第一道正儿八经的差分约束题
有排成一列的n个点，首先告诉你每个点的值最多是多少(最少显然要大于0），然后告诉你m段i，j,k，表示第i个点到第j个点的值的和至少有k，问你总和至少为多少。
要注意的是，告诉你的所有关系式都不要忘记建边，一开始漏了大于0的条件调半天o(╯□╰)o
不等式的形式是a-b<=c这样的= =
1 #include <cstdio> 2 #include <cstring> 3 #include <cmath> 4 #include <algorithm> 5 #include <climits> 6 #include <string> 7 #include <iostream> 8 #include <map> 9 #include <cstdlib> 10 #include <list> 11 #include <set> 12 #include <queue> 13 #include <stack> 14 15 using namespace std; 16 17 typedef long long LL; 18 const int maxn = 1000 + 5; 19 const int maxm = 50000 + 5; 20 const LL INF = LONG_LONG_MAX / 4; 21 int v[maxm],w[maxm],first[maxn],nxt[maxm]; 22 int n,m,C,cnt[maxn],ecnt; 23 LL d[maxn]; 24 bool inq[maxn]; 25 26 void solve() { 27 bool bad = false; 28 queue<int> q; 29 q.push(n); 30 for(int i = 0;i <= n;i++) { 31 inq[i] = false; 32 d[i] = INF; 33 cnt[i] = 0; 34 } 35 d[n] = 0; 36 inq[n] = true; 37 cnt[n] = 1; 38 while(!q.empty() && !bad) { 39 int x = q.front(); q.pop(); 40 inq[x] = false; 41 for(int i = first[x];i != -1;i = nxt[i]) { 42 if(d[v[i]] > d[x] + w[i]) { 43 d[v[i]] = d[x] + w[i]; 44 if(!inq[v[i]]) { 45 q.push(v[i]); 46 cnt[v[i]]++; 47 inq[v[i]] = true; 48 if(cnt[v[i]] > n + 1) { 49 bad = true; break; 50 } 51 } 52 } 53 } 54 } 55 if(bad) puts("Bad Estimations"); 56 else printf("%lld ",-d[0]); 57 } 58 59 void adde(int _u,int _v,int _w) { 60 v[ecnt] = _v; w[ecnt] = _w; 61 nxt[ecnt] = first[_u]; 62 first[_u] = ecnt; 63 ecnt++; 64 } 65 66 int main() { 67 while(~scanf("%d%d",&n,&m)) { 68 ecnt = 0; 69 memset(first,-1,sizeof(first)); 70 memset(nxt,-1,sizeof(nxt)); 71 for(int i = 1;i <= n;i++) { 72 scanf("%d",&C); 73 adde(i - 1,i,C); 74 adde(i,i - 1,0); 75 } 76 for(int i = 1;i <= m;i++) { 77 int a,b,c; scanf("%d%d%d",&a,&b,&c); 78 adde(b,a - 1,-c); 79 } 80 solve(); 81 } 82 return 0; 83 }