/* ZOJ 2770 Burn the Linked Camp 差分约束 */ #include <iostream> #include <cstring> #include <cstdio> #include <queue> using namespace std; const int MAXN = 1009; struct Edge { int v, ne, c; } G[MAXN*MAXN]; int head[MAXN], cnt; int C[MAXN], S[MAXN], dis[MAXN]; int vis[MAXN], sum[MAXN]; int n, m; inline void addE (int u, int v, int c) { G[++cnt].v = v, G[cnt].c = c, G[cnt].ne = head[u]; head[u] = cnt; } inline int SPFA() { queue<int> Q; dis[n] = 0, vis[n] = 1; Q.push (n); while (!Q.empty() ) { int k = Q.front(); Q.pop(); for (int i = head[k]; i; i = G[i].ne) { int v = G[i].v; if (dis[k] + G[i].c < dis[v]) { dis[v] = dis[k] + G[i].c; if (!vis[v]) Q.push (v), vis[v] = 1, sum[v]++; if (sum[v] >= n) return -1; } } vis[k] = 0; } return dis[n]-dis[0]; } inline void init() { memset (head, 0, sizeof head); memset (sum, 0, sizeof sum); memset (vis, 0, sizeof vis); memset (dis, 1, sizeof dis); cnt = 0; } int main() { while (scanf ("%d %d", &n, &m) == 2) { init(); for (int i = 1; i <= n; i++) { scanf ("%d", &C[i]); S[i] = S[i - 1] + C[i]; addE (i, i - 1, 0); } for (int i = 1; i <= n; i++) for (int j = i; j <= n; j++) addE (i - 1, j, S[j] - S[i - 1]); for (int i = 1, l, r, num; i <= m; i++) { scanf ("%d %d %d", &l, &r, &num); addE (r, l - 1, -num); } int ans = SPFA(); if (ans != -1) printf ("%d ", ans); else puts ("Bad Estimations"); } }
/* ZOJ 1508 Intervals 差分约束 */ #include <iostream> #include <cstring> #include <queue> #include <cstdio> using namespace std; const int MAXN = 500009; struct edge { int v, ne, c; } G[MAXN << 1]; int head[MAXN], cnt; int dis[MAXN], vis[MAXN]; int n, m, mr, ml; inline void addE (int u, int v, int c) { G[++cnt].v = v, G[cnt].c = c, G[cnt].ne = head[u]; head[u] = cnt; } inline int SPFA() { queue<int> ql; ql.push (mr); vis[mr] = 1, dis[mr] = 0; while (!ql.empty() ) { int x = ql.front(); ql.pop(); for (int i = head[x]; i; i = G[i].ne) { int v = G[i].v; if (dis[x] + G[i].c < dis[v]) { dis[v] = dis[x] + G[i].c; if (!vis[v]) vis[v] = 1, ql.push (v); } } vis[x] = 0; } return -dis[ml - 1]; } inline void init() { memset (head, 0, sizeof head); memset (dis, 1, sizeof dis); mr = 0, ml = 0x7fffffff; cnt = 0; } int main() { while (scanf ("%d", &n) == 1) { init(); for (int i = 1, l, r, num; i <= n; i++) { scanf ("%d %d %d", &l, &r, &num); addE (r, l - 1, -num); mr = max (mr, r), ml = min (ml, l); } for (int i = ml; i <= mr; i++) addE (i - 1, i, 1), addE (i, i - 1, 0); printf ("%d ", SPFA() ); } }
/* ZOJ 1060 King 差分约束 */ #include <iostream> #include <cstring> #include <queue> #include <cstdio> using namespace std; const int MAXN = 1009; struct edge { int v, c, ne; } G[MAXN*MAXN]; int head[MAXN], cnt; int dis[MAXN], vis[MAXN], sum[MAXN]; int n, m; inline void addE (int u, int v, int c) { G[++cnt].v = v, G[cnt].c = c, G[cnt].ne = head[u]; head[u] = cnt; } inline void init() { memset (head, 0, sizeof head); memset (dis, 1, sizeof dis); memset (vis, 0, sizeof vis); memset (sum, 0, sizeof sum); cnt = 0; } inline bool SPFA() { queue<int> ql; dis[n] = 0; for(int i=1;i<=n;i++){ ql.push(i),vis[i]=1; } while (!ql.empty() ) { int x = ql.front(); ql.pop(); for (int i = head[x]; i; i = G[i].ne) { int v = G[i].v; if (dis[x] + G[i].c < dis[v]) { dis[v] = dis[x] + G[i].c; if (!vis[v]) vis[v] = 1, ql.push (v), sum[v]++; if (sum[v] > n) return 0; } } vis[x] = 0; } return 1; } int main() { char s[10]; while (scanf ("%d", &n) == 1 && n) { scanf ("%d", &m); init(); for (int i = 1, l, r, c; i <= m; i++) { scanf ("%d %d %s %d", &l, &r, s, &c); r = l + r; if (s[0] == 'g') addE (r, l-1, -c - 1); else addE (l-1, r, c - 1); } if (SPFA() ) puts ("lamentable kingdom"); else puts ("successful conspiracy"); } }
以上三题都是根据题意直接建图的基础题。
/* ZOJ 1420 Cashier Employment 差分约束 */ #include <iostream> #include <cstring> #include <cstdio> #include <queue> using namespace std; const int MAXN = 1009; struct edge { int v, c, ne; } G[MAXN<<4]; int head[MAXN], cnt; int dis[MAXN], vis[MAXN], sum[MAXN]; int need[MAXN], join[MAXN], s[MAXN]; int n, m; inline void addE (int u, int v, int c) { G[++cnt].v = v, G[cnt].c = c, G[cnt].ne = head[u]; head[u] = cnt; } inline void init() { memset (head, 0, sizeof head); memset (sum, 0, sizeof sum); memset (vis, 0, sizeof vis); memset (dis, 1, sizeof dis); cnt = 0; } inline bool SPFA() { queue<int> ql; dis[24] = 0; for (int i = 0; i <= 24; i++) ql.push (i), vis[i] = 1; while (!ql.empty() ) { int x = ql.front(); ql.pop(); for (int i = head[x]; i; i = G[i].ne) { int v = G[i].v; if (dis[x] + G[i].c < dis[v]) { dis[v] = dis[x] + G[i].c; if (!vis[v]) vis[v] = 1, ql.push (v), sum[v]++; if (sum[v] >= 23) return 0; } } vis[x]=0; } return 1; } inline bool make (int ans) { init(); addE (24, 0, -ans); for (int i = 1; i <= 24; i++) { addE (i, i - 1, 0); addE ( i - 1, i, join[i]); } for (int i = 9; i <= 24; i++) addE (i, i - 8, -need[i]); for (int i = 1; i <= 8; i++) addE (i, i + 16, ans - need[i]); return SPFA(); } int main() { int Case; scanf ("%d", &Case); while (Case--) { for (int i = 1; i <= 24; i++) scanf ("%d", &need[i]); scanf ("%d", &n); for (int i = 1, x; i <= n; i++) { scanf ("%d", &x); join[x+1]++; } int l = 0, r = n, last = -1; while (l <= r) { int mid = (l + r) >> 1; if (make (mid) ) last = mid, r = mid - 1; else l = mid + 1; } if (last != -1) printf ("%d ", last); else puts ("No Solution"); } }
这题考虑
令 need[i] 为第i时刻需要多少出纳员，
join[i]为第i时刻有多少人应聘
s[i]作为节点,同时代表着从0时刻到i时刻需要多少出纳员
二分需要的出纳员数ans
可以列出关系:
s[i]-s[i-1]>=0;
s[i]-s[i-1]<=join[i];
s[24]-f[0]<=ans;
s[i]-s[i-8]>=need,i>8;
s[i]+s[24]-f[i+16]>=need[i],i<=8;
关键在于使图中所有结点连通。添加一个虚拟的任务0，要求在所有任务开始后开始。这时令0号任务的开时间为0，找到其它点的最短路径。
s[i]为任务i的开始时间，len[i]为任务I的持续时间
关系式
FAS u v s[u]+len[u]>=s[v];
FAF u v s[u]+len[u]>=s[v]+len[v];
SAF u v s[u]>=s[v]+len[v];
SAF u v s[u]>=s[v];
s[0]>=s[i]+len[i];
#include <iostream> #include <cstring> #include <cstdio> #include <queue> using namespace std; const int MAXN = 10009; struct edge { int v, c, ne; } G[MAXN << 4]; int head[MAXN], cnt; int sum[MAXN], vis[MAXN], dis[MAXN]; int n; int Len[MAXN]; inline void addE (int u, int v, int c) { G[++cnt].v = v, G[cnt].c = c, G[cnt].ne = head[u]; head[u] = cnt; } inline void init() { memset (head, 0, sizeof head); memset (sum, 0, sizeof sum); memset (vis, 0, sizeof vis); memset (dis, 1, sizeof dis); cnt = 0; } inline bool SPFA() { queue<int> ql; dis[0] = 0, vis[0] = 1, ql.push (0); while (!ql.empty() ) { int x = ql.front(); ql.pop(); for (int i = head[x]; i; i = G[i].ne) { int v = G[i].v; if (dis[x] + G[i].c < dis[v]) { dis[v] = dis[x] + G[i].c; if (!vis[v]) vis[v] = 1, ql.push (v), sum[v]++; if (sum[v] >= n) return 0; } } vis[x] = 0; } return 1; } int main() { char s[10]; int Cs = 0; while (scanf ("%d", &n) == 1 && n) { init(); for (int i = 1; i <= n; i++) { scanf ("%d", &Len[i]); addE (0, i, -Len[i]); } int u, v; while (scanf ("%s", s) == 1 && s[0] != '#') { scanf ("%d %d", &u, &v); if (!strcmp (s, "FAS") ) addE (u, v, Len[u]); if (!strcmp (s, "FAF") ) addE (u, v, Len[u] - Len[v]); if (!strcmp (s, "SAF") ) addE (u, v, -Len[v]); if (!strcmp (s, "SAS") ) addE (u, v, 0); } printf ("Case %d: ", ++Cs); if (SPFA() ) { int t=~(1<<31); for (int i = 1; i <= n; i++) t=min(t,dis[i]); for (int i = 1; i <= n; i++) printf ("%d %d ", i, dis[i]-t); } else puts ("impossible"); puts(""); } }
同样是简单的建图
令s[i]为第i头牛的位置:
有关系:
ML u v l s[v]-s[u]<=l;
MD u v l s[v]-s[u]>=l;
还有 s[i]-s[i-1]>=0;
/* POJ 3169 Layout 差分约束 */ #include <iostream> #include <cstdio> #include <queue> #include <cstring> using namespace std; const int MAXN = 1009; struct edge { int v, c, ne; } G[MAXN*MAXN]; int head[MAXN], cnt; int sum[MAXN], vis[MAXN], dis[MAXN]; int n, ML, MD; inline void addE (int u, int v, int c) { G[++cnt].v = v, G[cnt].c = c, G[cnt].ne = head[u]; head[u] = cnt; } inline void init() { memset (vis, 0, sizeof vis); memset (head, 0, sizeof head); memset (dis, 1, sizeof dis); memset (sum, 0, sizeof sum); cnt = 0; } inline int SPFA() { queue<int> ql; dis[1] = 0, vis[1] = 1; ql.push (1); while (!ql.empty() ) { int x = ql.front(); ql.pop(); for (int i = head[x]; i; i = G[i].ne) { int v = G[i].v; if (dis[x] + G[i].c < dis[v]) { dis[v] = dis[x] + G[i].c; if (!vis[v]) vis[v] = 1, ql.push (v), sum[v]++; if (sum[v] >= n) return -1; } } vis[x] = 0; } if (dis[n] == 0x01010101) return -2; return dis[n]; } int main() { int u, v, c; init(); scanf ("%d %d %d", &n, &ML, &MD); for (int i = 1; i <= ML; ++i) { scanf ("%d %d %d", &u, &v, &c); addE (u, v, c); } for (int i = 1; i <= MD; ++i) { scanf ("%d %d %d", &u, &v, &c); addE (v, u, -c); } for (int i = 2; i <= n; i++) addE (i, i - 1, 0); printf ("%d ", SPFA() ); }
题目说的也不是非常清楚,其实是求最小的非负边的最大值.
跟前面的题相比这道题的约束条件看起来就不那么明显了
先二分最短边为ans
令 s[i] 为点i改变的数
对每一条边u v c有 s[u]-s[v]>=ans-c;
对于用这样的关系建出的图,如果存在负环就是无解.
如果ans=最大边的情况下还没出现负环,那么最大值可以无限大
同理 ans=1的时候就有负环了,就是不能满足条件.
#include<iostream> #include<cstdio> #include<queue> #include<cstring> using namespace std; const int MAXN = 600; struct edge { int v, c, ne; } G[MAXN*MAXN]; int head[MAXN], cnt; int vis[MAXN], sum[MAXN], dis[MAXN]; int n, m; inline void add (int u, int v, int d) { G[++cnt].v = v, G[cnt].c = d, G[cnt].ne = head[u]; head[u] = cnt; } inline bool SPFA() { memset (dis, 0, sizeof dis); queue<int> q; for (int i = 1; i <= n; i++) vis[i] = 1, sum[i] = 1, q.push (i); while (!q.empty() ) { int u = q.front(); q.pop(); vis[u] = 0; for (int i = head[u]; i != 0; i = G[i].ne) { int v = G[i].v; if (dis[u] + G[i].c < dis[v]) { dis[v] = dis[u] + G[i].c; if (!vis[v]) q.push (v), vis[v] = 1; if (++sum[v] >= n) return 0; } } } return true; } inline bool make (int x) { for (int i = 1; i <= cnt; i++) G[i].c -= x; bool ok = SPFA(); for (int i = 1; i <= cnt; i++) G[i].c += x; return ok; } int main() { while (~scanf ("%d%d", &n, &m) ) { memset (head, 0, sizeof (head) ); cnt = 0; int u, v, x; for (int i = 0; i < m; i++) { scanf ("%d%d%d", &u, &v, &x); add (u, v, x); } if (make (10009) ) puts ("Infinite"); else if (!make (1) ) puts ("No Solution"); else { int l = 1, r = 10009, ans = 1; while (l <= r) { int mid = (l + r) >> 1; if (make (mid) ) ans = mid, l = mid + 1; else r = mid - 1; } printf ("%d ", ans); } } }
小结:
总的来说，这类题需要先找到题目中的约束条件,一般来说有求最值,判断可行性这几种问题。
对于求最值的情况，根据图的定义如果是直接可以通过路径长度得到的话，需要注意起点的设计，最短路还是最长路。如果是需要通过枚举或者二分求的最值，需要注意可行性需要判断的是负环还是正环。
有时候需要添加额外的点，使整个图连通。或者根据求的路径的性质（最长，最短）赋好距离初值，然后只要将所有点加入队列就行。