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

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

题意：有N个连成一排的营，给出每个营的人数限制条件，求出总人数。

题解：我们用s[i]表示从第一个营到第i个营的人数总和。由题意第i个营的人数不超过c[i],可得出s[i]-s[i-1]<=c[i]和s[i]-s[i-1]>=0；从i到j至少有k个兵 可得出s[j]-s[i-1]>=k。通过解这三个不等式来求出总人数。一道差分约束的题，直接用Bellman_Ford算法或者SPFA算法即可得出答案。

#include<cstdio>
#include<cstring>
using namespace std;
const int maxn=2000+7;
const int maxm=30000+7;
const int INF=0x3f3f3f3f;
struct edge {   //边
    int u;    
    int v;
    int w;
    edge(int u=0,int v=0,int w=0):u(u),v(v),w(w){};
};
edge e[maxm];  //边
int c[maxn];   
int s[maxn]; 
int low[maxn];   // low[i]  起点到i的最短路
int n,m,u,v,w;
int Bellman_Ford()
{
    memset(low,INF,sizeof(low));
    low[n]=0;   // 以n为源点
    for(int i=1;i<n+1;i++){   //对每条边松弛 n次
        for(int j=1;j<=m+n+n;j++){
            if(low[e[j].u]<INF&&low[e[j].u]+e[j].w<low[e[j].v]){
                low[e[j].v]=low[e[j].u]+e[j].w;
            }
        }
    }
    for(int j=1;j<=m+n+n;j++){    //如果还能松弛 则无解
        if(low[e[j].u]<INF&&low[e[j].u]+e[j].w<low[e[j].v]) return 0;
    }
    return 1;
}
int main()
{
    while(scanf("%d %d",&n,&m)==2){
        for(int i=1;i<=n;i++){
            scanf("%d",&c[i]);
        }
        for(int i=1;i<=m;i++){
            scanf("%d %d %d",&u,&v,&w);
            e[i]=edge(v,u-1,-w);
        }
        for(int i=1;i<=n;i++){
            e[m+i]=edge(i-1,i,c[i]);
            e[m+n+i]=edge(i,i-1,0);
        }
        if(Bellman_Ford()) printf("%d\n",-low[0]);  //解为n到0的最短路
        else printf("Bad Estimations\n");
    }
    return 0;
}