差分约束系统

变量之间的不等关系转成最短/长路求解。

若A - B <= x

则A <= B + x，此时有一条B->A边权为x的边。

一个形象的图：

不要忘了问题本身的约束条件。

求最小值->最长路

最大值->最短路

负环->无解

不连通->多解

参考资料。

例题：

HDU1384 Intervals 注意有多组数据。

 1 #include <cstdio>
 2 #include <queue>
 3 #include <cstring>
 4 
 5 const int N = 100010, INF = 0x3f3f3f3f;
 6 
 7 struct Edge {
 8     int nex, v, len;
 9     Edge(int N = 0, int V = 0, int L = 0) {
10         nex = N;
11         v = V;
12         len = L;
13     }
14 }edge[2000010]; int tp;
15 
16 std::queue<int> Q;
17 int d[N], e[N], a[N], b[N], c[N];
18 bool vis[N];
19 
20 inline void add(int x, int y, int z) {
21     tp++;
22     edge[tp] = Edge(e[x], y, z);
23     e[x] = tp;
24     return;
25 }
26 
27 inline void spfa(int s) {
28     memset(d, 0x3f, sizeof(d));
29     vis[s] = 1;
30     d[s] = 0;
31     Q.push(s);
32     while(!Q.empty()) {
33         int x = Q.front();
34         Q.pop();
35         vis[x] = 0;
36         for(int i = e[x]; i; i = edge[i].nex) {
37             int y = edge[i].v;
38             if(d[y] > d[x] + edge[i].len) {
39                 d[y] = d[x] + edge[i].len;
40                 if(!vis[y]) {
41                     vis[y] = 1;
42                     Q.push(y);
43                 }
44             }
45         }
46     }
47     return;
48 }
49 
50 int main() {
51     int n, lm = 0;
52     while(scanf("%d", &n) != EOF) {
53         memset(e, 0, sizeof(e));
54         tp = 0;
55         memset(vis, 0, sizeof(vis));
56         for(int i = 1; i <= n; i++) {
57             scanf("%d%d%d", &a[i], &b[i], &c[i]);
58             a[i]++; b[i]++;
59             lm = std::max(lm, b[i]);
60             /// sum[b[i]] - sum[a[i] - 1] >= c[i]
61             add(a[i], b[i] + 1, -c[i]);
62         }
63         for(int i = 1; i <= lm; i++) {
64             add(i, i + 1, 0);
65             add(i + 1, i, 1);
66         }
67         spfa(1);
68         printf("%d\n", -d[lm + 1]);
69     }
70     return 0;
71 }

AC代码

ZOJ2770 Burn the Linked Camp 一毛一样。多了一个判负环。

 1 #include <cstdio>
 2 #include <queue>
 3 #include <cstring>
 4 
 5 typedef long long LL;
 6 const int N = 100010;
 7 const LL INF = 0x3f3f3f3f3f3f3f3fll;
 8 
 9 struct Edge {
10     int nex, v;
11     LL len;
12     Edge(int N = 0, int V = 0, LL L = 0) {
13         nex = N;
14         v = V;
15         len = L;
16     }
17 }edge[2000010]; int tp;
18 
19 std::queue<int> Q;
20 int e[N], a[N], b[N], c[N], m, n, cnt[N];
21 LL d[N];
22 bool vis[N];
23 
24 inline void add(int x, int y, LL z) {
25     tp++;
26     edge[tp] = Edge(e[x], y, z);
27     e[x] = tp;
28     return;
29 }
30 
31 inline int spfa(int s) {
32     memset(d, 0x3f, sizeof(d));
33     memset(cnt, 0, sizeof(cnt));
34     memset(vis, 0, sizeof(vis));
35     vis[s] = 1;
36     d[s] = 0;
37     cnt[s] = 1;
38     Q.push(s);
39     while(!Q.empty()) {
40         int x = Q.front();
41         Q.pop();
42         vis[x] = 0;
43         for(int i = e[x]; i; i = edge[i].nex) {
44             int y = edge[i].v;
45             if(d[y] > d[x] + edge[i].len) {
46                 d[y] = d[x] + edge[i].len;
47                 cnt[y] = cnt[x] + 1;
48                 if(cnt[y] > n + 1) {
49                     return -1;
50                 }
51                 if(!vis[y]) {
52                     vis[y] = 1;
53                     Q.push(y);
54                 }
55             }
56         }
57     }
58     return 0;
59 }
60 
61 int main() {
62     while(scanf("%d%d", &n, &m) != EOF) {
63         memset(e, 0, sizeof(e));
64         tp = 0;
65         for(int i = 1; i <= n; i++) {
66             scanf("%d", &c[i]);
67             add(i + 1, i, c[i]);
68             add(i, i + 1, 0);
69         }
70         for(int i = 1; i <= m; i++) {
71             scanf("%d%d%d", &a[i], &b[i], &c[i]);
72             add(a[i], b[i] + 1, -c[i]);
73         }
74         int t = spfa(1);
75         if(t == -1) {
76             printf("Bad Estimations\n");
77         }
78         else {
79             printf("%d\n", -d[n + 1]);
80         }
81     }
82     return 0;
83 }

AC代码

HDU1531 King 注意可能不连通，而每个数也没有非负的限制，所以新建s向每个点连0边保证每个连通块都会被检查一遍。

 1 #include <cstdio>
 2 #include <queue>
 3 #include <cstring>
 4 
 5 typedef long long LL;
 6 const int N = 100010;
 7 const LL INF = 0x3f3f3f3f3f3f3f3fll;
 8 
 9 struct Edge {
10     int nex, v;
11     LL len;
12     Edge(int N = 0, int V = 0, LL L = 0) {
13         nex = N;
14         v = V;
15         len = L;
16     }
17 }edge[1000010]; int tp;
18 
19 std::queue<int> Q;
20 int e[N], a[N], b[N], c[N], m, n, cnt[N];
21 LL d[N];
22 bool vis[N];
23 
24 inline void add(int x, int y, LL z) {
25     tp++;
26     edge[tp] = Edge(e[x], y, z);
27     e[x] = tp;
28     return;
29 }
30 
31 inline int spfa(int s) {
32     memset(d, 0x3f, sizeof(d));
33     memset(cnt, 0, sizeof(cnt));
34     memset(vis, 0, sizeof(vis));
35     vis[s] = 1;
36     d[s] = 0;
37     cnt[s] = 1;
38     Q.push(s);
39     while(!Q.empty()) {
40         int x = Q.front();
41         Q.pop();
42         vis[x] = 0;
43         for(int i = e[x]; i; i = edge[i].nex) {
44             int y = edge[i].v;
45             if(d[y] > d[x] + edge[i].len) {
46                 d[y] = d[x] + edge[i].len;
47                 cnt[y] = cnt[x] + 1;
48                 if(cnt[y] > n + 2) {
49                     return -1;
50                 }
51                 if(!vis[y]) {
52                     vis[y] = 1;
53                     Q.push(y);
54                 }
55             }
56         }
57     }
58     return 0;
59 }
60 
61 char str[3];
62 
63 int main() {
64     //printf("%d \n", (sizeof(edge) + sizeof(cnt) * 7) / 1048576);
65     while(scanf("%d", &n), n) {
66         scanf("%d", &m);
67         memset(e, 0, sizeof(e));
68         tp = 0;
69         for(int i = 1; i <= m; i++) {
70             scanf("%d%d%s%d", &a[i], &b[i], str, &c[i]);
71             b[i] += a[i];
72             if(str[0] == 'g') {
73                 c[i]++;
74                 add(b[i] + 1, a[i], -c[i]);
75             }
76             else {
77                 c[i]--;
78                 add(a[i], b[i] + 1, c[i]);
79             }
80         }
81         for(int i = 1; i <= n + 1; i++) {
82             add(n + 2, i, 0);
83         }
84 
85         int t = spfa(n + 2);
86         if(t != -1) {
87             printf("lamentable kingdom\n");
88         }
89         else {
90             printf("successful conspiracy\n");
91         }
92     }
93     return 0;
94 }

AC代码

HDU1534 Schedule Problem 之前的都是序列模型，这个直接就是点之间的关系，不需要序列上前缀和。

每个事件的开始时间就是d。

  1 #include <cstdio>
  2 #include <queue>
  3 #include <cstring>
  4 
  5 typedef long long LL;
  6 const int N = 100010;
  7 const LL INF = 0x3f3f3f3f3f3f3f3fll;
  8 
  9 struct Edge {
 10     int nex, v;
 11     LL len;
 12     Edge(int N = 0, int V = 0, LL L = 0) {
 13         nex = N;
 14         v = V;
 15         len = L;
 16     }
 17 }edge[1000010]; int tp;
 18 
 19 std::queue<int> Q;
 20 int e[N], a[N], b[N], c[N], m, n, cnt[N], Time;
 21 LL d[N];
 22 bool vis[N];
 23 
 24 inline void add(int x, int y, LL z) {
 25     tp++;
 26     edge[tp] = Edge(e[x], y, z);
 27     e[x] = tp;
 28     return;
 29 }
 30 
 31 inline int spfa(int s) {
 32     memset(d, 0x3f, sizeof(d));
 33     memset(cnt, 0, sizeof(cnt));
 34     vis[s] = Time;
 35     d[s] = 0;
 36     cnt[s] = 1;
 37     Q.push(s);
 38     while(!Q.empty()) {
 39         int x = Q.front();
 40         Q.pop();
 41         vis[x] = 0;
 42         for(int i = e[x]; i; i = edge[i].nex) {
 43             int y = edge[i].v;
 44             if(d[y] > d[x] + edge[i].len) {
 45                 d[y] = d[x] + edge[i].len;
 46                 cnt[y] = cnt[x] + 1;
 47                 if(cnt[y] > n + 1) {
 48                     return -1;
 49                 }
 50                 if(vis[y] != Time) {
 51                     vis[y] = Time;
 52                     Q.push(y);
 53                 }
 54             }
 55         }
 56     }
 57     return 0;
 58 }
 59 
 60 char str[4];
 61 
 62 int main() {
 63     //printf("%d \n", (sizeof(edge) + sizeof(cnt) * 7) / 1048576);
 64     while(scanf("%d", &n), n) {
 65         memset(e, 0, sizeof(e));
 66         Time++;
 67         int x, y;
 68         tp = 0;
 69         for(int i = 1; i <= n; i++) {
 70             scanf("%d", &a[i]);
 71         }
 72         while(scanf("%s", str)) {
 73             if(str[0] == '#') break;
 74             scanf("%d%d", &x, &y);
 75             if(str[0] == 'F' && str[2] == 'S') {
 76                 add(y, x, a[x]);
 77             }
 78             else if(str[0] == 'F') {
 79                 add(y, x, -a[y] + a[x]);
 80             }
 81             else if(str[2] == 'F') {
 82                 add(y, x, -a[y]);
 83             }
 84             else {
 85                 add(y, x, 0);
 86             }
 87         }
 88         for(int i = 1; i <= n; i++) {
 89             add(n + 1, i, 0);
 90         }
 91         int t = spfa(n + 1);
 92         printf("Case %d:\n", Time);
 93         if(t == -1) {
 94             printf("impossible\n");
 95         }
 96         else {
 97             for(int i = 1; i <= n; i++) {
 98                 printf("%d %d\n", i, -d[i]);
 99             }
100         }
101         puts("");
102     }
103     return 0;
104 }

AC代码

HDU1529 Cashier Employment 把每个时间的选用人数做前缀和。发现有7个不等式有三个变量，于是二分其中一个，也就是总共选了多少人。

  1 #include <bits/stdc++.h>
  2 
  3 typedef long long LL;
  4 const int N = 110;
  5 
  6 struct Edge {
  7     int nex, v;
  8     LL len;
  9     Edge(int N = 0, int V = 0, LL L = 0) {
 10         nex = N;
 11         v = V;
 12         len = L;
 13     }
 14 }edge[100010]; int tp;
 15 
 16 int e[N], cnt[N], stk[N], top, n, m, a[N], vis[N], b[N];
 17 LL d[N];
 18 
 19 inline void add(int x, int y, LL z) {
 20     tp++;
 21     edge[tp] = Edge(e[x], y, z);
 22     e[x] = tp;
 23     return;
 24 }
 25 
 26 inline int spfa(int s) {
 27     static int Time = 0;
 28     Time++;
 29     memset(d, 0x3f, sizeof(d));
 30     memset(cnt, 0, sizeof(cnt));
 31     stk[++top] = s;
 32     vis[s] = Time;
 33     cnt[s] = 1;
 34     d[s] = 0;
 35     while(top) {
 36         int x = stk[top];
 37         top--;
 38         vis[x] = 0;
 39         for(int i = e[x]; i; i = edge[i].nex) {
 40             int y = edge[i].v;
 41             if(d[y] > d[x] + edge[i].len) {
 42                 d[y] = d[x] + edge[i].len;
 43                 cnt[y] = cnt[x] + 1;
 44                 if(cnt[y] > n + 1) {
 45                     return -1;
 46                 }
 47                 if(vis[y] != Time) {
 48                     vis[y] = Time;
 49                     stk[++top] = y;
 50                 }
 51             }
 52         }
 53     }
 54     return 0;
 55 }
 56 
 57 inline void clear() {
 58     memset(b, 0, sizeof(b));
 59     tp = 0;
 60     return;
 61 }
 62 
 63 inline bool check(int mid) {
 64     memset(e, 0, sizeof(e));
 65     tp = 0;
 66     for(int i = 1; i <= n; i++) {
 67         add(i, i + 1, 0);
 68         if(i < 8) {
 69             add(i + 17, i + 1, -a[i] + mid);
 70         }
 71         else {
 72             add(i - 7, i + 1, -a[i]);
 73         }
 74         add(i + 1, i, b[i]);
 75     }
 76     add(1, n + 1, -mid);
 77     add(n + 1, 1, mid);
 78     int t = spfa(1);
 79     return t != -1;
 80 }
 81 
 82 inline void solve() {
 83     n = 24;
 84     for(int i = 1; i <= n; i++) {
 85         scanf("%d", &a[i]);
 86     }
 87     scanf("%d", &m);
 88     for(int i = 1; i <= m; i++) {
 89         int x;
 90         scanf("%d", &x);
 91         b[x + 1]++;
 92     }
 93     /// calc
 94     int l = 0, r = m + 1;
 95     while(l < r) {
 96         int mid = (l + r) >> 1;
 97         if(check(mid)) {
 98             r = mid;
 99         }
100         else {
101             l = mid + 1;
102         }
103     }
104     if(r == m + 1) {
105         printf("No Solution\n");
106         return;
107     }
108     printf("%d\n", r);
109     return;
110 }
111 
112 int main() {
113     int T;
114     scanf("%d", &T);
115     while(T--) {
116         solve();
117         if(T) {
118             clear();
119         }
120     }
121     return 0;
122 }