POJ 2983 Is the Information Reliable(差分约束系统)

//差分约束系统//这题比较对差分约束的构图比较直观//eg.P 3 1 4//4 =< s3 - s1 <= 4-----------(1)//等价于 s3 - s1 <= 4--------(2)// s1 - s3 <= -4-------(3)//V 4 5//s4 - s5 >= 1 --------------(4)//等价于 s5 - s4 <= -1-------(5)//建立(2)(3)(5)的边就可以了，建立<=的差分系统，就必须求最短路，反之求最长路//只与信息是否可靠，就是判断图中是否存在负权回路即可//下面代码为SPFA做法#include<iostream>#include<queue>#define INF 1000000000#define MAX 1005#define MAXM 210000//注意边数的数量，最坏情况2*M+1000(超级源点的边)using namespace std;int head[MAX],V[MAXM],next[MAXM],W[MAXM];bool inq[MAX];int cnt[MAX],dis[MAX];int m,N,M;char cmd;int u,v,w;void addEdge(int u,int v,int w)//添加边{V[m] = v;W[m] = w;next[m] = head[u];head[u] = m++;}void buildGrapth()//构图函数{m = 0;memset(head,-1,sizeof(head));for(int i = 1;i <= N;++i)addEdge(0,i,0);while(M--){scanf("/n%c",&cmd);if(cmd == 'P'){scanf("%d%d%d",&u,&v,&w);addEdge(v,u,w);addEdge(u,v,-w);}else{scanf("%d%d",&u,&v);addEdge(u,v,-1);}}}bool SPFA(){memset(inq,0,sizeof(inq));memset(cnt,0,sizeof(cnt));queue<int> q;for(int i = 1;i <= N;++i)dis[i] = INF;dis[0] = 0;q.push(0);inq[0] = 1;while(!q.empty()){int u = q.front();q.pop();inq[u] = 0;for(int e = head[u];e != -1;e = next[e]){if(dis[u] + W[e] < dis[V[e]]){dis[V[e]] = dis[u] + W[e];if(!inq[V[e]]){q.push(V[e]);inq[V[e]] = 1;if(++cnt[V[e]] > N)return 0;//通过元素入队次数判断是否存在回路}}}}return 1;}int main(){//freopen("in.txt","r",stdin);while(scanf("%d%d",&N,&M) != EOF){buildGrapth();if(SPFA())printf("Reliable/n");else printf("Unreliable/n");}return 0;}