poj3683 2-SAT 求解
2013-11-18 16:00
211 查看
题目:http://poj.org/problem?id=3683
题目大意:在一个结婚高潮的时段,很多人想结婚,但是只有一个神父可以主持婚礼,给出每组婚礼的两个开始时间a,b和婚礼的持续时间d,问能否主持所有的婚礼,若能,则输出每个婚礼占用的时间段,不能就输出NO
考查点:2-SAT判定加求解
思路:因为每个婚礼只有两个开始时间a或b,由于只有两种状态,所以可以看作是x和非x,然后利用n方的时间判断两个婚礼的开始时间之间是否有冲突,若有,非(I and j) = 1
然后化为析取式建立边。
建完边以后,就可以求强连通分图。判断是否有冲突,如有,输出NO。若无,则进行求解。
求解是先缩图,然后反向建边。拓扑排序后,从拓扑序中第一个缩点开始,遇到没有染过色的,染成红色,然后把这个缩点内的所有点的非所在的缩点染成蓝色。然后再找下一个没有被染过色的缩点,然后往复执行前面的过程。知道所有点都被染了色
提交情况 Accepted
1次
经验:如何思考一个问题是不是2-SAT问题。每个状态有两种可能(看作是1,0),状态之间又有一些约束条件。
收获:拥有了2-SAT问题求解的模板
AC code:
#include <stdio.h>
#include <string.h>
#define MAXN 3000
#define MAXE MAXN * MAXN
struct NODE{
int st, to;
int next;
};
NODE edges[MAXE], minedges[MAXE],
incde[MAXE];
int head[MAXN], minhead[MAXN],
minmap[MAXN];
int ad, edge2, edge1, top, times, mintotal;
int time[MAXN][2], d[MAXN], start[MAXN], strack[MAXN],
In_strack[MAXN], vis[MAXN], dfn[MAXN], low[MAXN];
int ru[MAXN], belong[MAXN], colour[MAXN],
tp[MAXN];
void In_edges(int x,
int y){
edges[ad].st = x;
edges[ad].to = y;
edges[ad].next = head[x];
head[x] = ad ++;
}
void Built_Map(int
n){
ad = 0;
memset(head, -1, sizeof(head));
for(int i = 0; i < n; i
++)
for(int j = i + 1; j < n; j
++){
if((time[i][0] <=
time[j][0] && time[i][0] + d[i]
> time[j][0]) || (time[j][0] <=
time[i][0] && time[j][0] + d[j]
> time[i][0])){
In_edges(i * 2, j * 2 + 1);
In_edges(j * 2, i * 2 + 1);
}
if((time[i][0] <=
time[j][1] && time[i][0] + d[i]
> time[j][1]) || (time[j][1] <=
time[i][0] && time[j][1] + d[j]
> time[i][0])){
In_edges(i * 2, j * 2);
In_edges(j * 2 + 1, i * 2 + 1);
}
if((time[i][1] <=
time[j][0] && time[i][1] + d[i]
> time[j][0]) || (time[j][0] <=
time[i][1] && time[j][0] + d[j]
> time[i][1])){
In_edges(i * 2 + 1, j * 2 + 1);
In_edges(j * 2, i * 2);
}
if((time[i][1] <=
time[j][1] && time[i][1] + d[i]
> time[j][1]) || (time[j][1] <=
time[i][1] && time[j][1] + d[j]
> time[i][1])){
In_edges(i * 2 + 1, j * 2);
In_edges(j * 2 + 1, i * 2);
}
}
edge1 = ad;
}
void dfs(int
u){
int p;
low[u] = dfn[u] = times ++;
vis[u] = 1;
In_strack[u] = 1;
strack[ ++ top] = u;
for(p = head[u]; ~p; p =
edges[p].next)
if(!vis[edges[p].to]){
dfs(edges[p].to);
low[u] = low[u] < low[edges[p].to] ? low[u] :
low[edges[p].to];
}
else
if(In_strack[edges[p].to])
low[u] = low[u] < low[edges[p].to] ? low[u] :
low[edges[p].to];
if(low[u] == dfn[u]){
incde[ad].to = u; incde[ad].st = mintotal;
incde[ad].next = minhead[mintotal]; minhead[mintotal] = ad
++;
In_strack[u] = 0;
belong[u] = mintotal;
while((p = strack[top--]) !=
u){
incde[ad].to = p; incde[ad].st = mintotal;
incde[ad].next = minhead[mintotal]; minhead[mintotal] = ad
++;;
In_strack[p] = 0;
belong[p] = mintotal;
}
mintotal ++;
}
}
void Tarjan(int
n){
int i;
times = ad = 0, top = -1, mintotal = 0;
memset(vis, 0, sizeof(vis));
memset(minhead, -1, sizeof(minhead));
dfn[0] = 0;
for(i = 0; i < n *
2; i ++)
if(!vis[i]) dfs(i);
}
void built_min_map(int
ed){
memset(minmap, -1, sizeof(minmap));
memset(ru, 0, sizeof(ru));
edge2 = 0;
for(int i = 0; i < ed; i
++)
if(belong[edges[i].st] !=
belong[edges[i].to]){
minedges[edge2].to = belong[edges[i].st];
minedges[edge2].next = minmap[belong[edges[i].to]];
minmap[belong[edges[i].to]] = edge2 ++;
ru[belong[edges[i].st]] ++;
}
}
void tuopu(int
mintotal){
int i,j, count = 0;
top = -1;
for(i = 0; i <
mintotal; i ++ )
if(!ru[i])
strack[++top] = i, tp[count ++] = i;
while(top >=
0){
j = strack[top--];
for(i = minmap[j]; ~i; i =
minedges[i].next){
ru[minedges[i].to] --;
if(ru[minedges[i].to] ==
0){
strack[++top] = minedges[i].to;
tp[count++] = minedges[i].to;
}
}
}
}
int Getans(int
n){
int i, p;
for(i = 0; i < n; i
++)
if(low[i * 2] == low[i * 2 + 1])
return 0;
built_min_map(edge1);
tuopu(mintotal);
memset(colour, -1, sizeof(colour));
for(i = 0; i <
mintotal; i ++)
if(colour[i] == -1){
colour[i] = 1;
for(p = minhead[i]; ~p; p =
incde[p].next)
colour[belong[incde[p].to ^ 1]] = 0;
}
for(i = 0; i < n; i
++)
if(colour[belong[i *
2]])
start[i] = time[i][0];
else
start[i] = time[i][1];
}
int main(){
int n, i, x, y, z, w;
while(~scanf("%d", &n)){
for(i = 0; i < n; i
++){
scanf("%d:%d %d:%d %d",
&x, &y, &z,
&w, &d[i]);
time[i][0] = x * 60 + y;
time[i][1] = z * 60 + w - d[i];
}
Built_Map(n);
Tarjan(n);
if(Getans(n)){
printf("YES\n");
for(i = 0; i < n; i
++)
printf("%.2d:%.2d %.2d:%.2d\n",
start[i] / 60, start[i] % 60, (start[i] + d[i]) / 60, (start[i] +
d[i]) % 60);
}
else printf("NO\n");
}
return 0;
}
题目大意:在一个结婚高潮的时段,很多人想结婚,但是只有一个神父可以主持婚礼,给出每组婚礼的两个开始时间a,b和婚礼的持续时间d,问能否主持所有的婚礼,若能,则输出每个婚礼占用的时间段,不能就输出NO
考查点:2-SAT判定加求解
思路:因为每个婚礼只有两个开始时间a或b,由于只有两种状态,所以可以看作是x和非x,然后利用n方的时间判断两个婚礼的开始时间之间是否有冲突,若有,非(I and j) = 1
然后化为析取式建立边。
建完边以后,就可以求强连通分图。判断是否有冲突,如有,输出NO。若无,则进行求解。
求解是先缩图,然后反向建边。拓扑排序后,从拓扑序中第一个缩点开始,遇到没有染过色的,染成红色,然后把这个缩点内的所有点的非所在的缩点染成蓝色。然后再找下一个没有被染过色的缩点,然后往复执行前面的过程。知道所有点都被染了色
提交情况 Accepted
1次
经验:如何思考一个问题是不是2-SAT问题。每个状态有两种可能(看作是1,0),状态之间又有一些约束条件。
收获:拥有了2-SAT问题求解的模板
AC code:
#include <stdio.h>
#include <string.h>
#define MAXN 3000
#define MAXE MAXN * MAXN
struct NODE{
int st, to;
int next;
};
NODE edges[MAXE], minedges[MAXE],
incde[MAXE];
int head[MAXN], minhead[MAXN],
minmap[MAXN];
int ad, edge2, edge1, top, times, mintotal;
int time[MAXN][2], d[MAXN], start[MAXN], strack[MAXN],
In_strack[MAXN], vis[MAXN], dfn[MAXN], low[MAXN];
int ru[MAXN], belong[MAXN], colour[MAXN],
tp[MAXN];
void In_edges(int x,
int y){
edges[ad].st = x;
edges[ad].to = y;
edges[ad].next = head[x];
head[x] = ad ++;
}
void Built_Map(int
n){
ad = 0;
memset(head, -1, sizeof(head));
for(int i = 0; i < n; i
++)
for(int j = i + 1; j < n; j
++){
if((time[i][0] <=
time[j][0] && time[i][0] + d[i]
> time[j][0]) || (time[j][0] <=
time[i][0] && time[j][0] + d[j]
> time[i][0])){
In_edges(i * 2, j * 2 + 1);
In_edges(j * 2, i * 2 + 1);
}
if((time[i][0] <=
time[j][1] && time[i][0] + d[i]
> time[j][1]) || (time[j][1] <=
time[i][0] && time[j][1] + d[j]
> time[i][0])){
In_edges(i * 2, j * 2);
In_edges(j * 2 + 1, i * 2 + 1);
}
if((time[i][1] <=
time[j][0] && time[i][1] + d[i]
> time[j][0]) || (time[j][0] <=
time[i][1] && time[j][0] + d[j]
> time[i][1])){
In_edges(i * 2 + 1, j * 2 + 1);
In_edges(j * 2, i * 2);
}
if((time[i][1] <=
time[j][1] && time[i][1] + d[i]
> time[j][1]) || (time[j][1] <=
time[i][1] && time[j][1] + d[j]
> time[i][1])){
In_edges(i * 2 + 1, j * 2);
In_edges(j * 2 + 1, i * 2);
}
}
edge1 = ad;
}
void dfs(int
u){
int p;
low[u] = dfn[u] = times ++;
vis[u] = 1;
In_strack[u] = 1;
strack[ ++ top] = u;
for(p = head[u]; ~p; p =
edges[p].next)
if(!vis[edges[p].to]){
dfs(edges[p].to);
low[u] = low[u] < low[edges[p].to] ? low[u] :
low[edges[p].to];
}
else
if(In_strack[edges[p].to])
low[u] = low[u] < low[edges[p].to] ? low[u] :
low[edges[p].to];
if(low[u] == dfn[u]){
incde[ad].to = u; incde[ad].st = mintotal;
incde[ad].next = minhead[mintotal]; minhead[mintotal] = ad
++;
In_strack[u] = 0;
belong[u] = mintotal;
while((p = strack[top--]) !=
u){
incde[ad].to = p; incde[ad].st = mintotal;
incde[ad].next = minhead[mintotal]; minhead[mintotal] = ad
++;;
In_strack[p] = 0;
belong[p] = mintotal;
}
mintotal ++;
}
}
void Tarjan(int
n){
int i;
times = ad = 0, top = -1, mintotal = 0;
memset(vis, 0, sizeof(vis));
memset(minhead, -1, sizeof(minhead));
dfn[0] = 0;
for(i = 0; i < n *
2; i ++)
if(!vis[i]) dfs(i);
}
void built_min_map(int
ed){
memset(minmap, -1, sizeof(minmap));
memset(ru, 0, sizeof(ru));
edge2 = 0;
for(int i = 0; i < ed; i
++)
if(belong[edges[i].st] !=
belong[edges[i].to]){
minedges[edge2].to = belong[edges[i].st];
minedges[edge2].next = minmap[belong[edges[i].to]];
minmap[belong[edges[i].to]] = edge2 ++;
ru[belong[edges[i].st]] ++;
}
}
void tuopu(int
mintotal){
int i,j, count = 0;
top = -1;
for(i = 0; i <
mintotal; i ++ )
if(!ru[i])
strack[++top] = i, tp[count ++] = i;
while(top >=
0){
j = strack[top--];
for(i = minmap[j]; ~i; i =
minedges[i].next){
ru[minedges[i].to] --;
if(ru[minedges[i].to] ==
0){
strack[++top] = minedges[i].to;
tp[count++] = minedges[i].to;
}
}
}
}
int Getans(int
n){
int i, p;
for(i = 0; i < n; i
++)
if(low[i * 2] == low[i * 2 + 1])
return 0;
built_min_map(edge1);
tuopu(mintotal);
memset(colour, -1, sizeof(colour));
for(i = 0; i <
mintotal; i ++)
if(colour[i] == -1){
colour[i] = 1;
for(p = minhead[i]; ~p; p =
incde[p].next)
colour[belong[incde[p].to ^ 1]] = 0;
}
for(i = 0; i < n; i
++)
if(colour[belong[i *
2]])
start[i] = time[i][0];
else
start[i] = time[i][1];
}
int main(){
int n, i, x, y, z, w;
while(~scanf("%d", &n)){
for(i = 0; i < n; i
++){
scanf("%d:%d %d:%d %d",
&x, &y, &z,
&w, &d[i]);
time[i][0] = x * 60 + y;
time[i][1] = z * 60 + w - d[i];
}
Built_Map(n);
Tarjan(n);
if(Getans(n)){
printf("YES\n");
for(i = 0; i < n; i
++)
printf("%.2d:%.2d %.2d:%.2d\n",
start[i] / 60, start[i] % 60, (start[i] + d[i]) / 60, (start[i] +
d[i]) % 60);
}
else printf("NO\n");
}
return 0;
}
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