pku 2983 Is the Information Reliable? 差分约束

http://poj.org/problem?id=2983

题意：给定两种约束关系

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.

d[i]表示i与0的距离，在p关系可表示为x <=d[b] - d[a]<=x   V关系可以表示为 d[b] - d[a] >= 1。 化简约束关系： P: d[b] - d[a] <= x    d[a] - d[b] <= -c  V: d[a] - d[b] <= -1

求最短路，判断是否存在bellman_ford。在做的时候，用了个memset初始化g结果一直tle，去掉后就ac了，看来以后尽量自己写训话出事化。

 

#include <cstdio>

#include <cstring>

#include <iostream>

#define maxn 100007

using namespace std;



struct node

{

    int u,v,w;

}g[maxn*100];

int len;

int n,m,dis[1005];

const int inf = 99999999;



void add(int u,int v,int w)

{

    g[len].u = u;

    g[len].v = v;

    g[len].w = w;

    len++;

}

bool bellman_ford(int s)

{

    int i,j;

    for (i = 1; i <= n; ++i) dis[i] = inf;

    dis[s] = 0;

    for (i = 1; i < n; ++i)

    {

        bool flag = false;

        for (j = 0; j < len; ++j)

        {

            int u = g[j].u,v = g[j].v,w = g[j].w;

            if (dis[v] > dis[u] + w)

            {

                dis[v] = dis[u] + w;

                flag = true;

            }

        }

        if (!flag) return true;

    }

    for (j = 0; j < len; ++j)

    {

        int u = g[j].u,v = g[j].v,w = g[j].w;

        if (dis[v] > dis[u] + w) return false;

    }

    return true;

}

int main()

{

    int i;

    int a,b,c;

    char op[3];

    while (~scanf("%d%d",&n,&m))

    {

        len = 0;

        for (i = 0; i < m; ++i)

        {

            scanf("%s%d%d",op,&a,&b);

            if (op[0] == 'P')

            {

                scanf("%d",&c);

                add(b,a,-c);

                add(a,b,c);

            }

            else

            {

                add(b,a,-1);

            }

        }

        if (bellman_ford(1)) printf("Reliable\n");

        else printf("Unreliable\n");

    }

    return 0;

}