HDU 3776 Task

      这是一道查分约束方程的题。难点在于如何建图，task i starts at least A minutes later than task j得到的方程是：tj - ti <= -Ａ，task i starts within A minutes of the starting time of task j得到的方程是： tj - ti <= 0和ti - tj <= A，然后建图，最后用bellman_ford来判，注意输出时，为防止时间有负数或超出1000000，每个数要减去最小的数并加1。

程序代码：

#include<iostream>#include<cstdio>#include<string>#include<cctype>#include<sstream>using namespace std;#define MAXN 110#define INF (1 << 30)int d[MAXN], u[MAXN * MAXN], v[MAXN * MAXN], w[MAXN * MAXN];int e, n, vMin, num[4];int ballman_ford(){ for(int i = 0; i <= n; d[i++] = INF); d[0] = 0; for(int i = 0; i < n; i++){ for(int j = 0; j < e; j++){ if(d[v[j]] > d[u[j]] + w[j]) d[v[j]] = d[u[j]] + w[j]; } } for(int i = 0; i < e; i++) if(d[v[i]] > d[u[i]] + w[i]) return 0; vMin = (1 << 30); for(int i = 1; i <= n; i++) if(vMin > d[i]) vMin = d[i]; return 1;}int main(){ string input, s; int i, j, k, l, m, flag; //freopen("input.txt", "r", stdin); while(scanf("%d", &n)){ if(!n) break; scanf("%d", &m); cin.ignore(); e = 0; for(i = 0; i < m; i++){ getline(cin, input); istringstream in (input); //cout<<input<<endl; num[0] = num[1] = num[2] = 0; for(j = 0, l = 0; in>>s; j++){ if(isdigit(s[0])){ for(k = 0; k < s.size(); k++) num[l] = num[l] * 10 + s[k] - '0'; l++; } if(j == 3){ if(s == "at") flag = 1; else flag = 0; } } if(flag){ u[e] = num[0]; v[e] = num[2]; w[e++] = -num[1]; }else { u[e] = num[0]; v[e] = num[2]; w[e++] = 0; u[e] = num[2]; v[e] = num[0]; w[e++] = num[1]; } } for(i = 1; i <= n; e++, i++){ u[e] = 0; v[e] = i; w[e] = 0; if(ballman_ford()){ for(i = 1; i <= n; i++) printf("%d%c", d[i] - vMin + 1, i == n ? '/n' : ' '); }else printf("Impossible./n"); } return 0;}