ZOJ 2770 Burn the Linked Camp(最短路+SPFA+差分约束系统)

Burn the Linked Camp

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Time Limit: 2 Seconds Memory Limit: 65536 KB

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It is well known that, in the period of The Three Empires, Liu Bei, the emperor of the Shu Empire, was defeated by Lu Xun, a general of the Wu Empire. The defeat was due to Liu Bei's wrong decision that he divided his large troops into a number of camps, each of which had a group of armies, and located them in a line. This was the so-called "Linked Camps".

Let's go back to that time. Lu Xun had sent many scouts to obtain the information about his enemy. From his scouts, he knew that Liu Bei had divided his troops into n camps, all of which located in a line, labeled by 1..n from left to right. The ith camp had a maximum capacity of Ci soldiers. Furthermore, by observing the activities Liu Bei's troops had been doing those days, Lu Xun could estimate the least total number of soldiers that were lived in from the ith to the jth camp. Finally, Lu Xun must estimate at least how many soldiers did Liu Bei had, so that he could decide how many troops he should send to burn Liu Bei's Linked Camps.

Input:

There are multiple test cases! On the first line of each test case, there are two integers n (0<n<=1,000) and m (0<=m<=10,000). On the second line, there are n integers C1��Cn. Then m lines follow, each line has three integers i, j, k (0<i<=j<=n, 0<=k<2^31), meaning that the total number of soldiers from the ith camp to the jth camp is at least k.

Output:

For each test case, output one integer in a single line: the least number of all soldiers in Liu Bei's army from Lu Xun's observation. However, Lu Xun's estimations given in the input data may be very unprecise. If his estimations cannot be true, output "Bad Estimations" in a single line instead.

Sample Input:

3 21000 2000 10001 2 11002 3 13003 1100 200 3002 3 600

Sample Output:

1300Bad Estimations

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Author: ZHOU, Yuan
Source: ZOJ Monthly, October 2006

思路：典型差分约束。c[i]代表第I个营地最多容量。s[i]代表1-i的总容量

           求s[x]-s[0]的最小值。

            (k,k-1)<=c[k],   (k-1,k)<=0;

            (a-1,b)<=-c,  (b,a-1)<=s[b]-s[a-1]

       spfa(0)求最短路。

#include<iostream>#include<cstring>#include<queue>#include<vector>using namespace std;const int mm=1010;const int oo=1e9;class node{  public:int v,c;};int dis[mm],n,m,c[mm],s[mm],id[mm],pos;vector<node>e[mm];bool vis[mm];queue<int>q;bool spfa(int x){ memset(id,0,sizeof(id));  memset(vis,0,sizeof(vis));  for(int i=0;i<=n;i++)    dis[i]=oo;  dis[x]=0;int u,v,c;  q.push(x);int z;vis[x]=1;++id[x];  while(!q.empty())  {    z=q.front();q.pop();vis[z]=0;    for(int i=0;i<e[z].size();i++)    { u=z;v=e[z][i].v;c=e[z][i].c;      if(dis[v]>dis[u]+c)      {        dis[v]=dis[u]+c;        if(!vis[v])        { ++id[v];          if(id[v]>n)return 1;          vis[v]=1;q.push(v);        }      }    }  }  return 0;}int main(){  while(cin>>n>>m)  { memset(e,0,sizeof(e));    c[0]=0;s[0]=0;node z;    for(int i=1;i<=n;i++)    {      cin>>c[i];s[i]=s[i-1]+c[i];      z.v=i-1;z.c=c[i];e[i].push_back(z);///0为源点，(i,i-1)<=s[i]-s[i-1]      z.v=i;z.c=0;e[i-1].push_back(z);///(i,i-1)<=0    }    int a,b,c;    for(int i=0;i<m;i++)    {      cin>>a>>b>>c;      z.v=b;z.c=-c;e[a-1].push_back(z);///(b,a-1)>=c -> (a-1,b)<=-c      z.v=a-1;z.c=s[b]-s[a-1];e[b].push_back(z);///(b,a-1)<=s[b]-s[a-1];    }    bool flag=spfa(0);    if(flag)cout<<"Bad Estimations\n";    else cout<<dis[0]-dis[n]<<"\n";  }}