POJ 2983 （差分约束）

Is the Information Reliable? Time Limit: 3000MS   Memory Limit: 131072K Total Submissions: 11567   Accepted: 3661 Description The galaxy war between the Empire Draco and the Commonwealth of Zibu broke out 3 years ago. Draco established a line of defense called Grot. Grot is a straight line with N defense stations. Because of the cooperation of the stations, Zibu’s Marine Glory cannot march any further but stay outside the line. A mystery Information Group X benefits form selling information to both sides of the war. Today you the administrator of Zibu’s Intelligence Department got a piece of information about Grot’s defense stations’ arrangement from Information Group X. Your task is to determine whether the information is reliable. The information consists of M tips. Each tip is either precise or vague. Precise tip is in the form of P A B X, means defense station A is X light-years north of defense station B. Vague tip is in the form of V A B, means defense station A is in the north of defense station B, at least 1 light-year, but the precise distance is unknown. Input There are several test cases in the input. Each test case starts with two integers N (0 < N ≤ 1000) and M (1 ≤ M ≤ 100000).The next M line each describe a tip, either in precise form or vague form. Output Output one line for each test case in the input. Output “Reliable” if It is possible to arrange N defense stations satisfying all the M tips, otherwise output “Unreliable”. Sample Input 3 4 P 1 2 1 P 2 3 1 V 1 3 P 1 3 1 5 5 V 1 2 V 2 3 V 3 4 V 4 5 V 3 5 Sample Output Unreliable Reliable Source POJ Monthly--2006.08.27, Dagger

题意：n个点，m个约束条件。P: u v w 表示u点距v点w。V: u v 表示u点距v点至少为1。问你这些信息是否可靠，也就是说有没有矛盾的地方。

分析：写了一个Bellman-Ford算法。注意图的连通性。

code：

#include<stdio.h>
#define INF 1e9
int cnt;
struct Edge
{
    int from, to, cost;
}e[200010];

void add(int u, int v, int w)
{
    e[cnt].from = u;
    e[cnt].to = v;
    e[cnt++].cost = w;
}

int main()
{
    int n, m, d[1010];
    char c;

    while(~scanf("%d%d", &n,&m))
    {
        cnt = 0;
        for(int i=0; i<m; i++)
        {
            getchar();
            scanf("%c", &c);
            int A, B, D;
            if(c == 'P')
            {
                scanf("%d%d%d", &A,&B,&D);
                add(A,B,-D);
                add(B,A,D);
            }
            else
            {
                scanf("%d%d", &A,&B);
                add(A,B,-1);
            }
        }
        for(int i=1; i<=n; i++) d[i] = 0; //原点有可能不与负环连通
        bool update; //所以将初始值全赋值为0，这样负环就一定能被查出，因为此时图是连通的
        for(int j=0; j<n; j++)
        {
            update = false;
            for(int i=0; i<cnt; i++)
            {
                if(d[e[i].from]+e[i].cost < d[e[i].to])
                {
                    d[e[i].to] = d[e[i].from]+e[i].cost;
                    update = true;
                }
            }
            if(!update) break;
        }
        if(update) puts("Unreliable");
        else puts("Reliable");
    }
    return 0;
}