POJ 1932 XYZZY （差分约束+最长路）

解题思路：按照题目给定条件建图，判断能不能找到一条路到n，在这条路上权值和不会为负数，初始值为100.

只需要用spfa跑最长路，遇到正环的时候将当前节点到起点的距离置为 inf 即可。这样可以保证有正环的图一定可以得到解。

这道题的代码需要注意一些细节和剪枝。

AC代码：

/** @Author: wchhlbt* @Last Modified time: 2017-09-20*/#include <vector>#include <list>#include <map>#include <set>#include <queue>#include <stack>#include <bitset>#include <algorithm>#include <functional>#include <numeric>#include <utility>#include <sstream>#include <iostream>#include <iomanip>#include <cstdio>#include <cmath>#include <cstdlib>#include <ctime>#include <cstring>#include <limits>#include <climits>#include <cstdio>#define Fori(x) for(int i=0;i<x;i++)#define Forj(x) for(int j=0;j<x;j++)#define maxn 107#define inf 0x3f3f3f3f#define ONES(x) __builtin_popcount(x)#define pb push_back#define AA first#define BB second#define _  << "  " <<using namespace std;typedef long long ll ;const double eps =1e-8;const int mod = 998244353;const double PI = acos(-1.0);int dx[5] = {0,0,1,-1,0};int dy[5] = {1,-1,0,0,0};inline int read(){ int cnt;    scanf("%d",&cnt);   return cnt;}//每次使用前需要调用init函数初始化 可以处理负权边//最坏复杂度O(V*E)int d[maxn],inq[maxn];//inq数组储存当前点是否在队列中int deg[maxn];//记录每个节点入队次数vector<int> e[maxn]; //pair<节点, 边权> int val[maxn];int n,m;void init(){    for(int i = 0; i<maxn; i++){        e[i].clear();        d[i] = -inf;        inq[i] = 0;        deg[i] = 0;    }}void SPFA(int s)//s为起点{    queue<int> Q;    Q.push(s);  d[s] = 100; inq[s] = 1; deg[s] = 1;    while(!Q.empty()){        int hd = Q.front();        Q.pop();    inq[hd] = 0;            for(int i = 0; i<e[hd].size(); i++){            int u = e[hd][i];            int v = val[u];            if(d[u]<d[hd]+v && d[hd]+v>0){//注意这里的判断条件                d[u] = d[hd] + v;                if(inq[u]==1 || deg[u]>n)    continue;//注意这个地方的剪枝                inq[u] = 1;                deg[u]++;                if(deg[u]>n)   d[u] = inf;                Q.push(u);            }        }    }    if(d[n]>0)puts("winnable");   elseputs("hopeless");}int main(){    while(~scanf("%d",&n))    {    if(n==-1)break;    init();    for(int i = 1; i<=n; i++){    val[i] = read();    int num = read();    for(int j = 1; j<=num; j++){    int k = read();    e[i].pb(k);    }    }    SPFA(1);    }    return 0;}