Evacuation Plan
Time Limit: 1000MS   Memory Limit: 65536K
Total Submissions: 4617   Accepted: 1218   Special Judge


The City has a number of municipal buildings and a number of fallout shelters that were build specially to hide municipal workers in case of a nuclear war. Each fallout shelter has a limited capacity in terms of a number of people it can accommodate, and there's almost no excess capacity in The City's fallout shelters. Ideally, all workers from a given municipal building shall run to the nearest fallout shelter. However, this will lead to overcrowding of some fallout shelters, while others will be half-empty at the same time.

To address this problem, The City Council has developed a special evacuation plan. Instead of assigning every worker to a fallout shelter individually (which will be a huge amount of information to keep), they allocated fallout shelters to municipal buildings, listing the number of workers from every building that shall use a given fallout shelter, and left the task of individual assignments to the buildings' management. The plan takes into account a number of workers in every building - all of them are assigned to fallout shelters, and a limited capacity of each fallout shelter - every fallout shelter is assigned to no more workers then it can accommodate, though some fallout shelters may be not used completely.

The City Council claims that their evacuation plan is optimal, in the sense that it minimizes the total time to reach fallout shelters for all workers in The City, which is the sum for all workers of the time to go from the worker's municipal building to the fallout shelter assigned to this worker.

The City Mayor, well known for his constant confrontation with The City Council, does not buy their claim and hires you as an independent consultant to verify the evacuation plan. Your task is to either ensure that the evacuation plan is indeed optimal, or to prove otherwise by presenting another evacuation plan with the smaller total time to reach fallout shelters, thus clearly exposing The City Council's incompetence.

During initial requirements gathering phase of your project, you have found that The City is represented by a rectangular grid. The location of municipal buildings and fallout shelters is specified by two integer numbers and the time to go between municipal building at the location (Xi, Yi) and the fallout shelter at the location (Pj, Qj) is Di,j = |Xi - Pj| + |Yi - Qj| + 1 minutes.


The input consists of The City description and the evacuation plan description. The first line of the input file consists of two numbers N and M separated by a space. N (1 ≤ N ≤ 100) is a number of municipal buildings in The City (all municipal buildings are numbered from 1 to N). M (1 ≤ M ≤ 100) is a number of fallout shelters in The City (all fallout shelters are numbered from 1 to M).

The following N lines describe municipal buildings. Each line contains there integer numbers Xi, Yi, and Bi separated by spaces, where Xi, Yi (-1000 ≤ Xi, Yi ≤ 1000) are the coordinates of the building, and Bi (1 ≤ Bi ≤ 1000) is the number of workers in this building.

The description of municipal buildings is followed by M lines that describe fallout shelters. Each line contains three integer numbers Pj, Qj, and Cj separated by spaces, where Pi, Qi (-1000 ≤ Pj, Qj ≤ 1000) are the coordinates of the fallout shelter, and Cj (1 ≤ Cj ≤ 1000) is the capacity of this shelter.

The description of The City Council's evacuation plan follows on the next N lines. Each line represents an evacuation plan for a single building (in the order they are given in The City description). The evacuation plan of ith municipal building consists of M integer numbers Ei,j separated by spaces. Ei,j (0 ≤ Ei,j ≤ 1000) is a number of workers that shall evacuate from the ith municipal building to the jth fallout shelter.

The plan in the input file is guaranteed to be valid. Namely, it calls for an evacuation of the exact number of workers that are actually working in any given municipal building according to The City description and does not exceed the capacity of any given fallout shelter.


If The City Council's plan is optimal, then write to the output the single word OPTIMAL. Otherwise, write the word SUBOPTIMAL on the first line, followed by N lines that describe your plan in the same format as in the input file. Your plan need not be optimal itself, but must be valid and better than The City Council's one.

Sample Input

3 4
-3 3 5
-2 -2 6
2 2 5
-1 1 3
1 1 4
-2 -2 7
0 -1 3
3 1 1 0
0 0 6 0
0 3 0 2

Sample Output

3 0 1 1
0 0 6 0
0 4 0 1






 #define LL long long
 using namespace std;
 const int INF=1e9;
 inline int read(){
     ,flag=;char ch=getchar();
     return sum*flag;
 struct edge{
     int u,v,nxt,w;
 void add_edge(int u,int v,int w){
 //    printf("add:%d to %d :%d\n",u,v,w);
 int n,m,S,T;
 int mp[mxn][mxn];
 int dis[mxn];
 int pre[mxn];
 int cnt[mxn];
 bool inq[mxn];
 bool SPFA(){
     memset(dis,0x3f,sizeof dis);
     memset(inq,,sizeof inq);
     memset(cnt,,sizeof cnt);
     int v;
     while(!q.empty() && flag){
         for(int i=hd[u];i;i=e[i].nxt){
                     inq[v]=; cnt[v]++;
         memset(inq,,sizeof inq);
         int s=v;
             else break;
         memset(inq,,sizeof inq);
             int p=pre[s];
             if(p>n && s!=T) mp[s][p]--;
             else if(s>n && p!=T) mp[p][s]++;
         int ed=n+m;
                 )printf(" ");
 int x[mxn],y[mxn],w[mxn],in[mxn];
 void Build(){
     memset(hd,,sizeof hd);
     memset(,sizeof in);
     int i,j;
 //            printf("%d ",v);
 //    printf("\n");
 int main(){
     int i,j;

POJ2175 Evacuation Plan的更多相关文章

  1. POJ-2175 Evacuation Plan 最小费用流、负环判定

    题意:给定一个最小费用流的模型,根据给定的数据判定是否为最优解,如果不为最优解则给出一个比给定更优的解即可.不需要得出最优解. 解法:由给定的数据能够得出一个残图,且这个图满足了最大流的性质,判定一个 ...

  2. HDU 3757 Evacuation Plan DP

    跟 UVa 1474 - Evacuation Plan 一个题,但是在杭电上能交过,在UVa上交不过……不知道哪里有问题…… 将施工队位置和避难所位置排序. dp[i][j] 代表前 i 个避难所收 ...

  3. Codeforces Gym 100002 E &quot;Evacuation Plan&quot; 费用流

    "Evacuation Plan" Time Limit: 1 Sec Memory Limit: 256 MB 题目连接 http://codeforces.com/gym/10 ...

  4. POJ 2175 Evacuation Plan (费用流,负环,消圈法,SPFA)

    http://poj.org/problem?id=2175 Evacuation Plan Time Limit: 1000MS   Memory Limit: 65536K Total Submi ...

  5. POJ 2175 Evacuation Plan 费用流 负圈定理

    题目给了一个满足最大流的残量网络,判断是否费用最小. 如果残量网络中存在费用负圈,那么不是最优,在这个圈上增广,增广1的流量就行了. 1.SPFA中某个点入队超过n次,说明存在负环,但是这个点不一定在 ...

  6. POJ 2157 Evacuation Plan [最小费用最大流][消圈算法]

    ---恢复内容开始--- 题意略. 这题在poj直接求最小费用会超时,但是题意也没说要求最优解. 根据线圈定理,如果一个跑完最费用流的残余网络中存在负权环,那么顺着这个负权环跑流量为1那么会得到更小的 ...

  7. UVA 1474 Evacuation Plan

    题意:有一条公路,上面有n个施工队,要躲进m个避难所中,每个避难所中至少有一个施工队,躲进避难所的花费为施工队与避难所的坐标差的绝对值,求最小花费及策略. 解法:将施工队和避难所按坐标排序,可以看出有 ...

  8. poj 2175 Evacuation Plan 最小费用流判定,消圈算法

    题目链接 题意:一个城市有n座行政楼和m座避难所,现发生核战,要求将避难所中的人员全部安置到避难所中,每个人转移的费用为两座楼之间的曼哈顿距离+1,题目给了一种方案,问是否为最优方案,即是否全部的人员 ...

  9. Evacuation Plan-POJ2175最小费用消圈算法

    Time Limit: 1000MS Memory Limit: 65536K Special Judge Description The City has a number of municipal ...


  1. Devexress XPO xpPageSelector 使用

    在官网找到的.在这里做个备注. private void simpleButton1_Click(object sender, EventArgs e) { ) return; xpPageSelec ...

  2. js测试题

    (function(){ return typeof arguments;})();"object" var f = function g(){ return 23; };type ...

  3. Windows Iot:让Raspberry Pi跑起来(1)

    首先请大家原谅我的"不务正业",放着RabbitHub不写,各种系列的文章不写搞什么Iot,哈哈,最近心血来潮想搞个速度极快的遥控车玩,望着在角落的Raspberry Pi恶狠狠的 ...

  4. git 小结

    git cherry-pick de0ec64  可将另一个分支中的提交 cherry-pick到当前分支来

  5. linux msql

    安装mysql 1.使用rpm 安装mysql 或者使用yum安装 使用rpm 安装 下载 Centos 7 所需要的mysql包 tar -xf 解压整合包 根据依赖 安装 common>li ...

  6. IOC和Aop使用的扩展

    下面还有静态代理和动态代理 1.构造注入 lib包: 在entity包下新建一个实体类User 代码: package cn.happy.entity; public class User { pri ...

  7. 如何分隔两个base64字符串?

      如何分隔两个base64字符串?   用逗号或者任意的不在base64字符串内的字符都可以. All you have to do is to use a separator which is n ...

  8. JAVA 大数据内存耗用测试

    JAVA 大数据内存耗用测试import java.lang.management.ManagementFactory;import java.lang.management.MemoryMXBean ...

  9. array_intersect() php筛选两个数组共有的元素

    我们已经讲过如何筛选出连个数组中不共有的元素,今天就来看看php如何筛选出两个数组中共有的元素,例如筛选$array1和$array2共有的元素. 函数名:array_intersect(): 调用方 ...

  10. 动态链接库(dll) __declspec(dllimport) __declspec(dllexport)

    一. __declspec(dllexport) Microsoft 在 Visual C++ 的 16 位编译器版本中引入了 __export,使编译器得以自动生成导出名并将它们放到一个 .lib ...