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普通的最短路...dijkstra水过..
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#include<cstdio>
#include<algorithm>
#include<cstring>
#include<iostream>
#include<queue>
#define rep( i , n ) for( int i = 0 ; i < n ; i++ )
#define clr( x , c ) memset( x , c , sizeof( x ) )
using namespace std;
const int maxn = 2500 + 5;
const int maxm = 6200 + 5;
const int inf = 0x3f3f3f3f;
struct edge {
int to , dist;
edge* next;
};
edge* pt , EDGE[ maxm << 1 ];
edge* head[ maxn ];
void edge_init() {
pt = EDGE;
clr( head , 0 );
}
void add( int u , int v , int d ) {
pt -> to = v;
pt -> dist = d;
pt -> next = head[ u ];
head[ u ] = pt++;
}
#define add_edge( u , v , d ) add( u , v , d ) , add( v , u , d )
struct node {
int x , d;
bool operator < ( const node &rhs ) const {
return d > rhs.d;
}
};
int d[ maxn ];
priority_queue< node > Q;
void dijkstra( int S ) {
clr( d , inf );
d[ S ] = 0;
Q.push( ( node ) { S , 0 } );
while( ! Q.empty() ) {
node o = Q.top();
Q.pop();
int x = o.x , dist = o.d;
if( dist != d[ x ] ) continue;
for( edge* e = head[ x ] ; e ; e = e -> next ) {
int to = e -> to;
if( d[ to ] > dist + e -> dist ) {
d[ to ] = dist + e -> dist;
Q.push( ( node ) { to , d[ to ] } );
}
}
}
}
inline int read() {
char c = getchar();
int res = 0;
while( ! isdigit( c ) ) c = getchar();
while( isdigit( c ) ) {
res = res * 10 + c - ‘0‘;
c = getchar();
}
return res;
}
int main() {
freopen( "test.in" , "r" , stdin );
int n = read() , m = read() , s = read() - 1 , t = read() - 1;
edge_init();
while( m-- ) {
int u = read() - 1 , v = read() - 1 , d = read();
add_edge( u , v , d );
}
dijkstra( s );
cout << d[ t ] << "\n";
return 0;
}
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3408: [Usaco2009 Oct]Heat Wave 热浪
Time Limit: 3 Sec Memory Limit: 128 MB
Submit: 71 Solved: 59
[Submit][Status][Discuss]Description
Input
第1行：4个由空格隔开的整数T，C，Ts，Te.
第2到第C+1行：第i+l行描述第i条道路．有3个由空格隔开的整数Rs，Re，Ci.
Output
一个单独的整数表示Ts到Te的最小费用．数据保证至少存在一条道路．
Sample Input
7 11 5 4
2 4 2
1 4 3
7 2 2
3 4 3
5 7 5
7 3 3
6 1 1
6 3 4
2 4 3
5 6 3
7 2 1
Sample Output
7
HINT
Source
BZOJ 3408: [Usaco2009 Oct]Heat Wave 热浪( 最短路 )
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原文地址：http://www.cnblogs.com/JSZX11556/p/4575799.html
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