ZOJ 2770 Burn the Linked Camp【差分约束 SPFA】

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 2
1000 2000 1000
1 2 1100
2 3 1300
3 1
100 200 300
2 3 600

Sample Output:

1300
Bad Estimations

恩，题意大致是，军营中人数，输入中n　m代表n个军营，有m个关系，然后一行n个数代表对应军营中可能最多的人数，之后m个关系里 a  b   p 代表从a军营到b军营中至少有p个人，最后问这n个军营中至少有多少人。根据每个军营中可能 的最多的人数以及都要大于0 可得出两组不等式组，然后根据m个关系可得另一组不等式。以由m个关系构成的不等式组为例，第一组测试数据，Sn（前n个军营中总人数）S2-S0>=1100    S3-S1>=1300 也就是S0-S2<=-1100   S1-S3<=-1300也就是说从2到0权值为-1100    从3到1权值为-1300 如此建立有向图，用邻接表存，SPFA解，同时要判是否有负环。

#include <iostream>
#include<cstdio>
#include<cstring>
#include<vector>
#include<stack>
#include<queue>
#include<algorithm>
#define maxn 1010
#define inf 0x3f3f3f3f
using namespace std;

int n,m;
struct Edge
{
    int f,t,p,next;
};
Edge edge[20020];
int head[maxn];
int num[maxn];
int cnt;
void init()
{
    cnt=0;
    memset(head,-1,sizeof(head));
}
void add(int f,int t,int p)
{
    edge[cnt].f=f;
    edge[cnt].t=t;
    edge[cnt].p=p;
    edge[cnt].next=head[f];
    head[f]=cnt++;
}
void spfa()
{
    queue<int>q;
    int flag=0;
    int dis[maxn],used[maxn];
    bool vis[maxn];
    memset(used,0,sizeof(used));
    memset(vis,false,sizeof(vis));
    memset(dis,inf,sizeof(dis));
    dis[n]=0;
    vis[n]=true;
    q.push(n);
    used[n]++;
    while(!q.empty())
    {
        int u=q.front();
        q.pop();
        vis[u]=false;
        for(int i=head[u];i!=-1;i=edge[i].next)
        {
            int v=edge[i].t;
            if(dis[v]>dis[u]+edge[i].p)
            {
                dis[v]=dis[u]+edge[i].p;
                if(!vis[v])
                {
                    q.push(v);
                    used[v]++;
                    if(used[v]>n)
                    {
                        flag=1;
                        break;
                    }
                }
            }
        }
    }
    if(flag)
        printf("Bad Estimations\n");
    else
        printf("%d\n",-dis[0]);
}
int main()
{
    int a,b,p;
    while(~scanf("%d%d",&n,&m))
    {
        init();
        for(int i=1;i<=n;++i)
        {
            scanf("%d",&num[i]);
            add(i-1,i,num[i]);
            add(i,i-1,0);
        }
        while(m--)
        {
            scanf("%d%d%d",&a,&b,&p);
            add(b,a-1,-p);
        }
        spfa();
    }
    return 0;
}