poj2983Is the Information Reliable?——差分约束

poj2983Is the Information Reliable?：http://poj.org/problem?id=2983

Is the Information Reliable?

Time Limit: 3000MS Memory Limit: 131072KTotal Submissions: 12379 Accepted: 3898

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 4P 1 2 1P 2 3 1V 1 3P 1 3 15 5V 1 2V 2 3V 3 4V 4 5V 3 5

Sample Output

UnreliableReliable

题意：给出一些信息，P代表的是精确的信息；V代表的是不精确的信息。问得到的这些信息是否是可靠的。

思路，用差分约束来做，根据题意找出约束条件：d(u)-d(v)>=1;c>=d(u)-d(v)>=c.找一下最短路（当然不需要写出最短路）

这题最终要求的说通俗点就是判断一下有没有负环……

在做的时候由于题意没有理解好，导致在建边的时候建反了，wa了n发……

#include <iostream>#include <cstdlib>#include <cstdio>#include <cstring>#include <string>#include <algorithm>#include <set>#include <queue>using namespace std;#define INF 1000000007#define maxn 100000struct Edge{    int v;    int w;    int next;} edge[maxn * 3];int N,M,cnt = 0;int sum[maxn],head[maxn],vis[maxn],d[maxn];void Add(int u,int v,int w){    edge[cnt].v = v;    edge[cnt].w = w;    edge[cnt].next = head[u];    head[u] = cnt++;}bool spfa(){    memset(vis,0,sizeof(vis));    memset(sum,0,sizeof(sum));    for(int i = 0;i <= N;i++)d[i] = INF;    int top = 0;    queue<int>st;    st.push(0);    d[0] = 0;    vis[0] = 1;    while(!st.empty())    {        int x = st.front();        st.pop();        vis[x] = 0;        for(int i = head[x];i != -1;i = edge[i].next)        {            if(d[edge[i].v] > d[x] + edge[i].w)            {                d[edge[i].v] = d[x] + edge[i].w;                if(!vis[edge[i].v])                {                    vis[edge[i].v] = 1;                    st.push(edge[i].v);                    sum[edge[i].v]++;                    if(sum[edge[i].v] > N)return false;       // 当入度大于顶点数说明存在负环了                }            }        }    }    return true;}int main(){//    freopen("in.txt","r",stdin);    int u,v,w;    while(~scanf("%d%d",&N,&M))    {        cnt = 0;        memset(head,-1,sizeof(head));        while(M--)        {            char ss[10];            scanf("%s%d%d",ss,&u,&v);            if(ss[0] == 'P')            {                scanf("%d",&w);                Add(u,v,-w);       // 就是在这弄反了……                Add(v,u,w);       //            }            else Add(u,v,-1);        }        for(int i = 1;i <= N;i++)    // 添加了超级源点，共有N+1个点了            Add(0,i,0);        if(!spfa())printf("Unreliable\n");        else printf("Reliable\n");    }    return 0;}

对于求最短路还是最长路，一直混淆……

题目要求最小值，求最长路，约束条件化成d(v)-d(u)>=w(u,v)的形式；

题目要求最大值，求最短路，约束条件化成d(v)-d(u)<=w(u,v)的形式。