge[i].ed]){vis[edge[i].ed] = true;q[r] = edge[i].ed;r = (r + 1) % (n + 1);}}}}if(pre[t] == -1){    return false;}return true;}void MCMF(int s, int t, int n, int &flow, int &cost){//起点s，终点t，点数n，最大流flow，最小花费costint tmp, arg;flow = cost = 0;while(spfa(s, t, n))    {arg = INF;tmp = t;while(tmp != s){arg = min(arg, edge[pre[tmp]].flow);tmp = edge[pre[tmp]].st;}tmp = t;while(tmp != s){edge[pre[tmp]].flow -= arg;edge[pre[tmp] ^ 1].flow += arg;tmp = edge[pre[tmp]].st;}flow += arg;cost += arg * dis[t];}}int main(){    int t ,no ,num[15] ,flag[15];    int flow ,cost ,sum;    scanf("%d",&t);    for(int r = 1;r<=t;r++)    {        memset(flag,0,sizeof(flag));        memset(num,0,sizeof(num));        sum = 0;        scanf("%d%d%d",&n,&m,&k);        for(int i = 1;i<=m;i++)        {            scanf("%d",&b[i]);            sum += b[i];        }        no = 1;        for(int i = 1;i<=m;i++)        {            for(int j = 1;j<=n;j++)            {                scanf("%d",&like[i][j]);                if(like[i][j] && !flag[j])                {                    num[j] = no++;                }            }        }        init();        no = m + no - 1;        for(int i = 1;i<=m;i++)        {            add_edge(0,i,b[i]/k,0,-k);            if(b[i] % k)            {                no++;                add_edge(0,no,1,0,-(b[i] % k));            }            for(int j = 1;j<=n;j++)            {                if(like[i][j])                {                    add_edge(i,m + num[j],1,0,0);                    if(b[i] % k)                    {                        add_edge(no,m + num[j],1,0,0);                    }                }            }        }        no++;        for(int i = 1;i<=n;i++)        {            if(num[i])            {                add_edge(m + num[i],no,1,0,0);            }        }        MCMF(0,no,no + 1,flow,cost);        if(sum + cost > n - flow)        {            printf("Case #%d: NO\n",r);        }        else        {            printf("Case #%d: YES\n",r);        }    }    return 0;}