Task Schedule
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 4003 Accepted Submission(s): 1347
Problem Description
Our geometry princess XMM has stoped her study in computational geometry to concentrate on her newly opened factory. Her factory has introduced M new machines in order to process the coming N tasks. For the i-th task, the factory
has to start processing it at or after day Si, process it for Pi days, and finish the task before or at day Ei. A machine can only work on one task at a time, and each task can be processed by at most one machine at a time. However, a task can be interrupted
and processed on different machines on different days.
Now she wonders whether he has a feasible schedule to finish all the tasks in time. She turns to you for help.
Now she wonders whether he has a feasible schedule to finish all the tasks in time. She turns to you for help.
Input
On the first line comes an integer T(T<=20), indicating the number of test cases.
You are given two integer N(N<=500) and M(M<=200) on the first line of each test case. Then on each of next N lines are three integers Pi, Si and Ei (1<=Pi, Si, Ei<=500), which have the meaning described in the description. It is guaranteed that in a feasible schedule every task that can be finished will be done before or at its end day.
You are given two integer N(N<=500) and M(M<=200) on the first line of each test case. Then on each of next N lines are three integers Pi, Si and Ei (1<=Pi, Si, Ei<=500), which have the meaning described in the description. It is guaranteed that in a feasible schedule every task that can be finished will be done before or at its end day.
Output
For each test case, print “Case x: ” first, where x is the case number. If there exists a feasible schedule to finish all the tasks, print “Yes”, otherwise print “No”.
Print a blank line after each test case.
Print a blank line after each test case.
Sample Input
2 4 3 1 3 5 1 1 4 2 3 7 3 5 9 2 2 2 1 3 1 2 2
Sample Output
Case 1: Yes Case 2: Yes
Author
allenlowesy
Source
题意：有n个机器，m项任务，每一个任务须要Pi天时间，开工日期到收工日期为Si到Ei。一次仅仅能在一台机器上加工，能够挪到别的机器上，问是否能按期完毕全部任务。
题解：这题关键在构图，设置一个源点到每项任务有一条边，容量为该项任务所须要的天数，每项任务到合法加工日期内的每一个天数加一条边，容量为1，即每天工作量为1。然后每一个天数到汇点加入一条边，容量为机器数量n。表示一天最大加工量。
218ms
#include <stdio.h>
#include <string.h>
#define maxn 1200
#define maxm 700000
#define inf 0x3f3f3f3f
int head[maxn], n, m, id; // n machines
struct Node {
int u, v, c, next;
} E[maxm];
int source, sink, tar, maxDay, nv;
int que[maxn], Layer[maxn], pre[maxn];
bool vis[maxn];
void addEdge(int u, int v, int c) {
E[id].u = u; E[id].v = v;
E[id].c = c; E[id].next = head[u];
head[u] = id++;
E[id].u = v; E[id].v = u;
E[id].c = 0; E[id].next = head[v];
head[v] = id++;
}
void getMap() {
int i, j, u, v, p, s, e;
id = tar = maxDay = 0;
scanf("%d%d", &m, &n);
memset(head, -1, sizeof(head));
source = 0; sink = 705;
for(i = 1; i <= m; ++i) {
scanf("%d%d%d", &p, &s, &e);
tar += p;
if(e > maxDay) maxDay = e;
addEdge(source, i, p);
for(j = s; j <= e; ++j)
addEdge(i, m + j, 1);
}
sink = m + maxDay + 1; nv = sink + 1;
for(i = 1; i <= maxDay; ++i)
addEdge(m + i, sink, n);
}
bool countLayer() {
memset(Layer, 0, sizeof(int) * nv);
int id = 0, front = 0, u, v, i;
Layer[source] = 1; que[id++] = source;
while(front != id) {
u = que[front++];
for(i = head[u]; i != -1; i = E[i].next) {
v = E[i].v;
if(E[i].c && !Layer[v]) {
Layer[v] = Layer[u] + 1;
if(v == sink) return true;
else que[id++] = v;
}
}
}
return false;
}
int Dinic() {
int i, u, v, minCut, maxFlow = 0, pos, id = 0;
while(countLayer()) {
memset(vis, 0, sizeof(bool) * nv);
memset(pre, -1, sizeof(int) * nv);
que[id++] = source; vis[source] = 1;
while(id) {
u = que[id - 1];
if(u == sink) {
minCut = inf;
for(i = pre[sink]; i != -1; i = pre[E[i].u])
if(minCut > E[i].c) {
minCut = E[i].c; pos = E[i].u;
}
maxFlow += minCut;
for(i = pre[sink]; i != -1; i = pre[E[i].u]) {
E[i].c -= minCut;
E[i^1].c += minCut;
}
while(que[id-1] != pos)
vis[que[--id]] = 0;
} else {
for(i = head[u]; i != -1; i = E[i].next)
if(E[i].c && Layer[u] + 1 == Layer[v = E[i].v] && !vis[v]) {
vis[v] = 1; que[id++] = v; pre[v] = i; break;
}
if(i == -1) --id;
}
}
}
return maxFlow;
}
void solve(int cas) {
printf("Case %d: %s
", cas, tar == Dinic() ? "Yes" : "No");
}
int main() {
// freopen("stdin.txt", "r", stdin);
int t, cas;
scanf("%d", &t);
for(cas = 1; cas <= t; ++cas) {
getMap();
solve(cas);
}
return 0;
}62ms
#include <stdio.h>
#include <string.h>
#define maxn 1200
#define maxm 700000
int head[maxn], n, m, id; // n machines
struct Node {
int u, v, c, next;
} E[maxm];
int source, sink, tar, maxDay, nv;
const int inf = 0x3f3f3f3f;
int cur[maxn], ps[maxn], dep[maxn];
void addEdge(int u, int v, int c) {
E[id].u = u; E[id].v = v;
E[id].c = c; E[id].next = head[u];
head[u] = id++;
E[id].u = v; E[id].v = u;
E[id].c = 0; E[id].next = head[v];
head[v] = id++;
}
void getMap() {
int i, j, u, v, p, s, e;
id = tar = maxDay = 0;
scanf("%d%d", &m, &n);
memset(head, -1, sizeof(head));
source = 0;
for(i = 1; i <= m; ++i) {
scanf("%d%d%d", &p, &s, &e);
tar += p;
if(e > maxDay) maxDay = e;
addEdge(source, i, p);
for(j = s; j <= e; ++j)
addEdge(i, m + j, 1);
}
sink = m + maxDay + 1; nv = sink + 1;
for(i = 1; i <= maxDay; ++i)
addEdge(m + i, sink, n);
}
// 參数：顶点个数。源点，汇点
int Dinic(int n, int s, int t) {
int tr, res = 0;
int i, j, k, f, r, top;
while(true) {
memset(dep, -1, n * sizeof(int));
for(f = dep[ps[0] = s] = 0, r = 1; f != r; )
for(i = ps[f++], j = head[i]; j != -1; j = E[j].next) {
if(E[j].c && -1 == dep[k=E[j].v]) {
dep[k] = dep[i] + 1; ps[r++] = k;
if(k == t) {
f = r; break;
}
}
}
if(-1 == dep[t]) break;
memcpy(cur, head, n * sizeof(int));
for(i = s, top = 0; ; ) {
if(i == t) {
for(k = 0, tr = inf; k < top; ++k)
if(E[ps[k]].c < tr) tr = E[ps[f=k]].c;
for(k = 0; k < top; ++k)
E[ps[k]].c -= tr, E[ps[k]^1].c += tr;
res += tr; i = E[ps[top = f]].u;
}
for(j = cur[i]; cur[i] != -1; j = cur[i] = E[cur[i]].next)
if(E[j].c && dep[i] + 1 == dep[E[j].v]) break;
if(cur[i] != -1) {
ps[top++] = cur[i];
i = E[cur[i]].v;
} else {
if(0 == top) break;
dep[i] = -1; i = E[ps[--top]].u;
}
}
}
return res;
}
void solve(int cas) {
printf("Case %d: %s
", cas, tar == Dinic(nv, source, sink) ? "Yes" : "No");
}
int main() {
// freopen("stdin.txt", "r", stdin);
int t, cas;
scanf("%d", &t);
for(cas = 1; cas <= t; ++cas) {
getMap();
solve(cas);
}
return 0;
}