版权声明:如需转载,记得标识出处 https://blog.csdn.net/godleaf/article/details/88029353
题目链接:http://poj.org/problem?id=2983
原理在上篇博客已经讲过了,这里把要注意的讲下。
1,因为差分约束问题都是不等式约束,以不等式为基础给题目建图,但是如果题目给出的约束条件是 d[u] - d[v] == w,怎么处理?
w <= d[u] - d[v] <= w,这样就能表示d[u] - d[v] == w这个条件,一条正向正权值,一条反向负权值。
2,因为之前的题目做习惯了,习惯性的给点从1到N的顺序依次加一条权值为0的边,让构建出来的图是一个连通图,因为题目的点虽然是直线排列的,但是排列的顺序不是从1到N的,所以要么把顺序记下来,再根据顺序依次加一条权值为0的边,要么就是把所有的点都加到队列里面也是一样的,虽然图不连通,但是点全加到队列里面能保证每一条边都能够松弛得了。
#include <iostream>
#include <queue>
#include <cstring>
#include <string>
#include <map>
#include <cstdio>
#include <cmath>
#include <algorithm>
#include <vector>
using namespace std;
#define lson (cur<<1)
#define rson (cur<<1|1)
typedef long long ll;
typedef pair<int, int> pii;
const int INF = 0x3f3f3f3f;
const long long LINF = 1e18;
const int Maxn = 1e5+10;
const int Mod = 1e9+7;
struct Edge {
int v, w;
Edge(int x = 0, int y = 0):v(x), w(y) {};
} edge[Maxn*2+1000];
int d[1010], cnt[1010], Hash[Maxn<<1];
vector <int> G[1010];
bool vis[1010];
bool spfa(int E, int n) {
memset(vis, true, sizeof(vis));
memset(cnt, 0, sizeof(cnt));
memset(d, 0, sizeof(d));
queue<Edge> qu;
for(int i = 0; i < n; ++i)
qu.push(Edge(Hash[i], 0));
while(!qu.empty()) {
int v = qu.front().v; qu.pop();
vis[v] = false;
for(int i = 0; i < G[v].size(); ++i) {
Edge &e = edge[G[v][i]];
if(d[e.v] < d[v]+e.w) {
d[e.v] = d[v]+e.w;
if(!vis[e.v]) {
vis[e.v] = true;
cnt[e.v]++;
if(cnt[e.v] > E) return false;
qu.push(Edge(e.v, d[e.v]));
}
}
}
}
return true;
}
int main(void)
{
int N, M;
while(scanf("%d%d", &N, &M) != EOF) {
for(int i = 0; i <= N; ++i) G[i].clear();
memset(Hash, 0, sizeof(Hash));
int E = 0, n = 0, u, v, w;
char ch;
for(int i = 0; i < M; ++i) {
scanf(" %c", &ch);
if(ch == 'P') {
scanf("%d%d%d", &u, &v, &w);
edge[E].v = v; edge[E].w = w;
G[u].push_back(E);
E++;
edge[E].v = u; edge[E].w = -w;
G[v].push_back(E);
E++;
Hash[n++] = u; Hash[n++] = v;
} else {
scanf("%d%d", &u, &v);
edge[E].v = v; edge[E].w = 1;
G[u].push_back(E);
E++;
Hash[n++] = u; Hash[n++] = v;
}
}
sort(Hash, Hash+n);
n = unique(Hash, Hash+n)-Hash;
if(spfa(E, n)) printf("Reliable\n");
else printf("Unreliable\n");
}
return 0;
}