标签：poj1201
| Time Limit: 2000MS | Memory Limit: 65536K | |
| Total Submissions: 21591 | Accepted: 8122 |
Description
Input
Output
Sample Input
5 3 7 3 8 10 3 6 8 1 1 3 1 10 11 1
Sample Output
6因为这题不需要判断负环，所以不用添加虚拟节点，另外，求最短路径时，原点应该是最右边的点。
#include <stdio.h>
#include <string.h>
#include <queue>
#define maxn 50002
#define inf 0x3f3f3f3f
using std::queue;
int head[maxn], id, left, right, dist[maxn];
struct Node{
int to, w, next;
} E[maxn << 2];
bool vis[maxn];
void addEdge(int a, int b, int c)
{
E[id].to = b; E[id].w = c;
E[id].next = head[a]; head[a] = id++;
}
int SPFA()
{
int i, u, v, tmp;
for(i = left; i <= right + 1; ++i){
vis[i] = 0; dist[i] = inf;
}
queue<int> Q; Q.push(right);
dist[right] = 0; vis[right] = 1;
while(!Q.empty()){
u = Q.front(); Q.pop();
vis[u] = 0;
for(i = head[u]; i != -1; i = E[i].next){
v = E[i].to;
tmp = dist[u] + E[i].w;
if(tmp < dist[v]){
dist[v] = tmp;
if(!vis[v]){
vis[v] = 1;
Q.push(v);
}
}
}
}
return -dist[left];
}
int main()
{
//freopen("stdin.txt", "r", stdin);
int n, a, b, c, i;
while(scanf("%d", &n) == 1){
memset(head, -1, sizeof(head));
left = maxn; right = 0;
for(i = id = 0; i < n; ++i){
scanf("%d%d%d", &a, &b, &c);
addEdge(b+1, a, -c);
if(a < left) left = a;
if(b > right) right = b;
}
++right;
for(i = left; i < right; ++i){
addEdge(i+1, i, 0);
addEdge(i, i+1, 1);
}
printf("%d\n", SPFA());
}
return 0;
}正着求也对：
#include <stdio.h>
#include <string.h>
#include <queue>
#define maxn 50002
#define inf 0x3f3f3f3f
using std::queue;
int head[maxn], id, left, right, dist[maxn];
struct Node{
int to, w, next;
} E[maxn << 2];
bool vis[maxn];
void addEdge(int a, int b, int c)
{
E[id].to = b; E[id].w = c;
E[id].next = head[a]; head[a] = id++;
}
int SPFA()
{
int i, u, v, tmp;
for(i = left; i <= right + 1; ++i){
vis[i] = 0; dist[i] = -inf;
}
queue<int> Q; Q.push(left);
dist[left] = 0; vis[left] = 1;
while(!Q.empty()){
u = Q.front(); Q.pop();
vis[u] = 0;
for(i = head[u]; i != -1; i = E[i].next){
v = E[i].to;
tmp = dist[u] + E[i].w;
if(tmp > dist[v]){
dist[v] = tmp;
if(!vis[v]){
vis[v] = 1;
Q.push(v);
}
}
}
}
return dist[right];
}
int main()
{
// freopen("stdin.txt", "r", stdin);
int n, a, b, c, i;
while(scanf("%d", &n) == 1){
memset(head, -1, sizeof(head));
left = maxn; right = 0;
for(i = id = 0; i < n; ++i){
scanf("%d%d%d", &a, &b, &c);
addEdge(a, b+1, c);
if(a < left) left = a;
if(b > right) right = b;
}
++right;
for(i = left; i < right; ++i){
addEdge(i, i + 1, 0);
addEdge(i+1, i, -1);
}
printf("%d\n", SPFA());
}
return 0;
}POJ1201 Intervals 【差分约束系统】,布布扣,bubuko.com
标签：poj1201
原文地址：http://blog.csdn.net/chang_mu/article/details/38611263
